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Mathematics
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SARASWATI HOUSE PVT. LTD.(An ISO 9001:2008 Company)
EDUCATIONAL PUBLISHERS9, Daryaganj, Near Telephone Office, New Delhi-110002Ph: 43556600 (100 lines), 23281022 • Fax: 43556688
WORKSHEET–11. (B) For real and distinct roots, D > 0
⇒ b2 – 4ac > 0 ⇒ 16 – 4p > 0⇒ 4 > p ∴ p < 4.
2. (C) p(1)2 + p(1) + 3 = 0
⇒ 2p = – 3 ⇒ p = – 32
and (1)2 + 1 + q = 0 ⇒ q = – 2
∴ pq = − 32
(– 2) = 3.
3. (A)−
+ − −−
2( 5)2( 5 )
5k = 0
⇒ –10 – 2k = 5⇒ – 2k = 15
⇒ k =152
−.
4. Given equation: 4x2 – 3x – 5 = 0
Divide throughout by coefficient of x2, i.e.,4 to get
x2 – 3 54 4
x− = 0
12
of coefficient of x =3
8−
and square of it = 364
Therefore, required constant is 3
64.
5. TrueReason: The value of t for which givenequation has real and equal roots are± 2 21 which are irrational.
6. 22 + 7 + 5 2x x = 0
⇒ 22 + 5 + 2 5 2x x x + = 0
⇒ ( ) ( )2 5 2 2 5x x x+ ++ = 0
⇒ ( ) ( )2 2 5x x ++ = 0
⇒ + 2x = 0 or +2 5x = 0
⇒ 2x = − or 5
2x
−= .
7. − +22 5 3x x = 0
⇒ 2 5 32 2
x x− + = 0
⇒ −2 52
x x = − 32
Add both side
254
⇒2
2 5 52 4
x x − + =
23 52 4
− +
⇒ −
254
x =− +24 25
16 =
116
⇒ − 54
x = ± 14
⇒ = +5 14 4
x or −5 14 4
⇒ x =32
or 1.
8. x = 2 or 1Hint: The given eqn. can be written as:
( )( )− − −+ −
7 44 7
x xx x
=1130
⇒ x2 − 3x − 28 = − 30⇒ x2 − 3x + 2 = 0.
9. Let speed of stream = x km/h∴ upstream speed = (18 − x) km/h
downstream speed = (18 + x) km/hTime taken to cover upstream distance of
24 km = 24
18 x−
136 AM T H E M A T C SI X–
Time taken to cover downstream distance of
24 km = 2418 x+
According to question,
24 2418 18x x
−− +
= 1
⇒ 24+ − +
− 218 18
324x x
x= 1
⇒ 48x = 324 − x2
⇒ x2 + 48x − 324 = 0x2 + 54x − 6x − 324 = 0
⇒ x(x + 54) − 6(x + 54) = 0⇒ (x − 6) (x + 54) = 0⇒ x = 6 or x = − 54 (Reject)∴ x = 6 km/h.
ORLet size of square be x∴ No. of students in square = x2
∴ According to question,Case I. Total students = x2 + 24Case II. Also total students = (x + 1)2 – 25∴ x2 + 24 = (x + 1)2 – 25⇒ x2 + 24 = x2 + 2x + 1 – 25⇒ 2x = 48 ⇒ x = 24Number of students = (24)2 + 24
= 576 + 24 = 600.
WORKSHEET – 2
1. (A) D = 2( 2 2)− – 4 × 2 × 1 = 8 – 8 = 0.
2. (C) Two equal and real roots⇒ D = 0⇒ k2 – 4 × 2 × 3 = 0⇒ k2 = 4 × 6
⇒ k = ± 2 6 .
3. (A) x = a must satisfy the given equation⇒ a2 – (a + b)a + k = 0 ⇒ k = ab.
For real and distinct roots, D > 0⇒ 62 – 4 × p × 1 > 0 ⇒ p < 9.
2. (C) D = (– 2)2 – 4 4 33 4
= 4 – 4 = 0.
3. (D) For equal roots, D = 0⇒ k2 – 4 × 4 × 9 = 0 ⇒ k = ± 12.
4. For no real roots, D < 0∴ (5m)2 – 4 × 1 × 16 < 0
⇒ 25m2 < 64 ⇒ m2 < 6425
⇒ m2 < 28
5
⇒ – 85
< m < 85
.
5. 23 9 2 6 3x x x+ + + = 0
⇒ ( ) ( )3 3 3 2 3 3x x x+ + + = 0
⇒ ( ) ( )+ +3 2 3 3x x = 0
⇒ x =−2
3 or x = −3 3 .
6. 5 x2 + 9x + 4 5 = 0
Dividing both sides by 5 to get
x2 +95
x + 4 = 0
Adding 29
2 5
to both sides to get
x2 + 95
x + 4 + 2 29 9
2 5 2 5 =
⇒ x2 +95
x +29 81
202 5 =
– 4
⇒ 2 29 1
2 5 2 5x + =
⇒ 2 29 1
2 5 2 5x + −
= 0
⇒ 9 1 9 12 5 2 5 2 5 2 5
x x + + + − = 0
⇒ x = 5− , – 45
.
7. 13, 15Hint: Two consecutive odd numbers are oftype 2x + 1, 2x + 3∴ According to question,(2x + 1)2 + (2x + 3)2 = 394.Now solve.
8. Let y = + 1x
x
142 AM T H E M A T C SI X–
∴ Given equation becomes 2y2 − 5y + 2 = 2∴ D = b2 − 4ac⇒ D = 25 − 4 (2) (2) = 25 − 16 = 9
∴ y = D
2b
a− ±
⇒ y = ±5 94
= ±5 34
= 2, 12
∴ =+
21
xx
or =+
11 2
xx
⇒ x = 2x + 2 or 2x = x + 1⇒ x = − 2 or x = 1.
9. Let the present ages (in years) of fatherand son be x and y respectively.∴ x = y2 ... (i)1 year ago, father’s age = (x – 1) years andson’s age = (y – 1) years.According to the question,
x – 1 = 8(y – 1)⇒ x – 8y + 7 = 0 ... (ii)Substituting the value of x from equation (i)in equation (ii), we have
y2 – 8y + 7 = 0(y – 1)(y – 7) = 0
⇒ y = 1 or 7Case I. If y =1 year1 year ago, son’s age = 1 – 1 = 0So, 1 year ago the father could not 8 timesas old as his son.Case II. If y = 7 years1 year ago, son’s age = 7 – 1 = 6 yearsSo, 1 year ago, the father was 8 × 6 = 48years old.∴ x = y2 = 72 = 49Hence, present ages of the father and theson are 49 years and 7 years.
ORLet number of books purchased = x
∴ Cost of 1 book =80x
According to question,
80x
− +80
4x= 1
⇒ 80 + − +
4( 4)
x xx x
= 1
⇒ 320 = x2 + 4x⇒ x2 + 4x − 320 = 0⇒ x2 + 20x − 16x − 320 = 0⇒ x (x + 20) − 16 (x + 20) = 0⇒ (x − 16) (x + 20) = 0⇒ x = 16 or x = − 20 (Reject)∴ Number of books = 16.
WORKSHEET – 8
1. (C)Hint: Discriminant is: D = b2 − 4ac.
2. (B) For equal and real roots, D = 0.
∴ (– 5)2 – 4.k.k = 0 ⇒ 4k2 = 25 ⇒ k = ± 52
.
3. (D) – 4 must satisfy x2 + px – 4 = 0⇒ p = 3 ... (i)
x2 + px + q = 0 has equal roots⇒ p2 – 4q = 0 ... (ii)
From (i) and (ii), we have p = 3, q = 94
.
4. No.
LHS = − +24(3) 14(3) 16
= − +36 42 16 = 10
RHS = − +9 12 3 + −9 9 = 0
As LHS ≠ RHS ⇒ x = 3 is not a root.
5. − − +25 5 4 4 5z z z = 0
⇒ ( ) ( )− − −5 5 4 5z z z = 0
⇒ ( ) ( )− −5 4 5z z = 0
⇒ z = 45
or z = 5 .
6. 4x2 + 4 3 x + 3 = 0Divide both sides by 4.
x2 + 3x +34
= 0
143DAUQ AR CIT AUQE IT SNO
Area (A1) of 1st square = x2
Area (A2) of 2nd square = y2
Perimeter of 1st square = 4x Perimeter of 2nd square = 4yAccording to question,
9. Let the breadth of the rectangular park beb metres.Then its length = (b + 3) metresArea of the rectangular park = b(b + 3) sq. mArea of the triangular park
=12
× base × altitude
= 12
× b × 12
= 6bNow,area of rectangular park – area of triangularpark = 4
b(b + 3) – 6b = 4⇒ b2 + 3b – 6b – 4 = 0⇒ b2 – 3b – 4 = 0⇒ (b – 4)(b + 1) = 0 ⇒ b = –1, 4Reject b = –1 as breadth is not possible innegative.∴ b = 4 m and b + 3 = 7 mHence, length = 7 m and breadth = 4 m.
OR
Let first number be x and second numberbe x + 4.According to question,
1 14x x
−+
=421
⇒4
( 4)x xx x
+ −+
=421
146 AM T H E M A T C SI X–
⇒ 24
4x x+ =
421
⇒ 21
4x x+ =
121
⇒ x2 + 4x – 21 = 0⇒ x2 + 7x – 3x – 21 = 0⇒ x(x + 7) – 3(x + 7) = 0⇒ x = 3 or x = – 7.Reject x = – 7∴ First number = 3
Second number = 7.
WORKSHEET – 11
1. (A) We have, kx2 − 2kx + 6 = 0For real, equal roots D = 0⇒ 4k2 − 24k = 0⇒ 4k(k − 6) = 0⇒ k = 0 or k = 6∴ k = 6.
2. (D)Hint: D = b2 − 4ac.
3. (B) If x = – 2 is a root of the equation, thenk(– 2)2 + 5(– 2) – 3k = 0 ⇒ k – 10 = 0⇒ k = 10.
7. Discriminant for 3 x2 + 10x – 8 3 = 0 isgiven by
D = 102 – 4 × 3 × (– 8 3 )= 100 + 96 = 196
⇒ D > 0As D > 0, the given equation has real roots.
Now, x =10 196 10 14
2 3 2 3− ± − ±=
⇒ x = 4 3− , 23
Thus, the given equation has real roots
which are 4 3− and 23
.
8. Let P's obtained marks in Mathematicsbe x then obtained marks in Science be(28 – x).Consider the question,P’s new marks in Mathematics = (x + 3)and P’s new marks in Science = (28 – x – 4)
⇒ (x – 9)(x – 12) = 0⇒ x – 9 = 0, x – 12 = 0∴ x = 9, x = 12If x = 9; 28 – x = 19, so, marks obtained inMathematics and Science are 9 and 19respectively.If x = 12; 28 – x = 16, so, marks obtained inMathematics and Science are 12 and 16respectively.
ORLet the sides of the two squares be a and b.Now, sum of areas = 640⇒ a2 + b2 = 640 ... (i)Difference of perimeters = 64⇒ 4a – 4b = 64⇒ a – b = 16⇒ b = a – 16 ... (ii)From equations (i) and (ii), we have
a2 + (a – 16)2 = 640⇒ 2a2 – 32a – 384 = 0⇒ a2 – 16a – 192 = 0⇒ a2 – 24a + 8a – 192 = 0⇒ (a – 24)(a + 8) = 0⇒ a = 24 or a = – 8Rejecting a = – 8 (negative length),... a = 24 mUsing equation (ii), b = 8 m.Hence, sides of the two squares are 24 mand 8 m.
9. 3 hr 30 min.Hint: Let average speed = x km/h∴ Distance = 2800 km
∴ Original time (duration) = 2800
x
∴ New time = −
2800100x
∴ −
2800100x
− 2800
x =
12
Now solve.
WORKSHEET–12
1. (A) Let x = 12 12 12 ....+ + + ...(i)
⇒ x = 12 x+[Using equation (i)]
⇒ x2 = 12 + x (Squaring)⇒ x2 – x – 12 = 0⇒ x = 4, – 3 (Reject) ∴ x = 4.
2. (D) For no real roots, D < 0∴ k2 – 4 × 1 × 1 < 0 ⇒ k2 – 22 < 0⇒ (k – 2)(k + 2) < 0 ⇒ – 2 < k < 2.
3. (A) Let us find the discriminant of equation
x2 – 4x + 3 2 = 0.
D = (– 4)2 – 4 × 1 × 3 2 = 16 – 12 2= 16 – 16.97
⇒ D < 0.
Therefore, x2 – 4x + 3 2 has no real roots.4. Given equation is:
x2 + ax – 4 = 0D = b2 – 4ac
= a2 – 4(– 4)= a2 + 16 > 0
As D > 0, two real and distinct roots exist.
150 AM T H E M A T C SI X–
5. Let the required whole number be x.
∴ x – 20 = 69 ×1x
⇒ x2 – 20x = 69⇒ x2 – 20x – 69 = 0⇒ x2 – (23 – 3)x – 23 × 3 = 0⇒ (x – 23)(x + 3) = 0⇒ x = 23 or x = – 3But – 3 is not a whole number∴ x = 23.
6. x = 23
a b+ , 2
3a b+
Hint: See solved example 5(ii).
7. (x − 5) (x − 6) = 2
25(24)
⇒ x2 − 11x = 225
(24) − 30
Add 211
2
to both sides.
⇒ x2 − 11x +
2112
= 225
(24) − 30 +
2112
⇒ −
2112
x = 225
(24) − 30 +
1214
= 225
(24) +
14
= 25 144576+ =
21324
⇒ x112
− = 1324
± .
⇒ x = 112
± 1324
⇒ x = 132 13
24±
⇒ x = 14524
;11924
⇒ x = 1
624
; 234
24.
8. Let the tap of larger diameter takes x hoursto fill the tank. Therefore, the other tap willtake (x + 10) hours to fill the same tank.The tap of larger diameter will fill the tank1x part in one hour and the other one will
time, we havex = 15 hours and x + 10 = 25 hours.Hence the tap of larger diameter and ofsmaller diameter can separately fill the tankin 15 hrs and 25 hrs respectively.
OR
7 m, 4 mHint: See Worksheet −−−−− 10, Sol. 9.
9.1 1
a b x x−
+ +=
1 1a b
+
⇒( )
x a b xx a b x
− − −+ +
= b aab+
⇒ − ab = x (a + b) + x2
⇒ x2 + x (a + b) + ab = 0⇒ (x + a) (x + b) = 0
⇒ x = − a or − b.
OR
Hint:D = 4a(a3 + b3 + c3 – 3abc)∴ D = 0
a = 0 or a3 + b3 + c3 = 3abc.
151DAUQ AR CIT AUQE IT SNO
WORKSHEET – 15
1. (C) For real roots, D ≥ 0∴ (– k)2 – 4 × 5 × 1 ≥ 0 ⇒ k2 ≥ 20
⇒ k ≤ – 20 or k ≥ 20 .
2. (A) 3(2)2 – 2p(2) + 2q = 0
and 3(3)2 – 2p(3) + 2q = 0⇒ 4p – 2q = 12 and 6p – 2q = 27
⇒ p = 152
, q = 9.
3. (B) D = ( )24 3 – 4 × 3 × 4.
= 48 – 48 = 0⇒ Two roots are real and equal.
4. False.There can be quadratic equation which haveno real roots e.g. x2 + 2x + 7 = 0; This equationhas no real roots because D = – 24 < 0.
5. No.Let their ages be x years and y years.Then x + y = 20 ... (i)And (x – 4)(y – 4) = 48 ... (ii)Consider equation (ii).
xy = 112 ... (iii )From equations (i) and (iii), we have
x2 – 20x + 112 = 0Here, D < 0Hence, the given situation is not possible.
9. Yes, 25 m and 16 mHint: Let the two adjacent sides of the fieldbe a and b.Then 2(a + b) = 82 ⇒ a + b = 41And ab = 400.
OR3 cm and 9 cmHint: Let smaller leg = xFrom figure,
x2 + y2 = (3 10 )2 = 90⇒ y2 = 90 − x2 ... (i)
Also (3x)2 + (2y)2 = (9 5 )2
⇒ 9x2 + 4y2 = 405 ... (ii)Use (i) and (ii) and then solve.
ASSESSMENT SHEET – 1
1. (C) D = ( )25− – 4 × 2 × 1 = 5 – 8 = – 3 < 0.
Given equation has no real roots.
2. (C) 3x2 + 2 6x + 2 + x2 = 4x2 – 4x
i.e., ( )6 2+ x + 1 = 0which is not a quadratic equation.
A
B C
y
x
3 10 cm
158 AM T H E M A T C SI X–
3. (x – 1)(x + 2) + 2 = 0 ⇒ x2 + x = 0⇒ x(x + 1) = 0 ⇒ x = 0 or –1So, roots are 0 and –1.
4. False, because a quadratic equation havingnegative discriminant has no real root.
5. 3x2 + 5 5x – 10 = 0 ... (i)
We divide 5 5x into two parts such thatsum and product of them are 5 5x and– 30x2 respectively.
Such parts are 6 5x and 5x− so equation (i)forms as follows:
3x2 + 6 5x – 5x – 10 = 0
or 3x ( )2 5x + – ( )5 2 5x + = 0
or ( )( )2 5 3 5x x+ − = 0
i.e., x + 2 5 = 0 or 3x – 5 = 0
i.e., x = – 2 5 or x =5
3Hence, the roots of equation
3x2 + 5 5 10x − = 0 are 2 5− and 5
3.
6. If a quadratic equation has equal real roots,then its discriminant vanishes.i.e., D = 0or b2 – 4ac = 0From the given equation,a = k – 12; b = 2(k – 12); c = 2So, b2 – 4ac = 0 provides
7. a2x2 – 3abx + 2b2 = 0Dividing throughout by a2, we get
x2 – 3bx
a+
2
22ba
= 0
Here, 12
of coefficient of x is – 32
ba
Adding both the sides to 23
2ba
− , we get
x2 –3bx
a+
232
ba
− +
2
2
2ba
= 23
2ba
−
i.e., 2 2 2
2 23 9 22 4
b b bx
a a a − = −
=2
24ba
i.e., 23
2b
xa
− =
2
2ba
i.e., 32
bx
a− = ±
2ba
i.e., 32 2
b bx
a a− = or 3
2 2b b
xa a
− = −
i.e., 2bx
a= or x =
ba
Hence, the roots of the given equation are
2ba and
ba .
8. Let the speed of the faster train be x km/hrand that of the slower train be y km/hr.
Time =Distance
Speed
Time taken by the slower train – Time takenby the faster train = 3 hrs.
⇒ 600y
– 600
x = 3 ... (i)
The speed of the slower train is 10 km/hrless than that of the faster traini.e., x – y = 10or x = y + 10 ... (ii)Substitute x = y + 10 from equation (ii ) inequation (i) to get
⇒ y2 + 10y – 2000 = 0⇒ y2 + 50y – 40y – 2000 = 0⇒ y(y + 50) – 40 (y + 50) = 0⇒ (y + 50)(y – 40) = 0⇒ y = 40, – 50Neglecting y = –50 because speed is a non-negative quantity, we get
y = 40Substitute this value of y, i.e., y = 40in equation (ii) we get
x = 40 + 10 = 50Hence, speed of the faster train = 50 km/hrand speed of the slower train = 40 km/hr.
ASSESSMENT SHEET–2
1. (B) 722
3
+ t 23
– 3 = 0
⇒28 29 3
+ t – 3 = 0
⇒ t = 3 28
32 9
−
⇒ t = – 3 12 9
× = 16
− .
2. (A) Let us find the discriminant of:
x2 + 2 3x – 1 = 0
D = ( )22 3x – 4 × 1 × (–1) = 12 + 4 = 16 > 0
⇒ x2 + 2 3x – 1 = 0 has real roots.
3. Consider x2 + 5px + 16 has no real roots.⇒ D < 0⇒ (5p)2 – 4 × 1 × 16 < 0
⇒ 25p2 < 64 ⇒ p < ± 6425
⇒ 85
− < p < 85
.
4. True.Let equation is ax2 + bx + c = 0Case I. a > 0 and c < 0 ⇒ ac < 0 ⇒ – ac > 0
∴ D = b2 – 4ac > 0 ∴ b2 ≥ 0Case II. a < 0 and c > 0 ⇒ ac < 0 ⇒ – ac > 0
∴ D = b2 – 4ac > 0.
5. (a – b) x2 + (b – c) x + (c – a) = 0As this equation has equal roots, thediscriminant of it vanishes.i.e., D = 0⇒ (b – c)2 – 4 × (a – b) × (c – a) = 0⇒ b2 – 2bc + c2 – 4ac + 4a2
7. Given equation isa2b2x2 – (4b4 – 3a4)x – 12a2b2 = 0 ... (i)General quadratic equation isAx2 + Bx + C = 0 ... (ii)Comparing the coefficients of like powers ofx of equations (i) and (ii), we getA = a2b2; B = – (4b4 – 3a4); C = –12a2b2
When y = 5, x = 2 + 52 = 27Hence, present age of Sumita is 27 yearsand Riya is 5 years.
CHAPTER TEST1. (C) Let us consider option (C).
⇒ 2x2 + 3 + 2 6x + x2 = 3x2 – 5x
⇒ (5 + 2 6 ) x + 3 = 0which is not a quadratic equation.
2. (B) 9x2 + 34
x – 2 = 0
Let us add and subtract 1
64.
9x2 + 34
x + 1 1
264 64
− − = 0
⇒ 21
38
x + −
21 64 2
8
+ = 0
Clearly, the required number is 164
.
161DAUQ AR CIT AUQE IT SNO
3. (C) The given equation can be written asx4 + x2 + 1 = 0
Here, D = 12 – 4 × 1 × 1 = – 3 < 0As D < 0, there is no real root.
4. 2x2 – kx + k = 0 has equal roots, if discrimi-nant = 0.⇒ (– k)2 – 4 × 2 × k = 0 ⇒ k(k – 8) = 0⇒ k = 0 or 8.
5. True.Let us consider a quadratic equation
23 7 3 12 3 0x x− + =
Here, D = ( )2– 7 3 – 4 × 3 × 12 3
⇒ D = 147 – 144 = 3⇒ D > 0⇒ Roots are real and distinct.
So, x = 7 3 3
2 3±
∴ x = 4, 3 which are
rationals.OR
No.(x – 1)2 + (2x + 1) = 0
⇒ x2 – 2x + 1 + 2x + 1 = 0⇒ x2 + 2 = 0Here, D = 02 – 4 × 1 × 2 = – 8 < 0.Hence, the given equation has no real root.
6. 2111
2x x− + 1 = 0
Here, 1
= 2
a , b = 11− , c = 1
D = b2 – 4ac
= ( )211− – 4 × 1
2 × 1 = 9
x = D2
ba
− ± =
( )11 3
2 × 1
− − ±
α = 11 32
+, β =
11 32
−.
7. Let the required natural number be N.N2 – 84 = (N + 8) × 3 ⇒ N2 – 3N – 108 = 0⇒ (N – 12)(N + 9) = 0 ⇒ N = 12 or –9
But –9 is not a natural number.So, N = 12 is the required natural number.
8. (b – c)x2 + (c – a) x + (a – b) = 0A quadratic equation is a perfect square, ifits discriminant (D) is equal to zero.Here, A = b – c, B = c – a and C = a – bNow, D = 0⇒ D = B2 – 4AC = (c – a)2 – 4(b – c)(a – b)⇒ c2 + a2 – 2ca – 4 (ab – b2 – ca + bc) = 0⇒ c2 + a2 + 4b2 + 2ca – 4bc – 4ab = 0⇒ (c + a – 2b)2 = 0[∵ (x + y + z)2 = x2 + y2 + z2 + 2xy
+ 2yz + 2zx]
⇒ c + a – 2b = 0 ⇒ b = 2
a c+.
Hence proved.9. Let the thickness of the region having the
grass be x metres.From the adjoining figure, we haveAB = PQ = DC = SR = 50 mAP = BQ = KQ = x mKL = QR = JM = PS = (40 – 2x) m
Now, area of the grass field = 1184⇒ AB × AP × 2 + QR × KQ × 2 =1184⇒ 50 × x × 2 + (40 – 2x) × x × 2 = 1184⇒ 180x – 4x2 = 1184⇒ x2 – 45x + 296 = 0⇒ x2 – 37x – 8x + 296 = 0⇒ (x – 37)(x – 8) = 0⇒ x = 8 or x = 37But x = 37 is not possible as 2x > 50∴ x = 8 m∴ Length of the pond
= JK = 50 – 16 = 34 m And breadth of the pond = JM
= 40 – 16 = 24 m.
❑❑
162 AM T H E M A T C SI X–
2Chapter
ARITHMETIC PROGRESSIONS
WORKSHEET– 241. (D) Let us consider option (D).
2nd term – 1st term = – 6 – (–10) = 43rd term – 2nd term = – 2 – (– 6) = 44th term – 3rd term = 2 – (– 2) = 4∴ – 10, – 6, – 2, 2,........ is an A.P.
2. (B) 11th term of the A.P. – 62, – 59, – 56, ..., 7,10 is – 62 + (11 – 1) × 3, i.e., – 32.
5. FalseLet first term be a and common differencebe d of an A.P.Then
3rd term, a + 2d = 4 ... (i)9th term, a + 8d = – 8 ... (ii)
Subtracting equation (ii) from (i), we get⇒ – 6d = 12⇒ d = – 2Putting d = – 2 in (i), we get... a = 8... Given a + (n – 1)d = 0⇒ 8 + (n – 1) (– 2) = 0⇒ – 2n = – 10 ∴ n = 5
5th term = a + (n – 1)d= 8 + (5 – 1) (– 2)= 0
Thus its 5th term is 0. So the given statementis false.
6. First term = a = – 1
Second term = a + d = – 1 + 12
= – 12
Third term = a + 2d = – 1 + 2 × 12
= 0
Fourth term = a + 3d = – 1 + 3 × 12
= 12
Hence, the first four terms are: – 1, – 12
, 0, 12
.
7. Let the first term be a and the commondifference be d.A.P. = a, a + d, a + 2d,........According to question,T3 = 16 and T7 = 12 + T5
⇒ a + 2d = 16 and a + 6d = 12 + a + 4d⇒ a + 2d = 16 and d = 6⇒ a = 4 and d = 6So, the required A.P. will be 4, 10, 16, ......
8. (i) Here, a = 9, d = 17 – 9 = 8Let 636 be the sum of n term of this A.P.where (n ∈ N)
Let’s use Sn = 2n
[2a + (n – 1)d]
636 = 2n [2 × 9 + (n – 1)8]
⇒ 636 = 2n
[18 + 8n – 8]
⇒ 636 = 2n
[10 + 8n]
⇒ 636 = 2n × 2[5 + 4n]
⇒ 4n2 + 5n – 636 = 0
Using quadratic formula
n = 2 . .– 5 5 – 4 4 (– 636)
.2 4
±
= – 5 10201
8±
163TIRA MH ITE GORP ER SSC I NO S
= – 5 101
8±
= –10696
,8 8
∴ n = 12, – 13.25Rejecting n = –13.25 due to negativevalue.
4. p = 4Hint: Use: if a, b, c are in A.P. ⇒ 2b = a + c.
5. FalseIn an A.P., having nth term an = a + (n – 1)d,we know that an is a linear polynomial in n.Here an = n2 + n + 1 is not a linearpolynomial in n. So it can't be nth term ofan A.P.
6. Let the first term be a and the commondifference be d.
Middle term = th21 1
2+
term = 11th term.
Sum of three middle terms:a10 + a11 + a12 = 129
⇒ (a + 9d) + (a + 10d) + (a + 11d) = 129⇒ 3a + 30d = 129 ... (i)Sum of last three terms:
a19 + a20 + a21 = 237
⇒ (a + 18d) + (a + 19d) + (a + 20d) = 2373a + 57d = 237
...(ii)Subtracting equation (i) from equation (ii),we have
27d = 108 ⇒ d = 4Substituting d = 4 in equation (i), we have
7. Let the first term be a and the commondifference be d.
∵ Sn = 2n
[2a + (n – 1)d] ∴ 42 = 62
[2a + 5d]
⇒ 2a + 5d = 14 ...(i)
∵ an = a + (n – 1)d ∴ +
+9
29a d
a d =
13
⇒ 3a + 27d = a + 29d ⇒ 2a – 2d = 0 ⇒ a = dSubstituting a = d in equation (i), we have
d = 2 and so a = 2Now, a18 = 2 + 17d = 2 + 17 × 2 = 36.Hence the first term is 2 and 18th term is 36.
8. Hint: mam = nan⇒ m{a + (m – 1)d} = n{a + (n – 1)d}⇒ m{a + (m – 1)d} – n{a + (n – 1) d} = 0⇒ a(m – n) + {m(m – 1) – n(n – 1)} d = 0⇒ a(m – n) + {(m2 – n2) – (m – n} d = 0⇒ a(m – n) + {(m – n) (m + n – 1) d = 0⇒ a + (m + n – 1) d = 0 {... m ≠ n}⇒ am + n = 0.
9. The penalty (in `) is in A.P. as follows:200, 250, 300, ........... 30 termsSo, the contractor has to pay the penalty(in `) as the sum of the following series:200 + 250 + 300 + ............30 termsHere, a = 200, d = 250 – 200 = 50, n = 30
∴ Sum = 2n
[ 2a + (n – 1)d]
= 302
[ 2 × 200 + 29 × 50]
= 15(400 + 1450) = 27750Hence, the required penalty is ` 27750.
168 AM T H E M A T C SI X–
WORKSHEET–30
1. (B)Hint: The sequence is:
3, 9, 15, ......., 99.
Use: Sn = { }2n
a l+ .
2. (A) an = Sn – Sn–1
= (2n2 + 5n) – [2(n – 1)2 + 5(n – 1)]
= 2n2 + 5n – 2(n2 + 1 – 2n) – 5n + 5
= 2n2 + 5n – 2n2 – 2 + 4n – 5n + 5
= 4n + 3.
3. Here, –56
– (–1), i.e. 16
and –23
–
5–
6,
i.e., 16
are equal. So, the given sequence is
in A.P. with common difference 16
.
Therefore, the next three terms will be
2 1 2 1– + , – + 2 ×
3 6 3 6 and + ×2 1
– 33 6
These are 1 1 1– , – and –
2 3 6.
4. ∵ 8x + 4, 6x – 2 and 2x + 7 are in A.P.
∴ 6x – 2 – (8x + 4) = 2x + 7 – (6x – 2)
⇒ – 2x – 6 = – 4x + 9
⇒ 2x = 15
⇒ x = 152
.
5. 3, 9, 15Hint: Let the three numbers be
a − d, a, a + d.
OR
a17 = a10 + 7⇒ a + 16d = a + 9d + 7⇒ 7d = 7⇒ d = 1.
Let the first term and the common differenceof the given A.P. be a and d respectively.
5th term = 0 ⇒ a + 4d = 0⇒ a = – 4d ...(i)
23rd term: a23 = a + 22d⇒ a23 = – 4d + 22d
[From equation (i)]⇒ a23 = 18d ...(ii)
11th term: a11 = a + 10d⇒ a11 = – 4d + 10d
[From equation (i)]
⇒ a11 = 6d ⇒ a11 = 6 × 23
18a
[From equation (ii)]⇒ a23 = 3a11
⇒ 23rd term = 3 × 11th termHence proved.
9. Let the digits of the number be a – d, a anda + d such the required number is100(a – d) + 10a + a + d as the digits are inA.P.
171TIRA MH ITE GORP ER SSC I NO S
So, the required number = 111a – 99d ...(i)Sum of the digits = 15
⇒ a – d + a + a + d = 15⇒ a = 5 ...(ii)The number obtained by reversing the digits
= 100(a + d) + 10a + a – d= 111a + 99d ...(iii)
According the given condition, we have111a – 99d = 594 + 111a + 99d
[Using equation (i) and (iii)]⇒ –2 × 99d = 594⇒ d = –3 ...(iv)Using equations (i), (ii) and (iv), we arrivethat the original number is111 × 5 – 99 × (–3), that is 852.
OR
16 rows, 5 logs are placed in top row.Hint: Put Sn = 200, a = 20, d = –1
in formula Sn =2n
[2a + (n – 1)d]
So 41n – n2 = 400⇒ n = 16, 25... n = 25 not possiblebecause if n = 25 then the number of logs intop row
= – 4... n = 16 and a16 = 5.
WORKSHEET – 33
1. (A) a = 10, d = 7 – 10 = – 3a30 = a + 29d = 10 + 29(–3) = – 77.
2. (C) x + 10 – 2x = 3x + 2 – (x + 10)⇒ x – 2x – 3x + x = 2 – 10 – 10
⇒ x = 183
= 6.
3. (C) an = 2n + 1 ∴ a1 = 2 × 1 + 1 = 3
Now, Sn = 2n
(a1 + an) = 2n
(3 + 2n +1)
= n(n + 2)
4. No.Let an = 68
⇒ a + (n − 1) × d = 68
⇒ 7 + (n − 1) × 3 = 68⇒ 3n = 64
⇒ n =643
which is not a whole number so an = 68 notpossible.
7. Hint: Let the first term and the commondifference be a and d respectively.
a9 = 0 ⇒ a + 8d = 0 ⇒ a = – 8da29 = a + 28d = – 8d + 28d = 20da19 = a + 18d = – 8d + 18d = 10d.
8. 2, 6, 10, 14.Hint: Let the four parts be:
a − 3d, a − d, a + d, a + 3d.
OR
The sequence is: 150, 146, 142, ..........∴ Total number of workers who worked
all the n days
= 150 + 146 + ....... + n terms.∴ = n(152 − 2n)
Now had the workers not dropped thenthe work would have finished in (n − 8)days with 150 workers working on eachday.∴ Total number of workers who would
have worked all the n days is 150(n −8).∴ n(152 − 2n) = 150(n − 8)
⇒ n2 − n – 600 = 0⇒ (n − 25) (n + 24) = 0
⇒ n = 25 or n = – 24 (Reject)∴ n = 25.
ASSESSMENT SHEET – 3
1. (B) an = a + (n – 1)d
⇒ 210 = 21 + (n – 1) × 21
⇒ n – 1 = 210 – 21
21⇒ n = 10.
2. (A) a18 – a14 = 32 ⇒ a + 17d – (a + 13d) = 32
⇒ d = 324
⇒ d = 8.
3. Sn = 2n2 + 5n
⇒ Sn–1 = 2(n – 1)2 + 5(n – 1)
= 2(n2 + 1 – 2n) + 5n – 5
= 2n2 + n – 3
176 AM T H E M A T C SI X–
nth term = Sn – Sn – 1
= 2n2 + 5n – (2n2 + n – 3)= 4n + 3.
4. False, because nth term of an A.P. is alwaysa linear polynomial in n.
5. First term of the A.P. = a = 4
–3
Common difference of the A.P. = d
= – 1 –
4–
3=
13
Let the A.P. consists n terms.
∴ nth term = 14
3 = 13
3...(i)
But nth term is given byan = a + (n – 1)d
= 4
–3
+ (n – 1) 13
= 3n
– 53
...(ii)
From equations (i) and (ii), we get
3n
– 53
= 133
⇒ n = 18
Since n = 18 is even number so, the middle
most terms will be th18
2
and th18
12
+ terms, i.e., 9th and 10th terms.
Now, a9 = –43
+ (9 – 1) ×13
= –43
+83
=43
And a10 = –43
+ (10 – 1)13
= –43
+ 93
= 53
Therefore, the required sum
= a9 + a10 = 43
+53
=93
= 3.
6. First term = a = +–x y
x y
Common difference = d = +
3 – 2x yx y
– +–x y
x y
= +
2 –x yx y
Now, sum of n terms
=2n
[2a + (n – 1)d]
= 2n – 2 –
2 ( – 1)x y x y
nx y x y
× + + +
= 2 – 2 2 – – 2
2x y nx ny x yn
x y
+ + +
= 2 – –
2nx ny yn
x y +
= ( )+2n
x y {n(2x – y) – y}.
7. Let the first term and the common differenceof the given A.P. be a and d respectively.According to the given condition,
11
18
aa
= 102 2
3 17 3a da d
+⇒ =
+[Using an = a + (n – 1)d]
⇒ 3a + 30d = 2a + 34d⇒ a = 4d
Now, 5
21
aa =
++
420
a da d
= 4 44 20
d dd d
++
(Substituting a = 4d)
= 8 124 3
dd
=
i.e., a5 : a21 = 1 : 3.
Now, 5
21
SS
= +
+
52 4
221
2 202
a d
a d
( ){ }Using S 2 – 12nn
a n d = +
177TIRA MH ITE GORP ER SSC I NO S
= ( )( )
5 8 4
21 8 20
d d
d d
++[Substituting a = 4d]
= 60588
dd
= 549
i.e., S5 : S21 = 5 : 49.
8. Distances covered by a girl during 1st
minute, 2nd minute, 3rd minute,..... arerespectively 20 m, 18 m, 16 m,.......whichform an A.P. with first term (a) = 20 m andcommon difference (d) = – 2 m.(i) Distance covered in 10th minute
= 10th term of the A.P.= a + (10 – 1)d = 20 + 9 × (– 2) = 20 – 18= 2 m.
(ii) Distance covered in 10 minutes= sum of first 10 terms
= k(1 + 3 + 5 +..... n terms)⇒ 2(1 + 2 + 3 +........ n terms)= k(1 + 3 + 5 + ...... n terms)
⇒ 2+( 1)
2n n
= k ×2n
× {2 + (n – 1) × 2}
⇒ n2 + n = kn2
⇒ k = 1n
n+
.
2. (D) a + 8d = 449 and a + 448d = 9⇒ a = 457; d = – 1Further, 0 = 457 + (n – 1) (– 1)⇒ n = 458.
3. Let the A.P. has n terms. So, an = 2a.2a = a + (n – 1) × (b – a)
⇒ 2a = a + n(b – a) – b + a
⇒ n = –b
b a
Sn = ( )2 –b
b a (a + 2a)
⇒ Sn = ( )3
2 –ab
b a.
4. Yes; the first term is ` 2000 of an A.P. andthe common difference is ` 200 which isinterest per year.
5. The integers between 100 and 200, whichare divisible by 9 are:108, 117, 126,........, 198.Which is an A.P. with first term(a) = 108and common difference (d) = 9.Let this A.P. has n terms nth term is given by
an = a + (n – 1)d ⇒ 198 = 108 + (n – 1) × 9
⇒ n – 1 = 198 –108
9 ⇒ n = 1 +
909
⇒ n = 11
Now,
Sn = 2n
(a + l), l being last term
S11 = 112
(108 + 198) = 112
× 306
= 1683.Required sum
= 101 + 102 +....+ 199 – S11
= 992
(101 + 199) – 1683 = 14850 – 1683
= 13167.
6. We need to prove(x + 2y – z) (2y + z – x) (z + x – y) = 4xyz
...(i)If x, y, z are in A.P., then
y – x = z – y ⇒ y = 2
z x+...(ii)
Let us take LHS of (i)(x + 2y – z) (2y + z – x) (z + x – y)
= (x + z + x – z) (z + x + z – x)
–2
z xz x
+ + [Using (ii)]
= 2x × 2z × 2
z x+ = 2xz(z + x)
= 2xz × 2y [Using (ii)]
= 4xyz = RHS. Hence proved.
178 AM T H E M A T C SI X–
7. Let first term = a,Common difference = d
Sum of first n terms = 2n
[2a + (n – 1)d]
Sum of first 20 terms = 400
⇒ 202
[2a + (20 – 1)d] = 400
2a + 19d = 40 ...(i)Sum of first 40 terms = 1600
402
[2a + (40 – 1)d] = 1600
2a + 39d = 80 ...(ii)Subtracting equation (i) from equation (ii),we get
20d = 40 ⇒ d = 2Substituting d = 2 in equation (i), we get
2a + 38 = 40 ⇒ a = 1Now,
sum of first 10 terms = 102
[2a + (10 – 1)d]
= 5(2 + 9 × 2)= 100.
8. See Worksheet – 37, Sol. 8 (OR part).
CHAPTER TEST
1. (C) Sn = 2n
(a + an) ⇒ 399 = 2n
(1 + 20)
⇒ 21n = 2 × 399 ⇒ n = 38.
2. (B) a + d = 13 and a + 4d = 25⇒ a = 9, d = 4Now, a7 = a + 6d = 9 + 24 = 33.
3. (A) ∵ ap = 3 –1
6p
∴ an = 3 – 1
6n
and a1 = 13
Now, Sn = 2n +
1 3 – 13 6
n =
12n
(3n + 1).
4. Let the nth term be requireda9 = 449 ⇒ a + 8d = 449
a449 = 9 ⇒ a + 448d = 9
So, a = 457 and d = – 1Now, an = 300 ⇒ 457 + (n – 1) (–1) = 300⇒ (n – 1) = 157 ⇒ n = 158.
6. The first two digit number divisible by 4 is12 and the others are: 16, 20, 24,............., 96.Now, we have to find the sum of thefollowing series:12 + 16 + 20 +.............+ 96Here, a = 12, d = 16 – 12 = 20 – 16 = 4Let the number of terms in the series be n.nth term is given by
7. Let the profit to be ceased at nth day.Sale on first day = ` 8100
Sale on second day = ` (8100 – 150)= ` 7950
So, the sale (in `) will be day by day asfollows:8100, 7950, 7800,..........n termsHere, a = 8100, d = –150The profit will be ceased when it is equalto or less than ` 1500.Therefore, 8100 + (n – 1) × (– 150) ≤ 1500
[∵ an = a + (n – 1)d]
⇒ 8100 – 150n + 150 ≤ 1500⇒ 150n ≥ 6750 ⇒ n ≥ 45Hence, the profit to be ceased at 45th day.
179TIRA MH ITE GORP ER SSC I NO S
8. If x, y and z are in A.P., then
y = +2
x z...(i)
Now, in the given equation,LHS = (x + 2y – z) (2y + z – x) (z + x – y)
= (x + x + z – z) (x + z + z – x) (2y – y)[Using equation (i)]
= 2x × 2z × y= 4xyz= RHS. Hence proved.
9. Total number of rungs =
+
2.5 m1
25cm
= 25025
+ 1 = 11
Length of the largest rung = 45 cmLength of the shortest rung = 25 cm
Let the length of each rung decreases byx cm from bottom to top.So, lengths (in cm) of all rungs frombottom to top are respectively45, 45 – x, 45 – 2x, .......25.This is an A.P. of 11 terms∴ 45 + (11 – 1) × (– x) = 25
[Using a + (n – 1)d = an]⇒ – 10x = – 20 ⇒ x = 2 cmSo, the A.P. now becomes45, 43, 41,........., 25Now, required length of wood
= Sum of this A.P.
= 112
(45 + 25) = 112
× 70
= 385 cmHence, the required length of the wood is3.85 m.
❑❑
180 AM T H E M A T C SI X–
3Chapter
CIRCLES
6. See solved example 4.
7. AB = 13 cm, AC = 15 cmHint:Usear(∆OBC + ∆OAC + ∆OAB) = ar(∆ABC).
8. Hint: Join ABLet OA = r ⇒ OP = 2rIn ∆OAP,
sinθ = OA 1OP 2 2
rr= =
θ = 30°∴ ∠APB = 2 × 30° = 60°
In ∆ABP, AP = BP∴ ∠A = ∠B
But, ∠A + ∠B = 180° – ∠APB = 120°⇒ ∠A = ∠B = 60°∴ ∆APB is an equilateral triangle.
WORKSHEET– 421. (B) As ∠OQP = 90º∴ x = 90º – 30º = 60º.
2. (Α) ∠QOR = 180º – 30º = 150º
∠PRQ = 12
× 150º = 75º.
3. (C) BC = BP + PC= BR + CQ= 3 + [AC – AQ]= 3 + [11 – 4]= 10 cm.
WORKSHEET– 411. (B) Join OX.
In ∆XOY,∠OXY= 90°,
∴ XY = 2 2OY – OX
= 2 213 – 5 = 12 cm.
2. (D) OT2 + TQ2 = OQ2
⇒ OT2 + (24)2 = (25)2
⇒ OT2 = 625 – 576 = 49⇒ OT = 7 cm.
3. (B) As AB || PR⇒ ∠BQR = ∠ABQ = 70°
(Alternate interiar angles)
Also ∠ABQ = ∠BAQ = 70° {∵ ∆AMQ ≅ ∆BMQ}
∴ In ∆AQB, using Angle sum property∠AQB = 180° – 70° – 70° = 40°.
4. False,Perimeter of ∆ABC = AB + BC + AC= AB + BQ + CQ + CR + AR= AB + BP + CQ + CQ + AP= AB + (BP + AP) + 2CQ= 2(AB + CQ)= 2(8) = 16 cm.
5. LHS= AB + CD= (AP + PB) + (CR + RD)= AS + BQ + CQ + DS= (AS + DS) + (BQ + CQ)= AD + BC= RHS.
181CRIC SEL
4. True.Let M be the point ofcontact and O be thecentre of the circle.
OM = OM ... (iii) (Common)Using equations (i), (ii) and (iii) in ∆BMOand ∆CMO, we have
∆BMO ≅ ∆CMO (AAS corollory)∴ BM = CM (CPCT)⇒ BC is bisected at the point of contact.
5. As AD + BC = AB + CD⇒ AD + 7 = 6 + 4⇒ AD = 3 cm.
6. 11 cmHint: OQBP is a square∴ OQ = BP = 11 cm.
7. Let the tangents be PQ andPR corresponding to thechord QR of the circle withcentre O.Join OQ, OR and OP.In ∆PQO and ∆PRO,∠PQO = ∠PRO = 90°(Angles formed betweentangent and correspondingradius)
PO = PO (Common)QO = RO (Radii of same circle)
Therefore, we arrive at∆PQO ≅ ∆PRO(RHS axiom of congruence)So, PQ = PR (∴ CPCT)Thus, ∆PQR is an isosceles triangle.∴ ∠PQR = ∠PRQ. Hence proved.
8. Let the given parallelogrambe ABCD whose sidestouches a circles at P, Q, Rand S as shown in theadjoining figure.Since, length of twotangents drawn from anexternal point to a circle are equal.∴ AP = AS ...(i)Similarly, we have
PB = BQ ...(ii)DR = SD ...(iii)RC = QC ...(iv)
Adding these four equations, we haveAP + PB + DR + RC = AS + BQ + SD + QC
⇒ (AP + PB) + (DR + RC)= (AS + SD) + (BQ+ QC)
⇒ AB + DC = AD + BC∵ AB = DC and AD = BC
(ABCD is a parallelogram)∴ AB = BCThus, AB = BC = CD = DAHence, ABCD is a rhombus.
9. Let line l be thetangent at apoint P to thecircle with centreO. Let us takeany point Q onthe tangent l as shown in the figure.Join OQ to meet the circle at M.We know that if a point is met with thedifferent points of a line, then the shortestline segment is the perpendicular on thatline. Consider the adjoining figure:
OM = OP (Radii of same circle)OQ = OM + MQ
⇒ OQ = OP + MQ⇒ OQ > OPi.e., OP < OQClearly, OP is the shorter than OQ. Similarly,we can prove that OP is the shortest all OV,V being a variable point on the line otherthan P. Therefore, OP is the perpendicular toline l.
182 AM T H E M A T C SI X–
Hence, tangent l ⊥ radius OP.2nd Part: Join OY
∠OYX = 90ºand ∠OAY = b + a = ∠OYA
[∵ OA = OY = radius]⇒ b + a = 90º – a⇒ b + 2a = 90º.
WORKSHEET – 431. (B) ∠Q = ∠R = 90°
In quadrilateral PQOR,
∠P = 360° – (90° + 130° + 90°) = 50°
2. (C)Hint: AC = 10 cm∴ ar(∆ABC)= ar(∆AOB) + ar(∆BOC)
+ ar(∆AOC)
⇒ 24 =12
× (8 × r + 6 × r + 10 × r)
⇒ 48 = r × 24 ⇒ r = 2 cm.
3. ∠PTQ + ∠POQ = 180º⇒ ∠PTQ = 180º – 110º
= 70º.
4. CP = CQ = 11 cmBQ = CQ – CB
= 11 – 7 = 4 cm∴ BR = QB = 4 cm.
5. See Worksheet – 42, Sol. 8.
6. Hint: XP = XQ⇒ XA + AP = XB + BQ⇒ XA + AR = XB + BR
(∵ OP ⊥ PT)And ∠MOP = 90° – θ ... (v)In ∆TPM and ∆POM,
∠TPM = ∠POM = 90° – θ[From (iii) and (v)]
∠PTO = ∠MPO = θ[From (ii) and (iv)]
∠TMP = ∠PMO (Each 90°)
So, by AAA criterion of similarity, we have ∆TPM ~ ∆POM
⇒TPPO
= PMOM
⇒ TP7
= 5
2 6
⇒ TP = 35
2 6×
66
=35 6
12cm.
6. See Worksheet – 41, Sol. 5.7. Let the given two
tangents be PA and PBto the circle withcentre O.We need to prove∠APB + ∠AOB = 180°.We know that the angle formed by atangent to the circle and the radius passingthrough the point of contact is 90°.
∴ ∠PAO = ∠PBO = 90°Applying angle sum property in thequadrilateral AOBP, we get
(∵ OS ⊥ AD and OR ⊥ CD)⇒ ∠O = 90°⇒ OSDR is a square.⇒ DR = DS = rNow, ABCD is a subscribed quadrilateral∴ AB + CD = BC + DA⇒ x + 27 + 25 = 38 + r + x⇒ r = 14 cm.
6. See Worksheet – 44, Sol. 4.OR
We know that twotangents drawn froman external point to acircle are equal inlength.∴ PA = PB⇒ ∠PBA = ∠PAB = θ (say) ...(i)⇒ ∠APB = 180° – 2θ ...(ii)
(Using ∆APB)Further, PA is tangent and AO is corres-ponding radius⇒ PA ⊥ AO⇒ ∠PAO = 90°
185CRIC SEL
⇒ ∠OAB = 90° – ∠PAB⇒ ∠OAB = 90° – θ ...(iii)
[Using equation (i)]Dividing equation (ii) by equation (iii), weget
APBOAB
∠∠
= 180 – 2
90 –° θ° θ
= ( )2 90 –90 –
° θ° θ
= 2
⇒ ∠APB = 2∠OAB. Hence proved.
7. Let the given quadrilate-ral be ABCD subscribinga circle with centre O. Letthe sides AB, BC, CD andDA touch the circle at P,Q, R and S respectively(see figure).Join OA, OB, OC, OD, OP, OQ, OR and OS.We need to prove∠AOB + ∠COD = ∠BOC + ∠DOA = 180°.Proof: In ∆AOP and ∆AOS,
OP = OS (Radii of same circle)AP =AS (Tangents from external points)AO = AO (Common)
∴∆AOP ≅ ∆AOS(SSS axiom of congruence)∴∠1 = ∠8 ...(i) (CPCT)Similarly, we can prove that∠2 = ∠3, ∠4 = ∠5 and ∠6 = ∠7 ... (ii)As, ∠1, ∠2, ∠3, ∠4, ∠5, ∠6, ∠7 and ∠8 aresubtended at a point∴ ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8
⇒ 42 (x + 14)2 = (x + 14) × 6 × x × 8(On squaring both sides)
⇒ 16 (x + 14) (x + 14 – 3x) = 0⇒ x = 7 as x ≠ –14 (∵ x > 0)So, AB = x + 8 = 7 + 8 = 15 cmand AC = x + 6 = 7 + 6 = 13 cm.
ASSESSMENT SHEET–51. (B) BC = 2AB
= 2 169 – 25
= 24 cm.
2. (C) If a circle touches the sides of aquadrilateral, then the angles subtended byeach pair of opposite sides are supplementary.∴ ∠AOB + ∠COD = 180°⇒ 130° + ∠COD = 180°⇒ ∠COD = 50°.
3. In right triangle AOP,
tan (∠APO) = Perpendicular
Base
⇒ tan 30° = AOAP
⇒ 13
= 5
AP ⇒ AP = 5 3
∴ AP = BP = 5 3 cm.
4. True, because∠MAB = ∠ACB
and ∠NAC = ∠ABC, But ∠ABC = ∠ACB∴ ∠MAB = ∠ABCand ∠NAC = ∠ACBBut these are alternate pair of interior angles.∴ MN || BC.
5. Note that tangentsdrawn from anexternal point to acircle are equal inlength.Tangents are drawn from A, B and C, so werespectively get
AQ = AR ... (i)BQ = BP ... (ii)
And CR = CP ... (iii)Taking equation (i), we
AQ = AR = AC + CR = AC + CP[From (iii)]
= AC + BC – BP= AC + BC – BQ [From (ii)]= AC + BC – (AQ – AB)
187CRIC SEL
= AC + BC + AB – AQ⇒ 2 AQ = Perimeter of ∆ABC
⇒ AQ = 12
(Perimeter of ∆ABC)
Hence proved.
6. We are given twotangents AP and BPcorresponding tothe chord AB of acircle with centre O.We need to prove
∠PAB = ∠PBAJoin AO, BO and PO.In ∆AOP and ∆BOP,
∠PAO = ∠PBO = 90°(Angle between tangent and correspondingradius)
OP = OP (Common)AO = BO (Radii of same circle)
Therefore,∆AOP ≅ ∆BOP (RHS criterion)
∴ PA = PB (CPCT)Thus, ∆ PAB is an isosceles triangle.⇒ ∠PAB = ∠PBA. Hence proved.
7. Let O be thecentre of thecircle. Join OT tomeet PQ at R.Join OP and OQ.
In ∆PTR and ∆QTR,PT = QT
(Tangents from an external point to thesame circle)
∠PTR = ∠QTR(Line segment joining the centre and the
point of intersection of the tangents bisectsthe angle between the tangents)TR = TR (Common)
Therefore,∆PTR ≅ ∆QTR (SAS criterion)
⇒ PR = QR (CPCT)Also, ∠ORP = ∠ORQ = 90°(Angle between chord and the line segment
∴ AB = AD + DB = 4 + 4 = 8 cm.4. False, because the centres of the circles lie on
the perpendicular of PQ, which passesthrough A.
5. Let the sides AB,BC and CA of the∆ABC touch thecircle with centre Oat the point P, Qand R respectively.
In quadrilateral OQCR,∠OQC = ∠ORC = 90°
(Angles between tangent andcorresponding radius)
and ∠QCR = 90° (Given)⇒ ∠QOR = 90°⇒ OQCR is a square⇒ CQ = CR = r⇒ BQ = a – r, AR = b – r,⇒ AP = b – r, PB = a – rBut AB = c∴ b – r + a – r = c⇒ 2r = a + b – c
⇒ r = – c
2a b+
. Hence proved.
6. Draw a line QTpassing through Qand perpendicular toQP to meet SR at T.In ∆PQR,
PQ = PR(Tangents from an external point)
∴ ∠PRQ = ∠PQR ... (i)(Angles opposite to equal sides)
∠PQR + ∠PRQ + ∠QPR = 180° ... (ii)(Angle sum property for a triangle)
From equations (i) and (ii),∠PQR + ∠PQR + 30° = 180°
(Angle between tangent)⇒ TQR = 15° ... (iv) [Using (iii)]∴ SR || QP and QT ⊥ QP∴ QT ⊥ SR⇒ ST = TR ... (v)(∵ TQ passess through the centre of thecircle)In ∆STQ and RTQ,
and ∠OSD = ∠ORD = 90° ... (ii)(Angle between tangent and
corresponding radius)Therefore,
∠SOR = 90° ... (iii)⇒ Quadrilateral DROS is a rectangle
[From (i), (ii) and (iii)]But SO = SR (= r)∴ DROS is a square.⇒ r = SO = DR = SD = OR ... (iv)We know that tangents drawn from anexternal point to a circle are equal.∴ BQ = BP = 27 cm
(Tangents from point B)QC = BC – BQ = 38 – 27 = 11 cm
RC = QC = 11 cm(Tangents from point C)
DR = DC – RC = 25 – 11 = 14 cm∴ r = 14 cm [From (iv)]Hence, radius of the incircle is 14 cm.
8. Let AD = x. Weknow that thetangents drawnfrom an externalpoint to a circleare equal.
∴ AD = AF = x
BD = BE and CE = CFNow, BD = AB – AD = 8 – x = BEand CE = BC – BE = 10 – (8 – x)
= 2 + x = CF.AF = AC – CF = 12 – (2 + x)
= 10 – x = ADBut AD = x∴ 10 – x = x ⇒ x = 5 cm, i.e., AD = 5 cm
BE = 8 – x = 8 – 5 = 3 cmand CF = 2 + x = 2 + 5 = 7 cmThus, AD = 5 cm, BE = 3 cm and CF = 7 cm.
CHAPTER TEST
1. (C) ∵ OA ⊥ AT∴ ∠OAT = 90°In ∆OAT,
cos T = ATOT
⇒ cos 30° = AT4
⇒ AT = 2 3 cm.
2. (A) ∠PAO = 90°∠OAB = ∠PAO – ∠PAB
= ∠PAO – 12
(∠PAB + ∠PBA)
(∵ ∠PAB = ∠PBA)
= ∠PAO – 12
(180° – ∠APB) (ASP)
= 90° –12
(180° – 50°) = 25°.
3. In right ∆POQ,
PQ = 2 213 – 5 = 12 cm
PR = PQ = 12 cm
ar( PQOR)
= ar(∆PQO) + ar(∆PRO)
= 1
PQ QO2
× × +
1PR RO
2 × ×
190 AM T H E M A T C SI X–
= 1
12 52
× × +
112 5
2 × ×
= 2 × 30 = 60 cm2
4. True, as ∠BPA = 90°, (∵ AB is diameter)
∠PAB = ∠OPA = 60° (∵ OP = OA)
Also OP ⊥ PT.∴ ∠APT = 30°and ∠PTA = 60° – 30° = 30°.
5. Let the given chord be ABand two tangents to thecircle with centre O be APand BP.We need to prove
∠PAB = ∠PBAJoin OA and OB.Proof:In ∆AOB, AO = BO∴∠ABO = ∠BAO ... (i)As, the tangent is perpendicular to the radiuspassing through the point of contact,
7. We know that thetangents drawn from anexternal point to a circleare equal in length.
∴ AQ = AR ... (i)BQ = BP ... (ii)
and CP = CR ... (iii)Now,
AQ = AB + BQ= AB + BP [From (ii)]= AB + (BC – PC)= AB + BC – CR [From (iii)]= AB + BC – (AR – AC)= AB + BC + CA – AR= AB + BC + CA – AQ [From (i)]
⇒ AQ + AQ = AB + BC + CA
⇒ AQ = 12
(AB + BC + CA)
Hence proved.8. ∠BCA = 90° ... (i) (Angle in semicircle)
∠PCA = ∠PCB + ∠BCA
⇒ 110° = ∠PCB + 90°⇒ ∠PCB = 110° – 90° = 20° ...(ii)Since PC is a tangent and CB is a chord∴ ∠BAC = ∠PCB = 20° ... (iii) [Using (ii)]Now, applying angle sum property in ∆ABC,we have
1. (D) Since, the angle between two radii of a circle and the angle between corresponding twotangents are supplementary.∴ Required angle = 180° – 35° = 145°.
2. (B) Out of the numbers 8 and 5 in 85
, the greater one is 8, so the required number of
points is 8.3.
AB = 8.5 cm and ACCB
=37
AC = 2.55 cm, CB = 5.95 cm.
4. Steps of construction:
First, we draw a triangle ABC such that AB = 4 cm, AC = 5 cm and ∠BAC = 90°. Further, wewill draw a triangle A′BC′ similar to triangle ABC using the following steps.Step I: Draw a ray BX making an acute angle with BC on the side opposite to vertex A.Step II: Locate five points X1, X2, X3, X4 and X5 on BX so that BX1 = X1X2 = X2X3 = X3X4 = X4X5.Step III: Join X3C and draw X5C′ || X3C to intersect BC extended at C′.Step IV: Draw C′A′ || CA to intersect BA extended at A′.
192 AM T H E M A T C SI X–
4cm
5cm
90°
90°
B C C�
A�
A
X
X1
X2
X3
X4
X5
Then, ∆A′BC′ ~ ∆ABC.
5. Steps of construction:
Step I: First, draw a circle with radius as 5 cm and centre at O. Then take a point P so thatOP = 11 cm.
Step II: Bisect OP to find mid-point M of OP. Then take M as centre and MP = MO as radius,draw a circle to intersect the previous circle at Q and R.
Step III: Join PQ and PR which are the required tangents.
After measuring PQ and PR, we find PQ = PR = 9.8 cm (approximately).
Justification:
Join OQ and OR.
In ∆OPQ, OP = 11 cm, OQ = 5 cm and PQ = 9.8 cm
∴ OP2 – OQ2 = 112 – 52 = (11 + 5)(11 – 5) = 96
And PQ2 = (9.8)2 = 96.04
Clearly, OP2 – OQ2 ≈ PQ2
⇒ OP2 = OQ2 + PQ2
Also, OP2 = OR2 + PR2
Therefore, ∆POQ and ∆POR are right triangles with ∠PQO = ∠PRO = 90°.
193RTSNOC U C T I O N S
So, tangents are perpendicular to radii passing through their respective points of contact.i.e., PQ ⊥ OQ and PR ⊥ OR.
6. ∠C = 180° – (∠B + ∠A) = 180° – 150° = 30°.Steps of construction:In order to construct a triangle similar to ∆ABC, follow the following steps:Step I: First, construct a ∆ABC in which BC = 7 cm, ∠B = 45° and ∠C = 30°.
Step II: Make an acute angle CBX such that X is on the side opposite to vertex A.
194 AM T H E M A T C SI X–
Step III: Locate four points namely X1, X2, X3 and X4 on BX such that BX1 = X1 X2 = X2 X3 =X3 X4.
Step IV: Join X3C and draw a line X4C′ || X3C to intersect BC produced at C′.Step V: Draw a line C′A′ parallel to side CA of ∆ABC to intersect BA produced at A′. Then,∆A′BC′ is the required triangle.
WORKSHEET – 51
1. (B) Since 4 + 7 = 11, therefore, B will be joined to A11.2. (D) The required angle and the angle between the two tangents are supplementary.
∴ Required angle = 180° – 60° = 120°.
3. Here, 5 + 8 = 13AB = 7.6 cmAC : BC = 5 : 8AC = 2.92 cm and BC = 4.68 cm
195RTSNOC U C T I O N S
4. Steps of construction:
First, we draw ∆ABC with the givenmeasurements. Then we draw anothertriangle A′BC′ similar to ∆ABC and of
scalar factor 34
using the following steps:
Step I: Draw a ray BX such that ∠CBX isan acute angle.
Step II: Mark X1, X2, X3, X4 on BX suchthat BX1 = X1X2 = X2X3 = X3X4.
Step III: Join X4C and draw X3C′ || X4C.
Also, draw A′C′ || AC.
Thus, ∆A′BC′ ∼ ∆ABC.
5. Justification: In ∆PQR and ∆PQ′R′, ∠P = 45° is common and RQ || R′Q′.∴ ∆PQR ~ ∆PQ′R′We have draw PP1 = P1P2 = P2P3 = P3P4
∴ 3 4
3
P P 1=
PP 3
⇒ 3 4
3
P P 1+ 1 = 1
PP 3+
⇒ 3 4 3
3
P P PP 4=
PP 3+
⇒ 4
3
PP 4=
PP 3
And P3Q || P4Q′
∴ PQPQ 3
′ 4= .
Also, ∆PQR ~ ∆PQ′R′
Hence, PQPQ
′= PR
PR′ =
R QRQ′ ′
.
6. Steps of construction: In order to construct a pair of required tangents, follow the followingsteps:Step I: Draw a circle with radius OA = 3 cm and centre O.
196 AM T H E M A T C SI X–
Step II: Take any point P outside the circledrawn in step I and join OP.Step III: Obtain mid-point M of OP obtainedin step II and draw another circle with radiusOM = PM and centre M to intersect thecircle drawn in step I at A and B.Step IV: Join PA and PB.These PA and PB form the required pair oftangents.
WORKSHEET – 52
1. (C) The next step should be the line parallelto B5C should be passed through B4 as the
sides of required triangle are 45
of the
corresponding sides of ∆ABC.2. (C) Two distinct tangents to a circle can be
constructed from P only when P is situatedat a distance more than radius (here 2r)from the centre.
3. False. In the ratio 3 + 2 : 3 – 2 , i.e.,
11 + 6 2 : 7, 11 + 6 2 is not a positiveinteger, while 7 is.
4. Steps of construction: In order to construct a∆ABC and its similar triangle with givenmeasurements, follow the following steps:
Step I: Draw a ∆ABC inwhich BC = 7 cm, ∠B = 45° and∠C = 180° – (45° + 105°) = 30°.
Step II: Make an acute∠CBX such that X is on theopposite side of the vertexA and locate points B1, B2,B3 and B4 on BX such thatBB1 = B1B2 = B2B3 = B3B4.
Step III: Join B3C and drawB4C′ || B3C to intersect BCproduced at C′. Also drawC′A′ || CA to intersect BA produced at A′.Hence, ∆A′BC′ ~ ∆ABC.
5. Steps of construction: In order to draw a pair of tangents to the given circle, follow thefollowing steps:Step I: Draw a radius AO in the given circle with centre O and draw another radius makingan angle AOB of measure 180° – 60° = 120°.
197RTSNOC U C T I O N S
Step II: Make ∠OAP = 90° and ∠BOP = 90° to intersect each other at P.Such obtained AP and BP are the required tangents such that ∠APB = 60°.
6. We are given a ∆PQR with eachside of measure 6 cm.
Steps of construction: In order toconstruct ∆ABC follow thefollowing steps:
Step I: Make an acute angle RQXand locate seven points Q1, Q2,Q3, Q4, Q5, Q6 and Q7 on the rayOX such that QQ1 = Q1Q2 = Q2Q3= Q3Q4 = Q4Q5 = Q5Q6 = Q6Q7.
Step II: Join Q7 R and draw Q6 Cparallel to Q7 R to intersect QR atC.
Step III: Draw CA parallel to RPto intersect BP (B and Q coincide)at A.
Then, ∆ABC is the requiredtriangle.
198 AM T H E M A T C SI X–
WORKSHEET– 53
1. (A) Angle between the tangents is less than 180°.
2. (B) Line segment A5B7 divides the linesegment AB in the ratio 5 : 7.
3. False, because the point P lies inside the circle.
4.
Measuring the tangent AP, we get AP = 4.0 cm
5. We are given a circle of radius4 cm and centre O.Steps of construction: In order todraw the required pair of tangents,follow the following steps.Step I: Draw a pair of radius OAand OB inclined at an angle of180° – 120° = 60° to intersect thegiven circle at A and B respectively.Step II: Draw perpendiculars AP andBP which intersect each other at P.Then AP and BP are the requiredtangents.Justification: In quadrilateral,OBPA, applying Angle sum property,we have
Steps of construction:First, we draw ∆ABC with the givenmeasurements. Then we draw anothertriangle A′BC′ similar to ∆ABC and of
scalar factor 34
using the following steps:
Step I: Draw a ray BX such that ∠CBX isan acute angle.
Step II: Mark X1, X2, X3, X4 on BX suchthat BX1 = X1X2 = X2X3 = X3X4.
Step III: Join X4C and draw X3C′ || X4C.
Also, draw A′C′ || AC.
Thus, ∆A′BC′ ∼ ∆ABC.
1. (D) The minimum number of points shouldbe 9 as 9 > 5 out of the numerator and
denominator of 95
. The next step is to be
joined B5 to C.
2. (B) Angle of inclination, here θ, can liebetween 0° and 180°. So, the mostappropriate option is (B), i.e., 0 < θ < 180°.
3. In the adjoining figure,
∆AB′C′ ~ ∆ABC such that
ABAB
′=
B CBC′ ′
=ACAC
′ =
32
.
WORKSHEET – 54
200 AM T H E M A T C SI X–
4. Steps of construction:
First, we draw ∆ABC with the given measurements. Then we draw another triangle A′BC′
similar to ∆ABC and of scalar factor 34
using the following steps:
Step I: Draw a ray BX such that ∠CBX is an acute angle.
Step II: Mark X1, X2, X3, X4 on BX such that BX1 = X1X2 = X2X3 = X3X4.
Step III: Join X4C and draw X3C′ || X4C.
Also, draw A′C′ || AC.
Thus, ∆A′BC′ ∼ ∆ABC.
5.1 5
2 =2 2
Steps of construction: In order to construct an isosceles triangle and another triangle having52
of its corresponding sides, follow the steps given below:
Step I: Construct an isosceles triangle having any length of equal sides by drawing baseQR = 8 cm and altitude PM = 4 cm passing through the mid-point M of side QR.
Step II: Draw a ray QX such that ∠RQX is and acute angle; and divide the ray in five equalparts, namely QQ1, Q1Q2, Q2Q3, Q3Q4 and Q4Q5.
Step III: Join Q2R and draw Q5R′ || QR intersecting QR produced at R′.
201RTSNOC U C T I O N S
Step IV: Draw R′P′ || RP intersecting QP produced at P′.
Hence ∆P′QR′ is formed so that P Q QR P R 5
= = =PQ QR PR 2′ ′ ′ ′
.
202 AM T H E M A T C SI X–
6. Steps of construction: In order to draw the required pairs of tangents, follow the followingsteps:Step I: Draw a line segment AB of length 9 cm. Taking A as centre and radius 4 cm; andB as centre and radius 3 cm, draw circles.
Step II: Find the mid-point M of AB. Then, taking M as centre and radius as AM = MB,draw a circle to intersect circles drawn in step I at P, Q and R, S respectively.Step III: Join AR, AS, BP and BQ.Thus, obtained AR, AS and BP, BQ are the required pairs of tangents.
203RTSNOC U C T I O N S
ASSESSMENT SHEET – 71. (C) The required angle say θ and the angle between the tangents are supplementary.
∴ θ + 120° = 180° ⇒ θ = 60°.
2. (C) Since AB is divided in the ratio s : t.Therefore, the minimum number of points on AX would be s + t.
3. AB is the required tangent drawn frompoint A to the circle with centre Osuch that OA = 5 cm and OB = 3 cm.
4. True, because the angle between the tangents must be less than 180°.
5.
We are given a circle with centre O and radius =72
= 3.5 cm. We take points P and Q on
its extended diameter AB. Draw tangents PP1, PP2 from P and QQ1, QQ2 from Q, whichare the required tangents.
204 AM T H E M A T C SI X–
6. We are given a ∆PQR withPQ = QR = RP = 5 cm. Todraw ∆AQB ~ ∆PQR andwith given scale factor,draw ∠RQX < 90° and markQ1, Q2, Q3 and Q4 such thatQQ1 = Q1Q2 = Q2Q3 = Q3Q4.Join Q4R and draw Q3B || Q4Rand AB || PR.
Hence ∆AQB ~ ∆PQR with
AQPQ
= QBQR
=ABPR
= 34
.
7. First we draw ∆ABC using the given measurements. Further, we follow the steps givenbelow:
(a) Draw acute angle BCX and mark C1, C2 and C3 on it such that CC1 = C1C2 = C2C3. JoinBC3 and draw B′C2 || BC3 to meet BC at B′.
205RTSNOC U C T I O N S
(b) Draw B′A′ || BA to meet AC at A′.Thus ∆A′B′C ~ ∆ABC.Justification:
In ∆ABC and ∆A′B′C, AB || A′B′ and BC istransversal.∴ ∠ABC = ∠A′B′C′ = 90°Similarly, ∠BAC = ∠B′A′CTherefore, ∆A′B′C ~ ∆ABC.
⇒A B′ ′ΑΒ
=B CBC′
=A CAC
′...(i)
Let us take the ray CX.CC1 = C1C2 = C2C3
⇒ CC1+ C1C2 = 2CC1 and CC1 + C1C2
+ C2C3 = 3CC1
⇒ CC2 = 2CC1 and CC3 = 3CC1
⇒ 2
3
CCCC
= 23
⇒ CBCB
′ = 23
(Using Basic propor-
tionality theorem in ∆BCC3)
From (i) and (ii),
A B′ ′ΑΒ
=B CBC′
=A CAC
′ =
23
.
8. Let the common centreof the two circles be O.The point P is taken onthe outer circle. To drawa pair of tangents fromP to the inner circle, wefollow the instructionsgiven below:(a) Join PO and find itsmid-point M. Taking Mas centre, draw a circlepassing through P andO to intersect the innercircle at A and B.(b) Join PA and PB.PA and PB are therequired tangents.Length of PA: Onmeasuring, the length ofPA is 4.0 cm.Verification: Join AO.
⇒ PA2 = 25 – 9⇒ PA2 = 16 ⇒ PA = 4Clearly, PA is 4.0 cm.
206 AM T H E M A T C SI X–
ASSESSMENT SHEET– 8
1. (C) Q7 to R.
2. (D) 5 + 7 = 12.
3.
AP : PB = 4 : 5.
4. False, because in the ratio 3 – 1 : 3 1+ ,
i.e., 2 – 3 : 1, 2 – 3 is not a positiveinteger, while 1 is.
5. To draw a pair of tangents from P to thecircle with centre O, we follow the steps asgiven:
(a) Join OP and find its mid-point M.
(b) Taking M as centre and radius = MP =MO, draw a circle to intersect the givencircle at A and B.
(c) Join PA and PB.
PA and PB are the required tangents.
On measuring, PA = 6.35 cm and
PB = 6.35 cm. Clearly, PA and PB are ofsame length.
6. We know that the angle between a pair of tangents to a circle and the angle betweentheir corresponding radii are supplementary. Therefore, the angle between these radii= 180° – 90° = 90°.
PA and PB are the required tangents drawn from the external point P to the circle withcentre O and radius 4 cm.
In quadrilateral OAPB formed by tangents PA, PB and radii OA, OB, each internalangle is of 90° and each side is of length 4 cm. Therefore, OAPB is a square. Perimeterof square OAPB = 4 × 4 = 16 cm.
207RTSNOC U C T I O N S
7. First we construct a ∆ABC with thegiven measurements. Then weconstruct a ∆A′BC′ similar to ∆ABC
and scale factor 57
. For it, we follow
the steps given below.(a) Draw a ray BX such that O < ∠CBX< 90° and mark points B1, B2, B3, B4,B5, B6 and B7 on it such that BB1 =B1B2 = B2B3 = B3B4 = B4B5 = B5B6 =B6B7.
(b) Join B7C and draw B5C' || B7C tointersect BC at C′ and hence drawA′C′ || AC to intersect AB at A′.Thus ∆A′BC′ ~ ∆ABC.Justification: In ∆CBB7,
CB7 || C′B5 and 5
7
BBBB
=57
.
So, by Thale's theorem,BCBC
′=
57
...(i)
Similarly, in ∆ABC,A BAB
′=
57
...(ii)
208 AM T H E M A T C SI X–
Now, in ∆ABC,
BCBC
′=
A BAB
′=
57
[Using (i) and (ii); and ∠B = 90° (Given)]
∆A′BC′ ~ ∆ABC and A BAB
′=
BCBC
′ =
A CAC′ ′
=57
. Hence justified.
8. First we draw an isosceles triangle ABC with base BC = 7 cm and altitude AD = 4 cm.Altitude passes through the mid-point D of BC. Hence we construct a ∆A′BC′ similar to
∆ABC and of scalar factor 11
2, i.e.,
32
using following the steps given below:
(a) Draw an acute angle CBX opposite to the vertex A with respect to BC.
(b) Mark points X1, X2, X3 on ray BX such that BX1 = X1X2 = X2X3.
(c) Join X2C and draw X3C′ || X2C to meet BC produced at C′.
(d) Draw C′A′ || CA to meet BA produced at A′.
Thus formed ∆A′BC′ is similar to ∆ABC and of scalar factor 32
.
209RTSNOC U C T I O N S
CHAPTER TEST
1. (B) A line segment can't be divided in the ratio 6 1+ : 6 – 1 , i.e., 7 2 6+ : 5 as 7 2 6+ isnot a positive integer while 5 is.
2. (C) Line segment P3Q2 divides PQ in 3 : 2 at M. Therefore, P3M : Q2M = 3 : 2 and soQ2M : P3M = 2 : 3.
3. True, because the irrational ratio 1
3 :3
can be converted into the rational ratio that is 3 : 1.
5. Steps of construction: In order to draw pairs of tangents, follow the steps given below.Step I: Draw a line segment AB = 6 cm and then taking A and B as centres, draw the circlesof radii 3 cm and 2 cm respectively.
210 AM T H E M A T C SI X–
Step II: Find M as mid-point of AB and taking it as centre and radius as AM = MB, draw acircle intersecting the circle having the centre as A at P, Q and the circle having the centre asB at R, S.Step III: Join AR, AS, BP and BQ.Thus, AR, AS and BP, BQ are the required pairs of tangents.
6. Steps of construction: In order to construct triangles ABC and AQR, follow the steps givenbelow:Step I: Draw any line XY and take any point D on it.Step II: Draw any ray DZ such that ∠ZDY = 90°. Locate point C on DZ such that CD = 3 cm.Step III: Make an ∠DCB = 30° such that CB intersects XY at B.Step IV: Locate a point A onXB such that AB = 5 cm and byjoining AC, we find ∆ABC.Step V: Make an acute angleYAT and locate T1, T2 and T3on the ray AT such thatAT1 = T1T2 = T2T3.Step VI: Join T2B and drawT3Q || T2B to intersect line AYat Q. Also, draw QR to intersectAC extended at R.Thus, ∆AQR is obtained suchthat∆ABC ~ ∆AQR and
AQ QR AR 3= = =
AB BC AC 2.
211EMOS CILPPA A T I O N S O F NOGIRT O ...
5Chapter
SOME APPLICATIONS OF TRIGONOMETRY
WORKSHEET – 57
1. (A) cos 30° =3l
⇒ l = 2 3 m.
2. (B) sin 30° = 65l
⇒ l = 130 m.
3. Let the height of thetower be h.
tan 30° = Perpendicular
Base
⇒ 13
= 30h
⇒ h = 10 3 m.
4. Let OA be the horizontalground and K be theposition of the kite at aheight h m above theground, then AK = h m. Itis given that OK = 100 m,∠AOK = 60°.In ∆AOK, right angled at A, we have
sin 60° = 100
h ⇒ h = 100 sin 60°
⇒ h = 100 × 3
2 = 50 3 = 50 × 1.732
∴ h = 86.60 m.
5. 4.28 m, 2.14 m
Hint: sin 60° = 3.7l
tan 60° = 3.7x
.
6. Height = 94.64 m, Distance = 109.3 mHint:
tan 45° = QMYM
⇒ YM = QM
But XP = YM∴ XP = QM
tan 60° = 40 + QMQM
.
7. Let BD be the tower ofheight h m and CD bethe pole. In right-angledtriangle ABD,
tan 45° = BDAB
⇒ 1 = ABh
⇒ AB = h
In right-angled triangle ABC,
tan 60° = BCAB
⇒ 3 = BD + CD
AB
⇒ 5h +h
= 3
⇒ h = −
53 1
⇒ h =−
51.732 1
⇒ h = 6.83 m.
8. Let the chimney be AB with base B andanyone is walking from the point C to D.In ∆ABD, ∠B = 90° and ∠D = 45°∴ ∠DAB = 45° ⇒ BD = BA ...(i)In right angled ∆ABC,
tan 30° = ABBC
⇒ 13
= ABCD + BD
⇒13
= AB
50 + AB[Using equation (i)]
⇒ 3 AB = 50 + AB
212 AM T H E M A T C SI X–
⇒ AB = 3 150
×3 –1 3 1
++
= ( )+50 3 1
2
⇒ AB = 25 ( )+3 1 = 25 (1.732 + 1)
= 68.30 m.
WORKSHEET – 58
1. (A) ∠ACB = ∠XAC = 45°
sin (∠ACB) = 20x
and tan (∠ACB) = 20y
⇒ x = 20 2 m and y = 20 m.
2. (D) tan 60° =20h
.
⇒ 3 =20h
⇒ h = 20 3 m.
3. Let the length ofshadow of poleAB be BC = x, thenAB = x.Also let θ be theangle of elevation ofSun's altitude.In right-angledtriangle ABC,
tan θ = xx ⇒ θ = 45°
Hence the angle of elevation of the Sun'saltitude is 45°.
4. Let the angle ofelevation be θ. Letthe observer be ABwith his eye at Aand the tower beEC.
∴ CD = AB = 1.5 m
ED = 30 – 1.5 = 28.5 mAnd AD = BC = 28.5 mIn right-angled ∆ADE,
tan θ = = =DE 28.51
AD 28.5 ⇒ θ = 45°.
5. Let the balloon be at the point O, the threadbe OA and the required height be OB.Case I. The cable is inclined at 60°.
⇒ sin 60° = OBOA
⇒ 32
= OB215
⇒ OB = 215 3
2
= 215 × 1.732
2= 186.19 m.
Case II. The cable is inclined at 60° – 15°= 45°
sin 45° = OBOA
⇒ =1 OB2152
⇒ OB = × =215 2 215 222 2
= 215 × 1.414
2 = 152 m (approx.)
So, reduced height = 186.19 m – 152 m= 34.19 m.
6. Let AB is a hill and C and D be two citycentres subject to the angles of elevationof the top A of hill AB at C and D are 30°and 60° respectively, then ∠ACB = 30°,∠ADB = 60°, AC = 9 km.In right-angled ∆ABC,
sin 30° =AB9
⇒ AB = 9 × sin 30° = 9 ×12
= 4.5
In right-angled ∆ABD, we have
sin 60° =ABAD
⇒ AD = AB cosec 60°
⇒ AD = 4.5 ×23
= 9 × 33 × 3
= 3 3 = 3 × 1.732
= 5.196 ≈ 5.20 km.
213EMOS CILPPA A T I O N S O F NOGIRT O ...
3. True.
tan C = ABBC
= hx
If AB = 1110
h
and BC = 1110
x
Then, tan C = ABBC
= hx
.
4. Let the height of thetower CD be y metresand the horizontaldistance of point Afrom the building BCis AB = x metres.In right-angled triangleABC,
tan 45° = 20x
⇒ x = 20 m
Also, in right-angled triangle ABD,
tan 60° =+20 yx
⇒ 20 3 = 20 + y
⇒ y = ( )20 3 1− m
Thus, the height of the tower is ( )−20 3 1 m.
5. ( )7 3 1+ m
Hint: tan 45° =AEEC
⇒ EC = 7m
tan 60° = DEEC
.
6. Let the width of the river be AC such thatAC = AB + BC
Let P be the point on the bridge such thatBP = 3 m.
7. Let the ladder and thewall upto which theladder reaches be AC andBC respectively.(a) In ∆ABC,
sin 30° = BCAC
⇒ 12
= BC4
⇒ BC = 2 m.
(b) Also, cos 30° =ABAC
3
2 =
AB4
⇒ AB = 2 3 m.
8. 2 m
Hint: h = height of pedestal
tan 45° = hx
⇒ x = h
tan 60° = 1.46h +h
.
WORKSHEET– 59
1. (B) In ∆ABC,
tan A = BCAB
⇒ tan A = BC3 BC
(... AB = 3 BC)
tan A = 13
= tan 30°
∴ Angle of elevation is 30°.
2. (C) In ∆ABC,
tan 60° = ABBC
⇒ 3 = 30BC
BC = 30
3, ∴ BC = 10 3 m.
214 AM T H E M A T C SI X–
In right-angled triangle ABP,
tan 30° = 3
AB⇒ AB = 3 3 m
Also in ∆CBP, tan 45° =3
BC⇒ BC = 3 m
Now, AC = AB + BC= 3 3 + 3 = 3( 3 + 1) mHence, width of the river is 3( 3 + 1) m.
7. Let the tower be BC the flagstaffbe AB and the point on the planebe P.Let BC = hIn right-angled ∆BCP,
tan 30° = PCh
⇒ PC = h cot 30° ...(i)In right-angled ∆ACP,
tan 60° = 5+PC
h
⇒ PC = (5 + h) cot 60° ...(ii)Comparing equations (i) and (ii), we have
h cot 30° = (5 + h) cot 60°
⇒ 3h = (5 + h) 13
⇒ 3h = 5 + h⇒ h = 2.5Hence, the height of the tower is 2.5 m.
ORLet the two planes be at A and B respectively.Also P be the point on the groundIn right-angled triangle APC,
tan 30° = 3125PC
⇒ PC = 3125 3 m
Also in right-angled triangle BPC,
tan 60° = BCPC
⇒ BC = 3125 3 × 3 = 3 × 3125∴ AB = BC − AC = 3 × 3125 – 3125
= 2 × 3125 = 6250 m.Hence, distance between the two planes is6250 m.
8. Let the tower be PQ and the objects be Aand B.∵ ∠XQA = 45°and ∠XQB = 60°∴ ∠QAP = 45°and ∠QBP = 60° (Alternate angles)In right ∆APQ,
∠PAQ + ∠PQA = 90°⇒∠PQA = 90° – 45° = 45° (∵∠PAQ = 45°)∴ AP = PQ = 150 (∵ PQ = 150 m)⇒ AB + BP = 150 ...(i)In right ∆BPQ,
tan 60° = PQBP
⇒ 3 = 150BP
⇒ BP = 150
3...(ii)
Putting BP = 150
3 in equation (i), we get
AB + 150
3= 150 ⇒ AB = 150
11 –
3
⇒ AB = 150 × 3 – 1
3 ×
33
= 50 × ( )3 – 3
= 50 (3 – 1.73) = 50 × 1.27⇒ AB = 63.50 mThus, distance between the two objects is63.50 m.
WORKSHEET – 60
1. (C) tan 45° = OPPQ
= QPa
⇒ 1 = QPa
⇒ QP = a m
∴ ar(∆OPQ) = 12
× QP × OP
= 12
× a × a = 12
a2.
215EMOS CILPPA A T I O N S O F NOGIRT O ...
2. (B) tan 60° = TPPO
⇒ 3 = TP40
⇒ TP = 40 3 m.
3. True
tan 30° =ABBC
⇒ 13
= AB81
⇒ AB =81
3 = 81 3
3
⇒ AB =81×1.732
3 = 46.76 m.
4. Let the height of the pole AB = x mLength of the rope AC = 20 m
In ∆ABC,∠ACB = 30º
∴ sin 30° = ABAC
⇒12
= 20x
⇒ x = 10 m∴ Height of the pole = 10 m.
5. Let the tree be PQ, the width of the river beMP and the person moves from M to S.In right ∆PQS,
tan 45° = PQPS
⇒ 1 = PQPS
⇒ PS = PQ⇒ PM + MS = PQ⇒ PM + 40 = PQ ...(i)In right ∆PQM,
tan 60° = PQPM
⇒ = PQ3
PM⇒ PQ = 3 PM ...(ii)From equations (i) and (ii),
PM + 40 = 3 PM
⇒ PM = 3 + 140
×3 – 1 3 + 1
⇒ PM = ( )20 3 + 1 ... (iii)
From equations (ii) and (iii),
PQ = ( ) ( )3 × 20 3 + 1 20 3 + 3=
Hence height of the tree and the width of the
river are ( )20 3 + 3 m and ( )20 3 + 1 m
respectively.
6. 6.34 m
Hint: tan 45° =1 2+
15
y y
y1 + y2 = 15 …(i)
tan 30° =1
15
y
⇒ y1 =15
3…(ii)
∴ y2 = 15 − y1 = 6.34 m.
7. Distance = 17.32 m, Height = 40 m
Hint: tan 30° = 10x
⇒ x = 10 3 m
tan 60° =10 3
y
⇒ y = 30 m.
8. Let the plane moves from P to Q in 30seconds and the points H and K on theground be just below the points P and Qrespectively.
In right ∆AQK,
tan 30° = QKAK
⇒ 13
= 3600 3AK
216 AM T H E M A T C SI X–
⇒ AK = 10800 ...(i)In right ∆APH,
tan 60° = PHAH
⇒ 3 = 3600 3
AH
[∵ PH = QK = 3600 3 ]
⇒ AH = 3600 ...(ii)
Now, PQ = HK = AK – AH
⇒ PQ = 10800 – 3600
[Using (i) and (ii)]
⇒ PQ = 7200
∵ Speed = Distance travelled
Time
∴ Speed = PQ 7200=
30 30
= 240 m/s = 3600
240 ×1000
km/h
= 864 km/h.
WORKSHEET – 61
1. (A) tan C = ABBC
⇒ tan C = ABAB
(∵ BC = AB)
⇒ tan C = 1 = tan 45°
⇒ C = 45°.
2. (D) tan C = ABBC
⇒ tan 60° = 15BC
⇒ 3 = 15BC
⇒ BC = 15
3⇒ BC = 5 3 m.
3. False.Let height of the tower is h metres so theangle of elevation is 30°.
tan 30° = BCh
⇒13
= BCh
⇒ BC = h 3 ... (i)
When height = 2h,
tan θ = ABBC
⇒ tan θ = 2
3h
h[From (i)]
⇒ tan θ = 23
≠ tan 60°.
4. Let AB be the ladder leaning against a wallOB such that ∠OAB = 60° and OA = 9.6 m.In ∆OAB right angled at O, we have
cos 60° = OAAB
⇒ AB = OAcos 60°
⇒ AB = 9.60.5
= 19.2 m.
5. Let the point, cloud and reflection of thecloud be at P, Q and Q′ respectively.
Let PM = x, QM = y
We have to find QB, i.e., y + h
In right-angled triangle QPM,
tan α = yx
⇒ x = αtan
y… (i)
217EMOS CILPPA A T I O N S O F NOGIRT O ...
Also in right-angled triangle Q′PM,
tan β = + 2y hx
⇒ tantan
y βα
= y + 2h
[From equation (i)]
⇒ y β
− α
tan1
tan= 2h
⇒ y + h = h 2 tan
1tan tan
α+ β − α
=( )tan tan
tan tan
h β + αβ − α
.
Hence proved.
6. Hint:
ED = AC = length of ladder = l (say)
Now, cos α =BC y
l=
, sin α =+ b xl
,
cos β = + y al
, sin β = xl
,
Consider,cos cossin sin
α − ββ − α
=
y y al lx b xl l
+−
+−
=ab
−− =
ab
.
7. 250 m
Hint: tan α = hx
= 57
tan β = +150
hx
= 12
.
8. Let the tower, the flagstaffand the point on the planebe AB, BC and P respec-tively.
Let AB = y and AP = xIn ∆ABP,
tan α = yx
⇒ 1x
= αtan
y...(i)
In ∆ACP,
tan β = +h yx
⇒ 1x
= β
+tanh y
...(ii)
From equations (i) and (ii), we have
αtany
= β
+tanh y
⇒ y tan β = h tan α + y tan α⇒ y (tan β – tan α) = h tan α
⇒ y = α
β αtan
tan – tanh
. Hence proved.
WORKSHEET– 62
1. (C) sin 30° = ABAC
⇒12
= AB15
⇒ AB = 152
m.
2. (B) tan C = ABBC
= 3h
h
⇒ tan θ = 13
= tan 30°
⇒ θ = 30°.
218 AM T H E M A T C SI X–
3. True.
As tan θ = ABBC
⇒ tan θ = ABAB
(∵ AB = BC)
⇒ tan θ = 1 = tan 45°∴ θ = 45°.
4. Let AB be the tower and C be the point onthe ground.
In ∆ABC,
tan 30° = ABBC
⇒13
= AB30
⇒ AB = 30
3 = 10 3 m.
Hence, the height of the tower is 10 3 m.
5. Let the height (PQ) of mountain be h km.In right-angled ∆ABC,
sin 30° =BC1
⇒ BC =12
km
⇒ DP = 12
km …(i)
Also, cos 30° = AB1
⇒ AB =3
2km …(ii)
Similarly, from right-angled ∆APQ,AP = PQ = h …(iii)
And from right-angled ∆CDQ ,
3 = DQCD
⇒ 3 = −−
DPAB
hh
[From (iii)]
⇒ 3 =
123
2
h
h
−
−[From (i) and (ii)]
⇒ h =1
3 1−=
3 12+
= 1.366 km.
6. Let the aero-plane's firstsituation be atA and secondat B. Let thepoint of obser-vation be at O.From right-angled ∆AOD,
tan 45° = ADOD
⇒ OD = 3000 m
Again from right-angled ∆BOC,
tan 30° = 3000
3000 + DC
⇒ DC = 3000 ( )−3 1 m
Now speed of the plane
= Distance
Time=
( )3000 3 1
15
−
[ ]D C AB=∵= 146.42 m/sec
7. Let the height ofthe tower PQ andthe width of thecanal AP be h andx respectively.Let the anotherpoint be B such that
AB = 20 mIn right ∆APQ,
tan 60° = hx
h = 3 x ...(i)In right ∆BPQ,
tan 30° = +20h
x
219EMOS CILPPA A T I O N S O F NOGIRT O ...
⇒ h 3 = 20 + x ... (ii)
From equations (i) and (ii), we have
3 x × 3 = 20 + x ⇒ 2x = 20
⇒ x = 10
Substitute x = 10 in equation (i) to get
h = 10 3
Hence, the height of the tower is 10 3 mand the width of the canal is 10 m.
8. Let O be centre ofthe balloon of radiusr and P the eye ofthe observer. Let PA,PB be tangents fromP to the balloon.Then,
∠APB = α
... ∠APO = ∠BPO =2α
... OL ⊥ PX, ∠OPL = β
... In ∆OAP, sin2α
=OAOP
⇒ OP = r cosec2α
In ∆OPL, sin β = OLOP
⇒ OL = r cosec2α
sin β.
ASSESSMENT SHEET – 9
1. (C) Given:
AB : BC = 1 : 13
i.e., AB : BC = 3 : 1
i.e.,ABBC
= 3
1
∴ tan θ = ABBC = 3 = tan 60°
⇒ θ = 60°.
2. (A) sin 30° = BCAB
⇒12
=BC5
⇒ BC =52
m.
3. Wire is AB.CE = BD = 14 m.AE = AC + CE
⇒ 20 = AC + 14⇒ AC = 6 mIn ∆ABC, ∠C = 90°,
sin 30° = 6
AB
⇒12
= 6
AB ⇒ AB = 12 m.
4. False, because the tangent of the angle ofelevation doubles notthe angle of elevation.
5. BC is the multi-storeyedbuilding with the footB and the top C as thepoint of observation.AD is the building withbottom A and the top D. Draw DE || AB(see figure).Given angles are ∠XCD = 30° and∠XCA = 45°. ∠CDE and ∠XCD are alternateinterior angles.∴ ∠CDE = ∠XCD = 30°.Similarly, ∠CAB = ∠XCA = 45°
BE = AD = 8 m
In right triangle ABC,
tan 45° = CE+BE
AB⇒ 1 =
CE+8AB
⇒ AB = CE + 8 ...(i)
Also, in right triangle DCE,
tan 30° = CEDE
⇒ 13
= CEAB
(∵ DE = AB)
⇒ AB = 3 CE ...(ii)
220 AM T H E M A T C SI X–
From equations (i) and (ii), we get
( )3 – 1 CE = 8
⇒ CE = 8 3 + 1
×3 – 1 3 + 1
= 8 × (1.73 + 1)
3 – 1 = 10.92 m.
Substituting CE = 10.92 in (i), we getAB = 10.92 + 8 = 18.92 m
Further, BC = BE + CE = 8 + 10.92= 18.92 m
Hence, the height of the multi-storyedbuilding and the distance between the twobuildings is 18.92 metres each.
6. Let AB be the first towerwith bottom A and CDbe the second towerwith bottom C.
BE = 80 m
CD = 160 m
AB = CE
∵ XD || BE and BD is the transversal
∴ ∠DBE = ∠XDB = 30°
In right triangle BDE,
tan 30° = DEBE
= CD – CE
BE
⇒ 13
=160 – AB
80
⇒ 160 – AB = 80
3
⇒ AB = 160 – 80 80 3
160 –33
=
= 480 – 80 3
3
= 480 – 80 × 1.7323
= 113.81
Hence, the height of the first tower is 113.81metres.
7. Let the lower window, upper window andthe balloon be at A, B and C respectively.AB = AD = EF = FG = 2 m ...(i)BG = AF ...(ii)In right triangle BCG,
tan 30° = CGBG
⇒13 =
CGBG
⇒ CG = 13 BG
...(iii)
In right triangle ACF,
⇒ tan 60° = CFAF
⇒ 3 = CG+2
BG[Using (i) and (ii)]
⇒ BG = CG+2
3...(iv)
From equations (iii) and (iv),
CG = 13
× CG+2
3
⇒ 3CG = CG + 2⇒ CG = 1 ...(v)Now, CE = CG + FG + EF
= 1 + 2 + 2 [Using (i) and (v)]= 5
Hence, the height of the balloon is 5 metres.
8. Let the tower be AB, the flagstaff BC andthe point on the plane P.Let AB = y and AP = xIn ∆PAB,
tan α = ABAP
= yx
...(i)
In ∆PAC,
tan β = ACAP
⇒ tan β = +h yx
...(ii)
221EMOS CILPPA A T I O N S O F NOGIRT O ...
From equations (i) and (ii),
x = αtan
y =
+βtan
h y
⇒αtan
y = βtan
h +
βtany
⇒ α β
1 1–
tan tany =
βtanh
⇒β αα β
tan – tantan tan
y = βtan
h
⇒ y = α
β αtan
tan – tanh
Hence, the height of the tower AB isα
β αtan
tan – tanh
.
ASSESSMENT SHEET – 10
1. (A) cos 45° = 10 2AC
⇒ 12
= 10 2AC
⇒ AC = 20 m.
2. (B) ∠YAM = ∠XYA = 45°; tan 45° = 25
AM⇒ AM = 25 m
Time =255
= 5 seconds.
3. Let the kite be at A and the thread AB.AC = 75 m;
∠ABC = 60°
sin 60° = ACAB
⇒ 32
= 75AB
⇒ AB = ×150 33 3
= 50 3 = 50 × 1.732
⇒ AB = 86.6
⇒ The length of the string to the nearestmetre is 87 metres.
4. False, the angle of elevation will remainunchanged if the height of the tower isincreased by 10% too.
5. Let the trucks be at Aand B; the balloon at Pand the vertical line PQ.∠XPA = 45°∠XPB = 60°∠PAQ and∠PBQ are alternateinterior angles with∠XPA and ∠XPB respectively.∴ ∠PAQ = 45° and ∠PBQ = 60°In right ∆PAQ,
tan 45° = PQAQ
⇒ 1 = PQ
100 + BQ
⇒ BQ = PQ – 100 ...(i)In right ∆PBQ,
tan 60° = PQBQ
⇒ 3 = PQBQ
⇒ PQ = 3 BQ = 3 (PQ – 100)[Using (i)]
⇒ PQ ( )3 – 1 = 100 3
⇒ PQ = 100 3
3 – 1
⇒ PQ = 100 3
3 – 1 ×
++
3 1
3 1
= 50 ( )3 3+
Therefore, height of the balloon is
50 ( )+3 3 metres.
6. Let P be the point on the bridge; A and Bare the two points on the opposite banks ofthe river such that AB is the width of theriver. ∠X1PA = 45°, ∠X2PB = 30°.Draw PC ⊥ AB to meet AB at C.
∠PAB = ∠X1PA
= 45°
∠PBA = ∠X2PB
= 30°.
222 AM T H E M A T C SI X–
In right ∆ PAC,
tan 45° = PCAC
⇒ 1 = 30
AC ⇒ AC = 30
In right ∆PBC,
tan 30° = PCCB
⇒ 13
=30CB
⇒ CB = 30 3
Now, AB = AC + CB = 30 + 30 3
= 30 + 30 × 1.732
= 30 + 51.96 = 81.96Hence the width of the river is 81.96 metres.
7. Let the wall be AM.Let the ladder changes its position fromAB to A′B′. Let the length of the ladder bel such that
AB = A′B′ = l
Let BM = y and
A′M = z
In right triangleABM,
sin α = AMAB
⇒ sin α = +
3x
z
l
⇒ 3 sin α = + 3x zl
...(i)
And cos α = BMAB
⇒ cos α = yl
...(ii)
Add equations (i) and (ii) to get
3sin α + cos α = + + 3x y z
l...(iii)
In right triangle A′B′M,
sin β = ′′ ′
A MA B
= zl
⇒ 3 sin β = 3zl
...(iv)
And cos β = ′′ ′
B MA B
=+x yl
...(v)
Add equations (iv) and (v) to get
3sin β + cos β = + + 3x y z
l...(vi)
Divide equation (iii) by equation (vi) to get
α + αβ+ β
3sin cos3sin cos
= 1. Hence proved.
8. Let the observer be at P on the ground.
When the plane is at A vertically above C,∠APC = 60°.When the plane is at B vertically above D,∠BPD = 30°.
AC = BD = 1.2 km = 1200 metres.AB = CD
In right triangle APC,
⇒ tan 60° = ACPC
⇒ PC = AC
tan 60° =
12003
In right triangle BPD,
tan 30° = BDPD
= BD
PC+CD
⇒ 13
= +
12001200
AB3
⇒ 1200
3 + AB = 1200 3
⇒ AB = 1200 3 –1200
3
⇒ AB = 2
1200 ×3
AB = 800 × 3 metres
223EMOS CILPPA A T I O N S O F NOGIRT O ...
Speed = =Distance travelled AB
Time 15sec
= 800 3 m
15sec
=
1800 3 km
100015
h60 60
×
×
= 800 3 60 60
1000 15× × ×
× km/h
= 192 3 km/h = 192 × 1.73 km/h
= 332.16 km/hHence, the aeroplane is flying at a speed of332.16 km/hr.
CHAPTER TEST
1. (A) From the adjoin-ing figure, angle ofdepression of P is∠XOP = α and angleof depression of Q is
∠XOQ = 90° – β.
2. (B) tan 30° = 20 3h
⇒ h = 20 m.
3. Let the tower be BC and the length ofshadow be AB.
tan 60° = BCAB
⇒ 3 = 20AB
⇒ AB = 20
3m
⇒ AB = 20 33
m.
4. True, because the vertical tower, length ofthe shadow and the ray of the sun make aright angled isosceles triangle.
5. Let the ships be at A and B; and the towerbe PQ.
∠PAQ = ∠XPA = 30°∠PBQ = ∠XPB = 45°
In right ∆ BPQ,∵ ∠PBQ = 45°, ∴ ∠BPQ = 45°⇒ BQ = PQ = 75 ...(i)In right ∆ PAQ,
tan 30° = PQ
AB + BQ
⇒ AB + 75 = 75 3 [Using (i)]
⇒ AB = ( )75 3 – 1 m.
6. 8 3 m
Hint: A′C = AC
cos 30° = 8
AC
tan 30° = BC8
.
7. Let the window beat P and height ofthe opposite housebe h.In right ∆ APQ,
tan 45° = 60
AQ⇒ AQ = 60 ⇒ BP = 60
In right ∆ BCP,
tan 60° = – 6060
h ⇒ 60 3 = h – 60
⇒ h = 60 + 60 3 = 60 ( )1 3+
Thus, the required height is 60 ( )1 3+ m.
224 AM T H E M A T C SI X–
8. Let the cloud be at C, the point ofobservation be at P and the reflection ofthe cloud in the lake be at D. Let Q be anypoint just below the cloud, 60 m above thewater level.
In right ∆CPQ,
tan 30° = CQPQ
⇒13 =
– 60PQ
h
⇒ PQ = ( )3 – 60h ...(i)
In right ∆DPQ,
tan 60° = + 60PQ
h
⇒ PQ = + 60
3h
...(ii)
Using equations (i) and (ii), we have
⇒ ( )3 – 60h = + 60
3h
⇒ 3h – 180 = h + 60⇒ h = 120Hence, height of the cloud is 120 metres.
2. (D) Probability of non-happening of anevent = 1 – Probability of happening of
that event
=3
17
− = 47
.
3. (C) Sample space is {1, 2, 3, 4, 5, 6} andfavourable events are {2, 3, 5}
∴ Required probability = 36
= 12
.
4. No, because the number of favourable out-comes of getting ‘6’ and ‘not 6’ are respecti-vely 1 and 5; and so their probabilities are
16
and 56
.
5. Sample space: {1, 2, 3, ......... ,99}∴ n(S) = 99.The numbers divisible by 3 and 5 both arenumbers divisible by 15.So, favourable outcomes are: {15, 30, 45, 60,75, 90}Let E be the event getting a numberdivisible by 3 and 5.∴ n(E) = 6
∴ P(E) = (E) 6 2(S) 99 33
nn
= = .
6. (i) 123
(ii) 546
Hints:(i) Prime numbers are 5 and 7.
(ii) Perfect square numbers are
9, 16, 25, 36, 49.
7. Let A = The event that 5 will not come upeither time.Now sample space is given by
(i) Let E1 be the event that the card drawnis neither a hearts nor a king.Number of hearts = 13Number of kings = 4But one king is of hearts.∴ n(E1) = 52 – 13 – 4 + 1 = 36
Now, P(E1) = 1(E )(S)
nn
=3652
=9
13.
(ii) Let E2 be the event that the card drawnis an ace of spades.Since number of ace of spades is 1∴ n(E2) = 1
∴P(E2) = 152
(iii) Let E3 be the event that the card drawn is either a black card or a king.Number of black cards
= Sum of numbersof cards of clubsand spades
= 13 + 13 = 26Number of kings = 4
But 2 kings are black∴ n(E3) = 26 + 4 – 2 = 28
∴ P(E3) =28 752 13
= .
9. Number of outcomes in sample spacen(S) = 62 = 36
(iii) 1Hint: Favourable outcomes are thesame as the outcomes in sample space.
7. Let the number of black balls be x.So, total number of balls = x + 5Probability of drawing a black ball is
P1 = + 5x
x
Also, probability of drawing a red ball is
P2 = 5 + 5x
According to the question, we haveP1 = 2.P2
⇒ 2 × 5
= + 5 + 5x
x x ⇒ x = 10.
Hence, the number of black balls is 10.
8. Number of all cards = 50 – 5 + 1 = 46i.e., n(S) = 46(i) Let E1 be the event that the number
on the card taken out is a prime lessthan 10.Prime numbers from 5 to 9 are 5and 7.∴ n(E1) = 2
∴ P(E1) = 1(E ) 2 1(S) 46 23
nn
= = .
(ii) Let E2 be the event that the number onthe card taken out is a perfect square.The perfect square numbers from5 to 50 are 9, 16, 25, 36 and 49.
∴ n(E2) = 5
∴ P(E2) = 2(E )(S)
nn
=546
.
9. Number of all possible outcomes,n(S) = 52
(i) Number of spades = 13Number of aces = 4
But 1 ace is of spades.∴ Number of favourable outcomes
= 13 + 4 – 1 = 16∴ P(card drawn is a spades or an ace)
= 1652
= 4
13.
(ii) Number of red kings = 2∴ P(card drawn is a red king)
=2 152 26
= .
(iii) Number of kings = 4Number of queens = 4∴ Number of favourable outcomes
= 52 – 4 – 4 = 44∴ P (card drawn is neither a king nor aqueen)
=44 11
=52 13
.
(iv) Number of kings = 4Number of queens = 4
Number of favourable outcomes= 4 + 4 = 8
∴ P(card drawn is either a king or a
queen) =8 2
=52 13
.
ASSESSMENT SHEET – 11
1. (D) We know that probability (P) of any
event can be 0 ≤ P ≤ 1. Therefore, 1514
> 1
can’t be probability of an event.
2. (B) Let the number of bolts be x. Then
400x
= 0.035 ⇒ x = 400 × 0.035 ⇒ x = 14.
229BORP TILIBA Y
3. Number of cards = 50Prime numbers from 51 to 100 are: 53, 59,61, 67, 71, 73, 79, 83, 89, 97Therefore, number of all possible outcomes
= 50.And number of favourable outcomes = 10.
∴ Required probability = 10 1
=50 5
.
4. True,Ratio of probabilities
= Ratio of areas of regions a, b and c= Ratio of areas of corresponding sectors= Ratio of corresponding angles= 60° : 120° : 180° = 1 : 2 : 3.
5. A leap year contains 366 days, wherever52 weeks and 2 days. There are 52Thursdays in 52 weeks. Therefore, a leapyear consists 52 Thursdays and 2 days.These two days may be one choice out ofthe seven given below:
(i) Thursday and Friday(ii) Friday and Saturday
(iii) Saturday and Sunday(iv) Sunday and Monday(v) Monday and Tuesday
(vi) Tuesday and Wednesday(vii) Wednesday and ThursdaySo, two days are either Thursday andFriday or Wednesday and Thursday.Clearly, the number of all possibleoutcomes is 7 and the number of favourableoutcomes is 2.
Hence, required probability = 27
.
6. Let number of blue marbles= xNumber of green marbles = yNumber of white marbles = zTherefore, x + y + z = 54
P(selecting a blue marble) =x
x y z+ +
⇒ 13
=54x
⇒ x =54
= 183
P(selecting a green marble) =y
x y z+ +
⇒ 49
=54y
⇒ y =4 × 54
9 = 24
Now, substituting x = 18, y = 24 inx + y + z = 54, we get
18 + 24 + z = 54⇒ z = 54 – 42⇒ z = 12Hence, the jar contains 12 white marbles.
7. Number of blue triangles = 3Number of red triangles = 8 – 3 = 5Number of blue squares = 6Number of red squares = 10 – 6 = 4
Number of all possible outcomes= 8 + 10 = 18
(i) P(a lost piece is a triangle)
=8 4
18 9= .
(ii) P(a lost piece is a square)
=10 518 9
= .
(iii) P(a lost piece is a blue square)
=6 1
18 3= .
(iv) P(a lost piece is a red triangle)
= 518
.
8. Let E be the event of placing at least oneletter in the wrong envelope. Anyoneenvelope can be filled with a letter in 4ways. Second one can be filled by 3 ways.Third one can be filled by 2 ways. And thefourth one can be filled with the remainingone letter in 1 way. Therefore, the fourletters are placed in the four envelopes in4 × 3 × 2 × 1 = 24 ways. Out of these 24
230 AM T H E M A T C SI X–
ways, only 1 way is such that all the lettersare placed in the right envelopes and 23ways are such that at least one letter isplaced in the wrong envelope.
∴ P(E) =2324
.
ASSESSMENT SHEET – 12
1. (B) P(getting X or Y)= P(getting X) + P(getting Y)
3. The number of outcomes when a pair ofdice is rolled = 62 = 36.The outcomes such that the sum is divisibleby 3 are:(1, 2), (1, 5), (2, 1), (2, 4), (3, 3), (3, 6), (4, 2),(4, 5), (5, 1), (5, 4), (6, 3), (6, 6).These are 12 outcomes.The outcomes such that the sum is divisibleby 2 are:(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1),(3, 3), (3, 5), (4, 2), (4, 4), (4, 6), (5, 1), (5, 3),(5, 5), (6, 2), (6, 4), (6, 6).These are 18 outcomes.The outcomes such that the sum is divisibleby 6 are:(1, 5), (2, 4), (3, 3), (4, 2), (5, 1), (6, 6).These are 6 outcomes.Now, the number of outcomes which aredivisible by 3 or 2 is 12 + 18 – 6 = 24.
Hence, the required probability = 24 2
=36 3
.
4. No.An unbiased die has six equally likelyoutcomes. These are: 1, 2, 3, 4, 5 and 6.Each of them has equal probability.
Therefore, the probability of getting 6 is 16
and that of not 6 is 56
.
5. All possible outcomes are given byS = {1, 2, 3, ....., 1000}∴ n(S) = 1000(i) Let E1 be the event that the first player
wins a prize. Then,E1 = Perfect square numbers
greater than 500 and lessthan 1001.
= 529, 576, 625, 676, 729,784, 841, 900, 961.
∴ n(E1) = 9
Now, P(E1) =( )1E(S)
nn
= 9
1000
(ii) Let E2 be the event that the secondplayer wins a prize, if the first has won.
∴ n(E2) = n(E1) – 1 = 9 – 1 = 8 And number of all possible outcomes
= n(S) – 1 = 1000 – 1 = 999
Now, P(E2) = 2(E )999
n =
8999
.
6. Let E1 be the event ‘the mobile phone isacceptable to Varnika’ and E2 be the event‘the mobile phone is acceptable to thetrader’.∴ n(E1) = Number of good mobile
phones= 42
And n(E2) = Number of good mobilephones + Number of mobilephones having only minordefects
= 42 + 3 = 45Number of all mobile phones is given by
n(S) = 48
(i) P(E1) = 1(E )(S)
nn
= 4248
= 78
(ii) P(E2) = 2(E )(S)
nn
= 4548
= 1516
.
231BORP TILIBA Y
7. When two dice are thrown, the samplespace is given by
6. There are 52 cards in the pack. Therefore,the number of outcomes in sample space isgiven by
n(S) = 52Number of hearts cards = 13
Number of queens = 4Number of queens of hearts = 1
So, number of favourable outcomes is givenby
n(E) = 52 – (13 + 4 – 1) = 36
Now, the required probability will be givenby
P(E) =(E) 36 9
= =(S) 52 13
nn
.
7. Area of the rectangular region= Length × Breadth= 3 × 2 = 6 m2
Area of circular region= π × Radius2
=22 1
×7 4
=1114
m2
Now, required probability
=Area of circular region
Area of rectangular region
=
11146
= 11
14 6× =
1184
.
8. (i) 12
(ii) 14
Hint: A, B, C and D can be arranged in 24ways. A before B can be arranged in thefollowing ways:CADB, DACB, CABD, DABC, ABCD,ABDC, CDAB, DCAB, ACDB, ADCB,ACBD, ADBC.Out of these A just before B occurs in6 ways.
9. Let the number of white balls be x and thenumber of red balls be y.
Therefore,2x
=3y
⇒ 3x – 2y = 0 ... (i)and 2(x + y) = 3y + 8⇒ 2x – y = 8 ... (ii)Solving equations (i) and (ii ), we get
x = 16, y = 24Number of non-red balls = number of whiteballs = 16P(choosing ball is not red)
= 1616 24+
= 1640
= 25
.
❑❑
233OOC MOEGET EDR I AN T R Y
7Chapter
COORDINATE GEOMETRY
= { }1– 5(5 – 5) – 4(5 – 7) 4(7 – 5)
2+
= 1
(0 8 8)2
+ + = 8 sq. units.
7. Let A(1, 7), B(4, 2), C(−1, −1), D(− 4, 4)be the vertices.
∴ AB = ( ) ( )2 24 1 2 7− + − = 34
BC = ( ) ( )2 21 4 1 2− − + − − = 34
CD = ( ) ( )− + + +2 24 1 4 1 = 34
AD = ( ) ( )2 24 1 4 7− − + − = 34
Also,
AC = ( ) ( )2 21 1 1 7− − + − − = 68
BD = ( ) ( )2 24 4 2 4+ + − = 68
So, AB = BC = CD = AD and diagonals,i.e., AC = BD.⇒ ABCD is a square.
8. Draw DE ⊥ AB and join BD.Since, diagonals of a parallelogram bisecteach other.
...(iv)Solving results (i), (ii), (iii) and (iv), we get x1 = 3, y1 = 2, x2 = – 1, y2 = 2, x3 = 1, y3 = – 4.
Hence the required vertices are A(3, 2),B(– 1, 2) and C(1, – 4).
6. (7, 2) or (1, 0)
Hint: ar(∆ PAB) = 10
⇒ ar(∆ PAB) = + 10 or ar(∆ PAB) = −10
Let P(x, y) ∴ Use area of triangle
= ( ) ( ) ( )1 2 3 2 3 1 3 1 212
x y y x y y x y y − + − + − .
7. The given points would be collinear, if thearea of triangle formed by them as verticesis zero.
i.e., ( ){}
1– – ( – – )
2( – – ) 0
a c a a b b a b b c
c b c c a
+ + +
+ + =
Here, LHS
= ( ){ }1– ( – ) ( – )
2a c b b a c c b a+ +
= ( )1– – –
2ac ab ab bc bc ac+ + = 0
= 0 = RHS. Hence proved.
8. 24 sq. units
Hint: Area of rhombus = 21
d1 × d2
where d1, d2 = length of diagonals.
9. ar(∆ABC) = 5
⇒ ( ) ( ) ( ){ }16 – 1 – 2 1 – 2 3 2 – 6
2k k k+ = 5
⇒ ( )15 – 2 4 6 –18
2k k k+ + = 5
⇒ ( )115 – 20
2k = 5 ⇒ 15
– 102
k = 5
⇒ 15– 10
2k = ± 5 ⇒
152
k = 10 ± 5
⇒152
k = 15 or 5 ⇒ k = 2 or 23
.
WORKSHEET – 79
1. (B) A(0, 0), B(3, 3 ), C(3, λ)As AB = BC = AC
∴ AB = AC ⇒ AB2 = AC2
⇒ 12 = 9 + λ2
⇒ λ2 = 12 − 9
⇒ λ2 = 3 ∴ λ = ± 3
∴ Required value of λ = − 3 .
2. (D) Mid-point of (6, 8) and (2, 4) is P(4, 6).∴ If A(1, 2), then
AP = ( ) ( )2 24 1 6 2− + −
= 9 16+
= 5 units.
241OOC MOEGET EDR I AN T R Y
3. 2 or – 4
Hint: Use Pythagoras Theorem.
4.
∴PAPQ =
23
⇒ PA : AQ = 2 : 1
∴ Using section formula.
x = 23
; y = 13
.
∴ Coordinates of A are 2 1,
3 3
.
5. 11 52 ,
2 2
−
; 11 52 ,
2 2
+
Hint:Let AB = AC = 3.Use distance formula.
6. Let the points of trisection be P(x1, y1) andQ(x2, y2) such that P is the mid-point ofA(3, – 2), Q(x2, y2) and Q is the mid-pointof P(x1, y1), B(– 3, – 4).
⇒ 3x + y – 3 = 0 ...(ii)or 3x + y – 23 = 0 ...(iii)Now, we have to solve equations (i) and(ii) as well as equations (i) and (iii).Solving equations (i) and (ii), we get
x = 1, y = 0Solving equations (i) and (iii), we get
x = 7, y = 2Hence, the coordinates of P are (1, 0) or(7, 2).
OR
∵ P is mid-point of AB
∴ P ≡ +
1 3 5 – 7
,2 2
,
i.e., P ≡ (2, – 1)
∵ Q is mid-point of BC
∴ Q ≡ 3 0 – 7 4
,2 2+ +
, i.e., Q ≡
3 3
, –2 2
∵ R is mid-point of CA
∴ R ≡ + +
0 1 4 5
,2 2
, i.e., R ≡
1 9,
2 2
Now, ar(∆PQR)
= 1 3 9 3 9
2 – – 12 2 2 2 2
+ + + +
1 3–1
2 2
= + + 1 33 1
–122 4 4
= –7 74 4
= ...(i)
ar(∆ABC)
= { }11(– 7 – 4) 3(4 – 5) 0(5 7)
2+ + +
= 1
(–11– 3 0) – 7 72
+ = = ...(ii)
Dividing equation (i) by equation (ii), we
have ∆∆
( PQR)( ABC)
arar
=
747
⇒ ar(∆PQR) = 14
ar(∆ABC).
9. Since, A is onthe x-axis, so itscoordinates willbe of the form(x, 0). Similarly,the coordinatesof B will be ofthe form (0, y).Since P is the mid-point of AB.
∴ – 2 =0
2x +
and 3 = +02
y
∴ x = – 4 and y = 6∴ Coordinates of A are (– 4, 0) and coordi-nates of B are (0, 6).Now,
PO = 2 2(– 2) (3) 4 9 13+ = + =
PA = 2 2(– 4 2) (0 – 3) 4 9+ + = + = 13
Clearly, PA = PB = PO⇒ P is equidistant from A, B and theorigin O.
6. Let the required ratio be λ : 1.Here, we will use section formula as givenbelow.
x = 2 1mx nxm n
++
and y = 2 1my nym n
++
In this question,
– 3 = – 2 – 5
1λ
λ + and k =
3 – 41
λλ +
⇒ – 3λ – 3 = – 2λ – 5 and k =3 – 4
1λλ +
⇒ λ = 2 and i.e., k =3 2 – 4
2 1×
+
⇒ λ : 1 = 2 : 1 and k =23
Hence, the ratio is 2 : 1 and k =23
.
7. First, we find the length of each side ofquadrilateral ABCD.
AB = 2 2(1– 5) (5 – 6)+ = 2 2(– 4) (–1)+
= 16 1+ = 17
BC = 2 2(2 –1) (1– 5)+ = ( )221 – 4+
= 1 16+ = 17
CD = 2 2(6 – 2) (2 – 1)+ = 2 24 1+
= 16 1+ = 17
AD = 2 2(6 – 5) (2 – 6)+ = 2 21 (– 4)+
= 1 16+ = 17
Clearly, AB = BC = CD = ADAll the sides of quadrilateral ABCD areequal.Therefore, ABCD is a rhombus. It may be asquare if diagonals are equal. To confirm it,we have to find out the lengths of diagonalAC and BD.
AC = 2 2(2 – 5) (1– 6)+ = 2 2(–3) (–5)+
= 9 25+ = 34
250 AM T H E M A T C SI X–
BD = 2 2(6 – 1) (2 – 5)+ = 2 25 (– 3)+
= 25 9+ = 34
Clearly, AC = BD.Hence, quadrilateral ABCD is a square.
8. To find area of quadrilateral ABCD, wedivide it into two parts by either diagonal(see graph).Area of a triangle ABC
= 12
x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)
= 12
– 5 (– 5 + 6) – 4(– 6 – 7) – 1(7 + 5)
= 12
– 5 + 52 – 12 = 12
× 35
= 352
sq. units
ar(∆ACD) =12
– 5(– 6 – 5) – 1 (5 – 7) + 4(7 + 6)
= 12
55 + 2 + 52 =12
× 109
= 109
2 sq. units
Now, ar(quadrilateral ABCD)= ar(∆ABC) + ar(∆ACD)
= 352
+ 1092
= 1442
= 72 square units.
Alternative Method:ar(quadrilateral ABCD)
= 12
[(x1– x3) (y2 – y4) + (x2 – x4) (y3 – y1)]
= 12
[(– 5 + 1) (– 5 – 5) + (– 4 – 4) (– 6 – 7)]
= 12
[40 + 104] = 72 sq. units.
CHAPTER TEST
1. (D) Let the point of division be (x, y)
x = 1 × 3 + 2 × 7
1+2, y =
1 × 4 + 2(– 6)1+ 2
⇒ x = 173
, y = – 83
⇒
17 8, –
3 3 lies in the IVth quadrant.
2. (A) Mid-point of hypotenuse AB isequidistant from the vertices A, B and O.Therefore, the required point is
0 2 2 0,
2 2x y+ +
, i.e., (x, y).
3. Let the required point be P(0, y) such that
PA = PB
⇒ +2 2(0 – 1) ( – 5)y = +2 2(0 – 4) ( – 6)y
⇒ 1 + y2 – 10y + 25 = 16 + y2 – 12y + 36
(Squaring)
⇒ y = 13
Hence the coordinates of the point P be(0, 13).
251OOC MOEGET EDR I AN T R Y
4. False, because Q lies outside the circle asOQ > radius of circle.
5.
9a – 2 = 3 8 1 (3 1)
3 1a a× + × +
+
and – b = 3 5 1(– 3)
3 1× +
+
⇒ 36a – 8 = 24a + 3a + 1and – 3b – b = 15 – 3⇒ 9a = 9 and 4b = – 12Thus, a = 1 and b = – 3.
6. We know that area of a triangle is fourtimes the area of a triangle formed byjoining the mid-points of it.
D
1 5– ,
2 2, E(7, 3), F
7 7,
2 2
ar(∆DEF)
= 1 1 7 7 5 7 5
– 3 – 7 – – 32 2 2 2 2 2 2
+ +
= 1 1 77 –
2 4 4 +
= 114
sq. units.
Since, D, E and F are the mid-points ofsides of ∆ABC,∴ ar(∆ABC) = 4 × ar(∆DEF)
= 4 × 114
= 11 sq. units.
7. See Worksheet – 72, Sol. 8.
8. Let the coordinates of the vertices beA(x1, y1), B(x2, y2) and C(x3, y3)Observe the adjoining figure.
(Angle sum property for a triangle)⇒ ∠1 + ∠1 + 60° = 180° [Using (i)]⇒ ∠1 = ∠2 = 60° [Using (i)]⇒ ∆AOB is equilateral
∴ ar(∆AOB) =3
4r2 =
34
× 142
= 49 23 m
258 AM T H E M A T C SI X–
ar(sector AOBC) = πr2 × 60360
°°
= 22 114 14
7 6× × ×
= 2308 m
3Now, ar(segment ACB)
= ar(sector AOBC)– ar(∆AOB)
= 230849 3 m
3 −
.
7. Sides of ∆ABC are AB = 15 m, BC = 16 mand CA = 17 m. Areas grazed by threeanimals tied at A, B and C are respectivelyAPU, BQR and CST.The ungrazed area is shaded, i.e., PQRSTU(see figure) which to be required.In ∆ABC,
θ1 + θ2 + θ3 = 180° ... (i)
(Angle sum property of a triangle)
ar(sector APU) + ar(sector BQR)+ ar(sector CST)
= πr2 × 1
360θ
°+ πr2 × 2
360θ
°+ πr2 × 3
360θ
°
= 2
360rπ
° (θ1 + θ2 + θ3)
= 222 7
7 360×
° × 180° [Using (i)]
= 77 m2
For ∆ABC, semiperimeter
s =15 16 17
2+ +
= 24 m
∴ ar(∆ABC) = 24(24 15)(24 16)(24 17)− − −
= 24 9 8 7× × × = 24 221 m
Now, area of shaded region
= ( )24 21 77− m2
Hence, area of ungrazed region
= ( )24 21 77− m2.
8. ar(∆ABC) =12
× Base × Height
(∵ ∠A = 90°)
=12
× 10 × 10 = 50 cm2.
ar(sector APR) = π(7)2 × 90360
°°
=227
× 7 × 7 × 14
= 38.5 cm2.ar(PBCR) = ar(∆ABC) – ar(sector APR)
= 50 – 38.5 = 11.5 cm2.Cost of the silver plating on PBCR
= ar(PBCR) × Rate= 11.5 × 20= ` 230.00.
ASSESSMENT SHEET – 16
1. (B) πr2 = 1.54 ⇒ r2 =1.5422
× 7 ⇒ r = 0.7 m
Number of revolutions = 17622
2 0.77
× × = 40.
2. (A) This rhombus must be a square withdiagonals as the diameters of the circle.
Thus, cost of milk is ` 209 and cost of metalsheet is ` 156.75.
267FRUS AERACA S NA D ULOV EM SE
WORKSHEET– 97
1. (A) Surface area of cone= Surface area of hemisphere
⇒ πr 2 2+r h = 2πr2
⇒ r2 + h2 = 4r2
⇒ 3r2 = h2
⇒2
2rh
=13
⇒ r : h = 1 : 3 .
2. (D) N × 34 4.2
×3 2
π = 66 × 42 × 21
⇒ N = 66 × 42 × 21× 3 × 7
4 × 22 × 2.1× 2.1× 2.1 = 1500.
3. Let H = 120 cm = height of coneh = 180 cm = height of cylinder
r = 60 cm = radius of cone, cylinderand hemisphere
∴ Volume of water left in cylinder = Volume of cylinder – Volume of cone
– Volume of hemisphere
= 2 2 31 2H
3 3r h r rπ − π − π
= 2 H 23 3
r h r π − −
= 227
× 60 × 60 [180 – 40 – 40]
= 227
× 60 × 60 × 100 cm3 = 1.131 m3.
4. False, because 34R
3π = 8 × 34
3πr ⇒ r =
R2
.
5. r = 14 m
h = height of cone
= 13.5 – 3 = 10.5 m
H = height of cylinder
= 3 m
∴ l = 2 214 + 10.5 = 196 + 110.25
= 306.25 = 17.5 m
Area to be painted = C.S.A. of cone + C.S.A.
of cylinder.
= πr l + 2π rH = πr [l + 2H]
= 227
× 14 × 23.5 = 1034 m2
∴ Cost = 1034 × 2 = ` 2068.
6.
r = radius of cone =402
= 20 cm
h = height of cone = 24 cm
∴ Volume of cone (V1) = 13 π r2h cm3
Also, volume of water flows out of pipe in1 min
V2 = πR2H
= 2
351000 cm
20 π ×
Let conical vessel fills in 't' min.
∴ V1 = t × V2
268 AM T H E M A T C SI X–
⇒ t = 1
2
VV
= ( )
2
120 24
35
100020
2π ×
π × 400 × 8 × 400
25 ×1000= 51.2 min.
= 51 min 12 sec.
7. r1 = radius of small cylinder = 8 cmh1 = height of small cylinder = 60 cmr2 = radius of big cylinder = 12 cmh2 = height of big cylinder = 220 cm
Volume of pole= Volume of small cylinder
+ Volume of big cylinder.
= πr12h1 + πr2
2h2
= 3.14 × [64 × 60 + 144 × 200]
= 3.14 × [3840 + 31680]
= 111532.8 cm3
∴ Mass = 111532.8 × 8 g
= 892.26 kg.
8. 340π m2
Hint:
Quantity of canvas required
= C.S.A. of cone + C.S.A. of frustum.
WORKSHEET – 98
1. (A)
2. (C)
3. Number of bags = Volumeof circular drum
Volumeof each bag
= 23.14×(4.2) × 3.5
2.1≈ 92.
4. L = 15 cm, B = 10 cm and H = 5 cmVolume of block = L × B × H
= 15 × 10 × 5= 750 cm3.
The hole is only possible throughout thesurface having area 15 cm × 10 cm.Volume of circular hole
= π r2h = 222 7
× × 57 2
= 22 7
54× ×
=154 5
4×
= 192.5 cm3
∴ Volume of remaining solid= 750 − 192.5= 557.5 cm3
5. 6.521 kg (approx.)
Hint: Volume of metal in pipe = πh (R2 – r2)where, R = outer radius and r = inner radius.Also, mass = volume × density.
6. Canal is in the form of cuboid.So, width = 30 dm = 3 m
height = 12 dm = 1.2 mlength = distance travelled by water
in 30 min.= 5 km = 5000 m
∴ Volume of water in canal = Volume of cuboid
= 3 × 1.2 × 5000= 18000 m3
Let the required area for irrigation bex square metre. Then
8×
100x = 18000
{... Area × height = volume}⇒ x = 225000 m2.
269FRUS AERACA S NA D ULOV EM SE
7. Let BO = r
∴ OD = r
AC = 2 215 20+
= 225 400+
= 625 = 25 cm.
Area of ∆ABC = 1
× BC × AB2
=1
15 202
× × = 150 cm2 ...(i)
Again area of ∆ABC
= 1
× AC × BO2
= 1
× 25 ×2
r
...(ii)From (i) and (ii), we obtain
r = 12 cm.From ∆AOB,
AO = 2 220 12− = 16 cm.
Now, volume of double cone= Volume of cone BADOB + Volume
of cone BCDOB
= ( ) ( )2 21 1× 12 × 16 + × 12 × 9
3 3π π
= ( )1 22× × 144 16 + 9
3 7= 3771.43 cm3
Surface area of double cone
= ×12 × 20 + ×12 ×15π π
= ( )22× 12 20 + 15
7 = 1320 cm2.
8. Let height and radius of the original coneABC be AO = 3h and BO = 3r respectively.As planes PQR and STU trisect the coneABC,AQ = QT = TO = h
In ∆ABO, PQ BO,
∴PQBO
= AQAO
⇒ PQ3r
= 3hh
⇒ PQ = r
Similarly, ST = 2r
Now, VAPR = 213
πr h ...(i)
VPSUR = 3
πh[r2 + (2r)2 + r × 2r]
= 273
πr h ...(ii)
VSBCU = 3
πh[(2r)2 + (3r)2 + 2r × 3r]
= 2193
r hπ ...(iii)
Using equations (i), (ii) and (iii), we getVAPR : VPSUR : VSBCU := 1 : 7 : 19
Hence proved.OR
1.05 l; 1961.14 cm2
Hint: Capacity = Volume
= ( )2 21 2 1 23
hr r r r
π+ +
Surface area = ( )21 1 2r r r lπ + π + , r1 < r2
ASSESSMENT SHEET – 17
1. (C) π × 4x × 7x = 792
⇒ x2 = 792 7
22 4 7×
× × ⇒ x = 3
⇒ Radius = 4x = 12 cm.
2. (B) ∆AMN ~ ∆AOC
⇒h
h h′
+ ′= 2
1
rr
⇒ h′r1 = hr2 + h′r2⇒ h′(r1 – r2) = hr2
⇒ h′ = 2
2–hr
r r′
⇒ h′ + h = 1
1 2–hr
r r.
3. h = 12 m
π 21r = 9 and π 2
2r = 4
270 AM T H E M A T C SI X–
⇒ 21r =
9π and 2
2r = 4π
Volume = 3
πh× ( 2
1r + 22r + rl r2)
= × 123
π 9 4 9 4+ + ×
π π π π
= 4π × 1π
(9 + 4 + 6) = 76 m3 .
4. False, volume of the ball with radius 2a
=34
3 2 π
a= 31
6πa .
5. Edge of the cube = lRadius of the hemisphere
= r = 2l
After cutting out thehemispherical depres-sion, the surface area ofthe remaining solid
= Surface area of the cube + Innercurved surface area of hemisphere– Base area of the hemisphere
= 6l2 + 2πr2 – πr2 = 6l2 + πr2
= 6l2 + π2
2l
= 21
(24 )4
l + π sq. units.
6. Let the width of the embankment be x.Let radius of well (r)
= 72
m
Volume of the earth
dug out = πr2h
= π × 27
2
× 22.5
Volume of the embankment= π (r + x)2H – πr2H= πH (r + x + r) (r + x – r)= π × 1.5 × (7 + x) x
Since, the embankment is formed by usingthe earth dug out. Therefore, their volumesshould be equal.
i.e.,π × 1.5 (7 + x) x = π × 27
2
× 22.5
⇒ 7x + x2 = 15 × 7 × 7
2 × 2⇒ 4x2 + 28x – 735 = 0This is a quadratic equation in x. Now, wehave to solve it.
∴ ( )22 7+x = 735 + 49 = 282
⇒ 2x + 7 = ± 28⇒ x = 10.5, – 17.5Since the width cannot be negative.Therefore, x = 10.5Hence, width of the embankment is 10.5 m.
7. Radius of the pipe = r = 142
= 7 cm
Rate of flowing the water = 15 km/hr= 1500000 cm/hr
Volume of water flowing per hour from thepipe = πr2h
= 2227 1500000
7× ×
= 22 × 7 × 1500000 cm3
Volume of water required in the tanks= Length × Width × Height= 50 m × 44 m × 21 cm= 5000 × 4400 × 21 cm3
Required time
= Volume of water required
Volume of water flowing per hour
= 5000 × 4400 × 2122 × 7 × 1500000
= 2 hours.
8. In the adjoining figure,in ∆ABE and ∆ACD,BE CD, CD ⊥ AC andBE ⊥ AC⇒ ∆ABE ~ ∆ACD
⇒ABAC
= BECD
⇒2hh
= Rr
⇒ r = R2
⇒ r = 5 cm ... (i)
(∵ R = 10 cm as given)
271FRUS AERACA S NA D ULOV EM SE
V1 = Volume of upper part = 213
r hπ
= 1× 25
3hπ ...(ii)
[Using (i)]V2 = Volume of lower part
= 2 21( R R)
3π + +h r r
= ( )125 + 100 + 50
3πh
= 1
×1753
πh ... (iii)
∴ Required ratio 1
2
VV
=
1× 25
31
×1753
π
π
h
h
[Using(ii) and (iii)]
= 17
Hence, the ratio of the volumes of the twoparts is 1 : 7.
ASSESSMENT SHEET – 18
1. (B) C.S.A. = Inner C.S.A. + Outer C.S.A.
= 2π 21r + 2π 2
2r = 2π ( )2 21 2+r r .
2. (A) n ×43
π × 36
2
= ( )2112 24
3π × ×
⇒ n = 12 ×12 × 8 × 34 × 3 × 3 × 3
n = 32.
3. πr2h = 448 π ⇒ r2 × 7 = 448
⇒ r = 64 ⇒ r = 8 cm
Curved surface area = 2πrh = 2 ×227
× 8 × 7
= 352 cm2.
4. False, sides become a, a and 2a and so surfacearea will become2(a × a + a × 2a + 2a × a) = 2(a2 + 2a2 + 2a2)
= 10a2.
5. Let the edge of the cube be a and the radiusof the sphere be r.Surface area of the sphere
= Surface area of the cube
⇒ 4πr2 = 6a2 ⇒ r2 = 26
4πa ⇒ r =
32π
a
Volume of the sphere (V1) = 343
rπ
=
3234 3
3 2a π π
Volume of the cube (V2) = a3
∴ Ratio in their volumes 1
2
VV
=
3234 3
3 2a π π
: a3
=
–3 –32 2 2
–32
2 × 2 × ×
3 × 3
π πa3 : a3
=
12 ×
13
π a3 : a3 = 6π
: 1.
∴ V1 : V2 = 1 :6π
.
6. h = 7 cm; r = 52
mm = 0.25 cm
Volume of the barrel = πr2h
= 22
× 0.25 × 0.25 × 77
= 1.375 cm3
Volume of ink in the bottle
= 15
litre = 311000 cm
5× = 200 cm3
Number of barrels filled by the ink of bottle
= 200
1.375=
2000001375
= 1600
11
∵ Number of words written by 1 barrel = 330
272 AM T H E M A T C SI X–
∴ Number of words written by 1600
11 barrels
= 330 × 1600
11 = 48000.
7. The given hollowcone of base radius r2is cut at a distance h1from the vertex of it.The upper part is alsoa cone of height h1,base radius r1 and theslant height l1. Theremaining part is a frustum of height h2, baseradii r1, r2 and slant height l2. Thus, the heightand slant height of the whole cone will be h1+ h2 and l1 + l2 respectively (see figure).
∆ABC ~ ∆ADE (AA criterion of similarity)
⇒ADAB
= DEBC
= AEAC
⇒ 1 2
1
+h hh
= 2
1
rr
= 1 2
1
+l ll
... (i)
⇒ 2
11 + h
h= 1 + 2
1
ll
⇒ 1
2
hh
= 1
2
ll
... (ii)
Curved surface of the frustum = 89
× curved
surface of the whole cone
⇒ π (r1 + r2) l2 = 89
πr2 (l1 + l2)
⇒ 9r1l2 + 9r2l2 = 8r2l1 + 8r2l2
⇒ 8r2l1 – 9r1l2 = r2 l2
⇒ 1
2
8 ll
– 1
2
9 rr
= 1
(Dividing throughout by r2l2)
⇒ 8 1
2
hh
– 9 1
1 2
+
hh h
= 1
[Using (i) and (ii)]
⇒21 1 2 1 2
2 1 2
8 + 8 – 9( + )
h h h h hh h h
= 1
⇒ 8 21h – h1 h2 = h1 h2 + 2
2h
⇒ 8 21h – 2
2h = 2h1h2
⇒ 8 1 2
2 1–
h hh h
= 2
(Dividing throughout by h1h2)
⇒ Put 1
2
hh
= x to get
⇒ 8x – 1x
= 2
⇒ 8x2 – 2x – 1 = 0
⇒ (2x – 1) (4x + 1) = 0
⇒ x = 12
or x = 1
–4
Since ratio of heights can't be negative,
so x ≠ 1–
4
∴ x = 12
⇒ 1
2
hh
= 12
⇒ h1 : h2 = 1 : 2
Hence the plane divides the altitude of thecone in the ratio 1 : 2.
8. Radius of cylindrical vessel = R = 212
cm and
height of it = H = 38 cmVolume of each cylindrical vessel = πR2HVolume of ice cream in 4 vessels = 4πR2H
Radius of cone = r = 72
cm
Height of cone = h = 12 cmVolume of ice cream in the
each conical shape = 213
πr h
Radius of hemisphere = r = 72
cm
Volume of ice cream in each hemispherical
shape = 323
πr
273FRUS AERACA S NA D ULOV EM SE
∴ Volume of ice cream in one cone
= 2 31 23 3
π + πr h r
= 21( 2 )
3r h rπ +
Number of required cones
=
Volume of ice cream in 4cylindrical vessels
Volume of ice cream in 1 conewith hemispherical top
= ( )
2
2
4 R H1
23
r h r
π
π + =
( )2
212R H
2+r h r
= ( )
21 2112 × × × 38
2 27 7× × 12 + 72 2
=12 × 3 × 3 × 38
19
= 216.
CHAPTER TEST
1. (D) 1
2
VV
= 6427
⇒
31
32
4343
r
r
π
π=
3
343
⇒ 1
2
rr = 4
3
1
2
SS
= 2
12
2
44
ππ
rr
= 2
1
2
rr
=24
3
=169
∴ S1 : S2 = 16 : 9.
2. (A) 49 × 33 × 24 = 34 22× ×
3 7r
⇒ r = 3 9261 ⇒ r = 21 cm.
3. r1 = 442
= 22 cm,
r2 = 242
= 12 cm, h = 35 cm
Capacity = 3
πh 2 21 2 1 2( )r r r r+ +
= ( )22 35× × 484 +144 + 264
7 3
= 32706.67cm3 = 32706.67
1000l
= 32.7 l.
4. True, because capacity
= 2 32–
3π πr h r
= 2
(3 – 2 )3
πrh r .
5. Radius of each cone = r = 62
= 3 cm
Let the heights of the cone be h1 and h2respectively.∴ h1 + h2 = 21 cm ...(i)
Given: 1
2
VV
= 21
∴
21
22
1313
r h
r h
π
π= 2
1 ⇒ h1 = 2h2
Substitute h1 = 2h2 in equation (i) to geth2 = 7 cm ∴ h1 = 14 cm
Now, V1= 21
13
r hπ = 21 22× × 3 × 14
3 7= 132 cm3
and V2 = 22
13
πr h = 21 22× ×3 ×7
3 7= 66 cm3
Volume of remaining portion= Volume of the cylinder – (V1 + V2)= πr2 × 21 – (132 + 66)
= ( )2223 21 – 132 66
7× × +
= 594 – 198 = 396 cm3.
6. Let the edge of the new formed cube be a.Note that volume of a cube = (Edge)3
Sum of the volumes of three cubes= Volume of the new formed cube
⇒ 33 + 43 + 53 = a3
⇒ a3 = 27 + 64 + 125
⇒ a = ( )13216
⇒ a = 6 cmHence, the required edge is 6 cm.
274 AM T H E M A T C SI X–
7. See Worksheet – 96, Sol. 6.
8. Let height of the building be h.∴ Radius of dome or cylinder
= r =23
h × 12
= 13
h
Capacity of the building = 67 31m
21
⇒ πr2 (h – r) + 323
πr = 1408
21
⇒ πr2h – πr3 + 323
πr = 1408
21
⇒13
πr2 (3h – r) = 1408
21
⇒13
×227
× 219
h 13 –
3
h h =1408
21
⇒ h3 = 1408 × 21× 9 × 3
22 × 8 × 21 ⇒ h = 3 216
⇒ h = 6 m
Hence, height of the building is 6 m. ❑❑
275R SIT C E P A EP A CR P
Practice Paper–1
SECTION-A
1. (C) D = b2 – 4ac = (– 4)2 – 4 ( ) ( )2 – 2
= 16 + 8 = 24.
2. (B) a = – 5, d = 12
∴ an = a + (n – 1)d
⇒ an = – 5 + (n – 1) 12
= – 5 + 2n –
12
= – 11
2 2n+ = ( )1
– 112
n .
3. (C) M is the nearest point from P because
PMO is a straight line.
PO2 = OQ2 + PQ2
(∵ ∠Q = 90°)
⇒ (PM + 6)2 = 62 + 82
⇒ PM + 6 = 10⇒ PM = 4 cm.
4. (C) ∵ tan P = ABBP
= 2
2 3 =
13
= tan 30°
∴ P = 30°.
5. (D)As we know AB + DC= AD + BC⇒ 18 + x = 16 + 10⇒ 18 + x = 26⇒ x = 8 cm.
6. (C)∵ Arc length =180π θ
°r
⇒ 3π =(6)
180π θ
°⇒
3 1806
π × °π ×
= θ
∴ θ = 90°.
7. (D)∵∠OPQ = ∠OPT – ∠QPT= 90° – 60°= 30°.
8. (C) N × 43
× π ×36
2
= π ×24
2
× 45
⇒ N × 36 × π = π × 4 × 45
⇒ N =18036
= 5.
9. (A) In a single throw of die, the evennumber may be 2, 4 or 6.
∴ P(getting an even number) = 36
= 12
.
10. (D) Let the required angle be θ.We know that angle between tangentsand the angle between correspondingradii are supplementary
∴ θ + 35° = 180°⇒ θ = 145°.
SECTION-B
11. V1 = Volume of cone = 213
πr h
= ( )212.1 8.4
3× π × ×
V2 = Volume of sphere = 34R
3π
As V1 = V2
∴ 34R
3π = ( )21
× 2.1 ×8.43
π
⇒ R3 = (2.1)3
⇒ R = 2.1 cm.
12. As ∠APB = 80°⇒ ∠APO = 40°
(... OP bisect ∠APB)Also ∠PAO = 90°∴ In ∆OAP,
∠A + ∠P + ∠O = 180°⇒ ∠O = 50°⇒ ∠POA = 50°.
PRACTICE PAPERS
276 AM T H E M A T C SI X–
13. We have kx2 – 5x + k = 0
∴ Comparing it with ax2 + bx + c = 0,
we get a = k, b = – 5 and c = k.For real and equal roots,
b2 – 4ac = 0 ⇒ 25 – 4k2 = 0
⇒ 4k2 = 25 ⇒ k = ±52
.
14. Let the common difference of the twoA.P.’s be d. Then, their nth terms are:
16. Since A(x, y), B(1, 2), C(7, 0) are collinear.So ar(∆ABC) = 0
⇒ 12
x(2 – 0) + 1(0 – y) + 7(y – 2) = 0
⇒ 2x – y + 7y – 14 = 0⇒ x + 3y – 7 = 0.
17. Total number of cards = 1000The cards each bearing a perfect squarenumber greater than 500 are: 529, 576,625, 676, 729, 784, 841, 900 and 961.∴ Number of such cards = 9(i) P(the first player wins a prize)
= 9
1000(ii) After winning the first player:
Number of total cards = 999and number of card each bearing aperfect square number greater than500 = 8
∴ P(the second player wins a prize, if
the first has won) =8
999.
18. Let OAB be the given sector.
Then perimeter of OAB = 22 cm
⇒ OA + OB + l = 22 cm
⇒ 6 + 6 + l = 22 cm
⇒ l = 10 cm
∴ Area of sector =12
lr
=12
× 10 × 6 = 30 cm2.
OR
ND = CD
2=
122
= 6 cm
MB = AB2
= 162
= 8 cm
In ∆ODN, ON = 2 2OD – ND
∴ ON = 2 210 – 6 = 64 = 8 cm
In ∆ OBM, OM = 2 2OB – MB
∴ OM = 2 210 – 8 = 36 = 6 cm
∴ Required distance,
i.e., MN =ON – OM = 8 – 6 = 2 cm.
SECTION-C
19. tan θ = 158
= ABBC
⇒ AB = 15k and BC = 8k
⇒ AC = 2 2225 64+k k
= 17k.
∴ 17k = 90
⇒ k = 9017
∴ AB = 15k = 15 × 90
17 = 79.41 m.
⇒ Height of kite above the ground is79.41 m.
277R SIT C E P A EP A CR P
20. Given equation is 4x2 + 4bx – (a2 – b2) = 0Comparing with Ax2 + Bx + C = 0, we getA = 4, B = 4b, C = – (a2 – b2)∴ D = B2 – 4AC⇒ D = 16b2 + 16(a2 – b2)
= 16a2 > 0∴ Two distinct real roots are given by
x = – B ± D
2 A =
2– 4 168
±b a
= – 4 4
8±b a
∴ x = –2
a b or –2
a b+ .
21. Let first term of an A.P. = aand common difference = d
∴ am = 1n
⇒ a + (m – 1)d = 1n ...(i)
and an = 1m
⇒ a + (n – 1)d = 1m ...(ii)
Subtract equation (ii) from (i),
(m – n)d = 1n –
1m
⇒ (m – n)d = –m n
mn
⇒ d = 1
mn
Using d in (i), we get
a + (m –1) 1
mn=
1n
⇒ a + 1n
– 1
mn= 1
n
⇒ a = 1
mn
∴ Smn = 2
mn {2a + (mn – 1)d}
= 2
mn ( )– 12 +
mnmn mn
Smn = 1
( 1)2
+mn .
Hence proved.22. We have
SP = ( ) ( )2 22 – 2 – 0+at a at
= a ( )22 2– 1 4t t+
= 4 2 21 – 2 4a t t t+ +
= 4 22 1a t t+ +
= ( )22 1a t + = a ( )2 1+t
and SQ = 2 2
22
– – 0a a
att
+
= ( )22 2 2
4 2
1 – 4a t at t
+
⇒ SQ = 2at ( )22 21 – 4+t t
= 2at ( )221 + t = 2
at
(1 + t2)
∴1 1
SP SQ+ = ( ) ( )
2
2 2
1
1 1+
+ +t
a t a t
= ( )2
2
1
1
+
+t
a t =
1a
;
which is independent of t.Hence proved.
23. As we know that opposite sides of thecircumscribing quadrilateral sub-tendsupplementary angles at the centre of thecircle.
Total number of balls = 10 + 5 + 7 = 22(i) Probability of drawing a red ball
= Number of red balls
Total number of balls=
1022
= 5
11
(ii) Probability of drawing a green ball
= Number of green ballsTotal number of balls
= 722
(iii) Probability of drawing a blue ball
= Number of blue ballsTotal number of balls
= 522
∴ Probability of drawing not a blue ball
= 1 – Probability ofdrawing a blue ball
= 1 – 522
=1722
.
SECTION-D
29. Height of the cylinder h = 2.4 cm
and radius r = 1.42
= 0.7 cm
Also height of the conical cavity= 2.4 cm
and radius = 0.7 cm∴ Slant height of the cavity l
= 2 2+l h
= 2 2(0.7) +(2.4)
= 6.25 = 2.5 cm.
280 AM T H E M A T C SI X–
Now, T.S.A. of remai-ning solid = C.S.A. ofcylinder + C.S.A. ofcone + Area of baseof cylinder.⇒ 2πrh + πrl + πr2
⇒ πr (2h + l + r)
=227
× 0.7
(4.8 + 2.5 + 0.7)= 17.6 cm2 ≅ 18 cm2.
30. Let V = Volume of poolLet x = Number of hours required by
second pipe to fill the pool∴ x + 5 = Number of hours taken by first
pipe to fill the poolx – 4 = Number of hours taken by third
pipe to fill the pool.
The parts of poolfilling in one hourby first, second andthird pipe would be
V+ 5x
, Vx
and V– 4x
respectively.∴ According to question,
V V+
+ 5x x= V
– 4x
⇒ 1 1+
+ 5x x=
1– 4x
⇒ (2x + 5)(x – 4) = x2 + 5x⇒ x2 – 8x – 20 = 0⇒ x2 – 10x + 2x – 20 = 0⇒ (x – 10)(x + 2) = 0⇒ x = 10 or x = – 2∴ x = 10.(Value with negative sign is rejected)Hence times required by first, second andthird pipe to fill the pool individually are15 hours, 10 hours and 6 hoursrespectively.
ORLet Nisha’s present age be x years, so atpresent Asha’s age be (x2 + 2) years whenNisha grows to her mother’s present age,i.e., (x2 + 2) years then Asha’s age wouldbe (10x – 1) years.That means Nisha grows (x2 + 2x – x) yearand hence Asha also grows the sameyears. That means Asha’s age would be{(x2 + 2) + (x2 + 2 – x)} years.Now equating Asha’s both of abovementioned ages, we get:
Therefore, Nisha’s present age = 5 yearsand Asha’s present age = x2 + 2 = 27 years.
31. Join OA, OC, OB;and also join OE,OF; E and F beingpoint of contacts.Now, radius ofcircle= OD = OE= OF = 4 cmand BC = CD + DB = 14 cmAlso CD = CF = 6 cmand BD = BE = 8 cm.
Let AF = AE = x⇒ AC = 6x; AB = 8 + x
∴ a =14, b = 6 + x, c = 8 + x Now, using Heron’s formula
ar(∆ABC) = ( – )( – )( – )s s a s b s c
where, s =+ +
2a b c
=14 + 6 + + 8 +
2x x
= 14 + x
281R SIT C E P A EP A CR P
∴ar(∆ABC) = (14 + ) × × 8 × 6x x
= 48 (14 + )x x ... (i)
Also,ar(∆ABC) = ar(∆OBC) + ar(∆OAB)
+ ar(∆OAC)
=1 1
× BC × OD + AB × OE2 2
1+ × AC × OF
2
=1
× 4 × (BC + AB + AC)2
= 2(14 + 6 + x + 8 + x)
= 56 + 4x ... (ii)
∴ From (i) and (ii),
48 (14 + )x x = 56 + 4x
4 3 (14 + )x x = 56 + 4x
3 (14 + )x x = 14 + x
Squaring both sides, we get⇒ 3x (14 + x) = (14 + x)2
⇒ 3x = 14 + x⇒ 2x = 14⇒ x = 7∴ AB = 8 + x = 15 cmAnd AC = 6 + x = 13 cm.
OR
We have, radius (OP) = 5 cm = OE andOT = 13 cm.OP ⊥ PT so in right-angled ∆OPT,
OP2 + PT2 = OT2 (Pythagoras Theorem)
⇒ 52 + PT2 = 132
⇒ PT = 169 – 25 = 12 cm.
Let AE = x so PA = x.
⇒ TA = PT – PA = 12 – x
As OE ⊥ AB, in right-angled ∆AET,
AE2 + ET2 = AT2
⇒ x2 + (8)2 = (12 – x)2
⇒ x2 + 64 = 144 – 24x + x2
⇒ 24x = 80 ∴ x = 8024
= 103
cm
Similarly, BE = 103
cm
Therefore, AB = AE + BE = 203
cm.
32. r1 = Radius of top = 28 cmr2 = Radius of bottom = 7 cm
V = Capacity
= 21560 cm3
h = Height of bucket
∴ V= 2 21 2 1 2
1( + + )
3h r r r rπ
⇒ 21560 =1 22
×3 7
h (784 + 49 + 196)
⇒ 20580 = h × 1029
⇒ h = 20 cm. Now, whole surface area (S)
= πr22 + π(r1 + r2)l
Here, l = 2 2
1 2+ ( – )h r r
= 400 + 441 = 841 = 29 cm
∴ S = 227
× (7)2 + 227
(28 + 7) × 29
= 227
× (49 + 35 × 29)
= 3344 cm2.
33. Cash prizes are:
` 320, ` 280, ` 240, ` 200, ` 160, ` 120, ` 80.
282 AM T H E M A T C SI X–
Hint: S7 = 1400; d = – 40; n = 7
∴ 1400 = 72
[2a – 6 × 40]
(∵ Sn = 2n
{2a + (n – 1)d})
⇒ 200 = a – 120
⇒ a = 320.
34. In Figure, PCdenotes the multi-storeyed buildingand AB denotesthe 8 m tallbuilding. We areinterested to deter-mine the height of the multi-storeyedbuilding, i.e., PC and the distance betweenthe two buildings, i.e., AC.Look at the figure carefully. Observe thatPB is a transversal to the parallel lines PQand BD. Therefore, ∠QPB and ∠PBD arealternate angles, and so are equal. So∠PBD = 30°. Similarly, ∠PAC = 45°.In right-angled ∆PBD, we have
PDBD
= tan 30° = 13
or BD = PD 3
In right-angled ∆PAC, we have
PCAC
= tan 45° = 1
i.e., PC = ACAlso, PC = PD + DC.Therefore, PD + DC = AC.Since, AC = BD and DC = AB = 8 m,we get
PD + 8 = BD = PD 3This gives
PD = ( )
( )( )8 3 + 18
=3 – 1 3 + 1 3 – 1
= ( ) m4 3 + 1 .
So, the height of the multi-storeyed
building = ( ){ }4 3 + 1 + 8 m = 4 ( )3 + 3 m
and the distance between the two buildings
is also 4 ( )3 + 3 m.
Practice Paper –2
SECTION-A
1. (A) D > 0
⇒ 62 – 4 × 1 × k > 0
⇒ k < 9.
2. (D) 5th term from the end= 201 + (5 – 1) × (– 2)= 201 – 8= 193.
3. (B) 7
Hint: To draw a triangle similar to a given
triangle with a scalar factor pq , p > 0, q > 0,
we should locate the number of pointswhich is greater of p and q.
4. (C) Sum of opposite sides of aquadrilateral having a circle inscribed itare equal.
∴ PQ + RS = PS + QR.
5. (B) Join AC and PC.PC bisects ∠APB⇒ ∠APC = 45°⇒ ∠ACP = 45°In ∆APC,
∠APC = ∠ACP⇒ AC = AP = 4 cm⇒ Length of each tangent = 4 cm.
283R SIT C E P A EP A CR P
6. (A)43
π R3 = 49 × 33 × 24
⇒ 34 22R
3 7× × = 49 × 33 × 24
⇒ R3 = 49 × 33 × 24 × 3 × 7
4 × 22
= 49 × 7 × 3 = 73 × 33 = (21)3
∴ R = 21 units.
7. (B) Average
=3 5 5 7 7 7 9 9 9 9
10+ + + + + + + + +
= 7
P(selected number is 7) = 3
10.
8. (B) In quadrilateral POQT,
∠P + ∠O + ∠Q + ∠T = 360°
⇒ 90° + 110° + 90° + ∠T= 360°
⇒ ∠PTQ = 70°.
9. (D) r =4.22
= 2.1 cm;
h = 4.2 cm
Volume= 213
r hπ
h = ×1 22× 2.1× 2.1 × 4.2
3 7
= 19.404 cm3 ≈ 19.4 cm3.
10. (A) In right ∆ABC,
sin 30° = 6
AC
⇒ 12
= 6AC
⇒ AC = 12 m.
SECTION-B
11. r1 = 82
= 4 m , r2 = 42
= 2 m; h = 6 m
Volume of frustum
= π + +2 21 2 1 2
1( )
3h r r r r
= 1 22
× × 6(16 + 4 + 8)3 7
= 447
× 28 = 176 m3.
Thus, the capacity of the reservoir is 176 m3
or 176 kl.
12. Let E be the event of getting the sum as aperfect square.∴ E = {(1, 3), (2, 2), (3, 1), (3, 6), (4, 5),
(5, 4), (6, 3)}∴ n(E) = 7,
Now, P(E) = (E)(S)
nn
= 736
. [∵ n(S) = 36]
13. Since –4 is a root of x2 + px – 4 = 0⇒ (– 4)2 + p(– 4) – 4 = 0⇒ 16 – 4p – 4 = 0⇒ 4p = 12 ⇒ p = 3.Also as the equation x2 + px + k = 0 hasequal roots.∴ D = 0 ⇒ b2 – 4ac = 0⇒ p2 – 4k = 0⇒ 9 – 4k = 0
(∵ p = 3)
⇒ k = 94
.
ORFalse, because consider the equation2x2 + 3x – 1 = 0; its coefficients are all rational
but its roots are x = – 3 9 + 8
2 2±
× =
– 3 174
±
both of them are irrational.
14. The numbers are: 3, 9, 15, 21,.......,99.∴ It is an A.P.⇒ Let a = 3, d = 6
an = 99⇒ a + (n – 1)d = 99⇒ 3 + (n – 1) × 6 = 99 ⇒ n = 17
284 AM T H E M A T C SI X–
∴ Required sum = Sn =2n
(a + an)
S17 =172
(3 + 99) = 867.
15. True.Let PQ be the tangent at A to the circum-circle of isosceles ∆ABC.Given: AB = AC.∴ ∠ABC = ∠ACB
Construction: Draw AD; perpendicularbisector of BC as AB = AC.∴ Perpendicular bisector of BC will passthrough A as well as centre of circle O.
Let in ∆ABC;D(1, 2), E(0, – 1)and F(2, – 1) bethe mid-pointsof sides BC, ACand AB respec-tively.Let A(x1, y1), B(x2 , y2), C(x3, y3) be verticesof ∆ABC.Now, using mid-point formula
Adding equations (i), (ii) and (iii), we getx1 + x2 + x3 = 3 ...(vii)And from equations (iv), (v) and (vi), we gety1 + y2 + y3 = 0 ...(viii)By solving equation (vii) with equations (i),(ii) and (iii) respectively pairwise, we have
x3 = –1, x1 = 1, x2 = 3Similarly, we can find that
y3 = 2, y1 = – 4, y2 = 2Thus, A = (1, – 4), B = (3, 2), C = (– 1, 2)
∴ Coordinates of the vertices are(1, – 4); (3, 2) and (– 1, 2).
20. Let ABCD bethe square inwhich A(–1, 2)and C(3, 2) aregiven vertices.Let coordinatesof B are (x, y).As AB = BC ⇒ AB2 = BC2
⇒ (x + 1)2 + (y – 2)2 = (x – 3)2 + (y – 2)2
⇒ (x + 1)2 = (x – 3)2
⇒ x2 + 2x + 1 = x2 + 9 – 6x
⇒ 8x = 8 ⇒ x = 1Also ∠ B = 90°.So, AB2 + BC2 = AC2
24. The ratio between sides is 3 : 2Let ∆ ABC; AB = AC and AM ⊥ BC;
AM = 4 cm∴ A′BC′ is required triangle such that
∆ A′BC′ ~ ∆ ABC.
287R SIT C E P A EP A CR P
25. As AB = 28 cm;BC = 21 cm;∠B = 90°
⇒ AC2 = AB2 + BC2
⇒ AC2 = 282 + 212
= 784 + 441= 1225
⇒ AC = 1225 = 35 cm∴ Area of semicircle at AC as diameter
A1 = π
21 352 2
= 35 × 351 22
× ×2 7 2 × 2
= 55 × 35
4 = 481.25 cm2
Area of ∆ABC
A2 = 1
× BC × AB2
= 1
× 21 × 282
= 294 cm2
Area of quadrant PCB
A3 = 21(21)
4π
= × × ×1 2221 21
4 7= 346.50 cm2.
∴ Required area= Area of ∆ ABC + Area of semicircle – Area of quadrant= A2 + A1 – A3
= 294 + 481.25 – 346.50= 428.75 cm2.
26. The given equation is
3x2 + 5 5 x – 10 = 0The discriminant = b2 – 4ac
= (5 5 )2 – 4 × 3 × (– 10)= 125 + 120 = 245 > 0
So, the given equation has two distinctreal roots.
Now, 3x2 + 5 5 x – 10 = 0
⇒ 3x2 + 6 5 x – 5 x – 10 = 0
⇒ 3x(x + 2 5 ) – 5 (x + 2 5 ) = 0
⇒ (x + 2 5 ) (3x – 5 ) = 0
Therefore, x + 2 5 = 0 or 3x – 5 = 0
∴ x = – 2 5 or 3x = 5 ⇒ x = 5
3
∴ The two real roots are – 2 5 and 5
3.
27. The total surface area of the cube = 6 × (edge)2
= 6 × 5 × 5 cm2 = 150 cm2.Note that the part of the cube where thehemisphere is attached is not included in thesurface area.So, the surface area of the block
= T.S.A. of cube – base area of hemisphere + C.S.A. of hemisphere= 150 – πr2 + 2πr2 = (150 + πr2) cm2
= + × × 2 222 4.2 4.2
150 cm cm7 2 2
= (150 + 13.86) cm2 = 163.86 cm2.
OR
∆AOB ~ ∆AO’B’
⇒AO'AO =
O'B'OB
⇒6
12=
8r
⇒ r = 4 cm
Volume of the whole cone (V )
=13
π × 82 × 12
= 256π cm3
Volume of the upper part (small cone)
V1 = 13
π × r2 × 6
= 13
π × 42 × 6
= 32π cm3
288 AM T H E M A T C SI X–
Now,volume of upper partvolume of lower part
= 1
1
VV – V
= 32
256 – 32π
π π= 1
7.
Hence, the required ratio is 1 : 7.
28. Total number of coins = 100 + 50 + 20 + 10= 180
(i) Total number of 50 paise coins = 100∴ Probability of getting a 50 paise coin
=100 5
=180 9
.
(ii) Total number of 5 rupee coins = 10∴ Total number of coins other than 5 rupee
= 180 – 10 = 170∴ Probability of getting a coin other than
of 5 rupee = 170180
= 1718
.
SECTION-D
29. First instalment = ` 1000Second instalment = ` 1100
Third instalment = ` 1200...........................................................................................................................................................................
30th instalment = ? (to be calculated)Here, 1000, 1100, 1200,........ forms an APwith a = 1000
d = 100n = 30
Using an = a + (n – 1)d,a30 = a + 29d
= 1000 + 29 × 100= 1000 + 2900= 3900
∴ 30th instalment will be of ` 3900
Now, using Sn = 2n
(a + an)
Also S30 = 302
(1000 + 3900) = 15 (4900)
= ` 73500∴ Total amount left to be paid is 118000 – 73500 = ` 44500.
30. Let unit’s digit = xand ten’s digit = y
Then, original number = 10y + xNow, xy = 15 ... (i) (Given)According to question,
(10y + x) + 18 = 10x + y⇒ – 9x + 9y = – 18⇒ y – x = – 2⇒ y = x – 2Using it in equation (i), we get
⇒ x = – 3 or x = 5As negative digits are not acceptable here,∴ x = 5∴ Unit’s digit = 5And ten’s digit = 3Hence, number is 35.
31. Let C(O, r) be a circlewith centre at O andradius r. The circletouches the sides AB,BC, CD and DA of aquadrilateral ABCDat the points P, Q, Rand S respectively.
To show: ∠AOB + ∠COD = 180° ∠AOD + ∠BOC = 180°
Construction: Join OP, OQ, OR and OS.Proof: Since two tangents drawn from anexternal point to a circle subtend equalangles at the centre∴ ∠1 = ∠2; ∠3 = ∠4;
We are given a circlewith centre O, anexternal point T andtwo tangents TP andTQ to the circle,where P, Q are the points of contact (seefigure) . We need to prove that
∠PTQ = 2∠OPQLet ∠PTQ = θAs we know that tangent segmentsdrawn from an external point to the circleare equal. So, TP = TQ
7. (D) A non-leap year having 365 dayshas 52 weeks and one day out of 7 days.That one day could be Sunday, Monday,Tuesday, Wednesday, Thursday, Friday orSaturday.∴ Total probabilities = 7. Favourable outcomes = 1
∴ Required probability
(i.e., having 53 sundays) = 17
.
8. (C) Let radius = 4x; slant height = 7xConsider, π × 4x × 7x = 792⇒ x2 = 9⇒ x = 3∴ Radius = 4x = 12 cm.
9. (C) From figure,AB = AP + BP
= AQ + BS= AQ + (BC – CS)= AQ + (BC – CR)= 5 cm + (7 cm – 3 cm)= 9 cm. ∴ x = 9 cm.
10. (D) Let height of pole be x mIn ∆ABC,
sin 30° = ABAC
⇒ 12
= ABAC
⇒ 12
= 20x ⇒ x = 10 m.
SECTION-B
11.
Using distance formula,
2 2(10 2) ( 3)y− + + = 10
⇒ 64 + (y + 3)2 = 100⇒ (y + 3)2 = 100 – 64 = 36⇒ y + 3 = ± 6 ⇒ y = 3 or – 9.
12. Let number of blue balls = x
∴ Total balls = x + 6
∴ P(drawing a blue ball) = 6
xx +
and P(drawing a red ball) = 6
6x +According to questions,
6x
x + = 6
36x
+
⇒ x = 18.OR
No, the sample space has 8 outcomes. Outof them only one outcome say, TTT,represents no heads, so the probability of
no heads is 18
.
292 AM T H E M A T C SI X–
13. For real roots D ≥ 0 ⇒ k2 – 4 ≥ 0
⇒ (k – 2)(k + 2) ≥ 0⇒ k ≤ –2 or k ≥ 2.∴ Required value of k is 2.
14. No.Each side of the outer square
= diameter of the circle= d
So, area of square = d2
Each diagonal of theinner square = d
i.e., 2 × side = d
∴ Side = 2
d ⇒ Area =
2
2d
.
Hence area of inner square
=12
× area of outer square.
15. an = 2n + 1 ⇒ a1 = 3
∴ Sn =2n
(a1 + an) ⇒2n
(3 + 2n + 1)
= n(n + 2).
16. Let third vertex be C(x, y).
We have A(2, 3), B(– 2, 1) and G(1,23
)
Using Centroid formula,
1 = 2 – 2 +
3x
; 23
= 3 +1+
3y
⇒ x = 3, y = – 2⇒ Vertex C is (3, – 2).
17. As XP = XQ ... (i)AP = AR ... (ii)BQ = BR ... (iii)
From equation (i),⇒ XA + AP = XB + BQ
{... XP = XA + AP; XQ = XB + BQ}⇒ XA + AR = XB + BR [Using (ii ) and (iii)]
Hence proved.
18. As A =12
lr
⇒ 20π =12
× 5π × r
⇒ r = 8 cm.
SECTION-C
19. The possible outcomes of the experimentare listed in the given table the first numberin each ordered pair is the numberappearing on the blue die and the secondnumber is that on the grey die.
So, the number of possible outcomes= 6 × 6 = 36.
(i) The outcomes favourable to the event‘the sum of the two numbers is 8’denoted by E, are: (2, 6), (3, 5), (4, 4),(5, 3), (6, 2) (see figure).i.e., the number of outcomesfavourable to E = 5.
Hence, P(E) =536
(ii) As you can see from figure, there isno outcome favourable to the eventF, ‘the sum of two numbers is 13’.
So, P(F) =036
= 0
(iii) As you can see from figure, all theoutcomes are favourable to the eventG, ‘sum of two numbers ≤ 12’.
So, P(G) =3636
= 1.
293R SIT C E P A EP A CR P
20. Comparing given equation withax2 + bx + c = 0, we geta = p2, b = p2 – q2, c = – q2
∴ D = b2 – 4ac⇒ D = (p2 – q2)2 – 4(p2)(– q2)
= p4 + q4 + 2p2q2
= (p2 + q2)2 > 0∴ Given equation has two real roots given by:
x =– D
2b
a±
= 2 2 2 2
2
–( – ) ( + )
2
p q p q
p
±
⇒ x =2
2
q
p or –1.
OR23 2 5 2x x− − = 0
We need to divide –5x into two parts suchthat sum of them is –5x and product ofthem is 23 2 ×(– 2)x = – 6x2. Such partsare – 6x and x.∴ 23 2 6 2x x x− + − = 0or 3 2 ( – 2) 1 ( 2)x x x+ − = 0
or ( 2)(3 2 1)x x− + = 0
i.e., x – 2 = 0 or 3 2x + 1 = 0
i.e., x = 2 or x = 1
3 2− = –
26
Hence, the required roots are 2 and 2
6− .
21. As a3 + a7 = 6 and a3.a7 = 8
⇒ (a + 2d) + (a + 6d) = 6 ... (i)and (a + 2d)(a + 6d) = 8 ... (ii)From (i), 2a + 8d = 6⇒ a + 4d = 3 ⇒ a = 3 – 4d∴ Using it in (ii), we get
ORWe can divide the given figure into twoparts, one part as a rectangle with lengtha = 8 m, breadth b = 4 m and the other oneas a semicircle with radius r = 2 m.
28. Let BPC be the hemisphere and ABCbe the cone standing on the base of thehemisphere (see figure). The radius BO ofthe hemisphere (as well as of the cone)
296 AM T H E M A T C SI X–
=12
× 4 cm = 2 cm.
So, volume of the toy = 3 22 13 3
r r hπ + π
= 3 22 1× 3.14 × (2) × 3.14 × (2) × 2
3 3 +
= 23 × 2 1
× 3.14 × 3.143 3
+ = 8 × 3.14
= 25.12 cm3.
Now, let the right circular cylinder EFGHcircumscribe the given solid. The radius ofthe base of the right circular cylinder = HP= BO = 2 cm, and its height isEH = AO + OP = (2 + 2) cm = 4 cm
So, the volume required= Volume of the right circular cylinder
– Volume of the toy= (3.14 × 22 × 4 – 25.12) cm3
= 25.12 cm3
Hence, the required difference of the twovolumes = 25.12 cm3.
SECTION-D
29. Let PQ be the leaning tower.
Let PM = x and QM = y,
In right ∆PMQ,
cot θ = xy
⇒ x = y cot θ .... (i)
Similarly, right triangles AMQ and BMQrespectively provides.
a + x = y cot α ... (ii)and b + x = y cot β ... (iii)Eliminating x from equation (i) and (ii), weget
a = y (cot α – cot θ) ... (iv)Again eliminating x from equations (i) and(iii), we get
b = y (cot β – cot θ) ... (v)Eliminating y from equations (iv) and (v),we get
ab
=cot cot cot cot
α − θβ − θ
⇒ b cot α – b cot θ = a cot β – a cot θ⇒ a cot θ – b cot θ = a cot β – b cot α⇒ cot θ (b – a) = b cot α – a cot β
⇒ b – a =cot cot
cot b aα − β
θHence proved.
ORLet AB = height of building = 30 m
CD = height of boy = 1.5 m∴ AG = AB – GB
= AB – CD {∵ GB = CD}= 30 – 1.5 = 28.5 m
Let the distance travelled by the boy towardsthe building be x m.
In ∆AEG,
tan 60° =AGGE
⇒ 3 = AGGE
⇒ GE =AG 28.5
=3 3
... (i)
297R SIT C E P A EP A CR P
Again in ∆AGC,
tan 30° =AGGC
= AG
GE + x
⇒ 13
=28.5
GE + x
⇒ GE + x = 28.5 × 3
From equation (i),
⇒28.5
+3
x = 28.5 × 3
⇒ 28.5 + 3 x = 28.5 × 3
⇒ 3 x = 85.5 – 28.5
⇒ x =57 3
×3 3
=57
33
= 19 3 m.
30. Let side of 1st square = x mand side of 2nd square = y m
(Assuming y > x)∴ Area of 1st square = x2
And area of 2nd square = y2
According to question,x2 + y2 = 468 ... (i)
Also, perimeter of 1st square = 4xAnd perimeter of 2nd square = 4y
a – d + a + a + d = 180°(Angles sum property of a triangle)
∴ 3a = 180°or a = 60° ...(i)
Also, a – d = 12
(a + d)
(Given condition)2a – 2d = a + d
or a = 3dUsing (i), we get
3d = 60°⇒ d = 20°∴ a – d = 60° – 20° = 40°and a + d = 60° + 20° = 80°Hence, the required angles are 40°, 60°and 80°.
22. 24 sq. units.
Hint: PQ = 26 ; QR = 26 ;
RS = 26 ; SP = 26⇒ PQRS is a rhombus.
As PR = 4 2 ; QS = 6 2 .
∴ PR ≠ QS⇒ diagonals are unequal.⇒ PQRS is not a square.
302 AM T H E M A T C SI X–
Area of PQRS = 12
× (Product of diagonals)
= 24 sq. units.
23. We are given a circlewith centre O, anexternal point T andtwo tangents TP andTQ to the circle,where P, Q are the points of contact (seefigure). We need to prove that
∠PTQ = 2∠OPQLet ∠PTQ = θAs we know that tangent segmentsdrawn from an external point to the circleare equal. So, TP = TQ
⇒ 6x – 6 – 3y + 5y – 20 = ± 20⇒ 6x + 2y – 26 = ± 20⇒ 6x + 2y – 46 = 0 or 6x + 2y – 6 = 0⇒ 3x + y – 23 = 0 ...(ii)or 3x + y – 3 = 0 ...(iii)Solving (i) and (ii), we get x = 7, y = 2Solving (i) and (iii), we get x = 1, y = 0∴ Required point is (7, 2) or (1, 0).
28. Radius of the hemispherical tank
= 32
m
Volume of the tank = 32 22 3
× ×3 7 2
m3
= 9914
m3
304 AM T H E M A T C SI X–
Also, tn = t1 + (n – 1)d⇒ c = a + (n – 1) (b – a)⇒ c – a = (n – 1) (b – a)
⇒ ––
c ab a
= n – 1
⇒ n = ––
c ab a
+ 1 = ––
c a + b – ab a
∴ n = +c 2–
b – ab a
.
As we know Sn = 2n
(a + l) where a is first
term and l is last term.
i.e., Sn =
( – 2 )–2
b c cb a+
. (a + c)
∴ Sn =( )( – 2 )
2( – )a c b c a
b a+ +
Hence proved.30. Let the speed of the stream be x km/h.
Therefore, the speed of the boatupstream = (18 – x) km/h and the speedof the boat downstream = (18 + x) km/h.The time taken to go upstream
= =distance 24speed 18 – x
hours.
Similarly, the time taken to return downstream
= 24
18+ x hours.
According to the question,
24 24–
18 – 18+x x= 1
i.e., 24(18 + x) – 24(18 – x) = (18 – x) (18 + x)i.e., x2 + 48x – 324 = 0Using the quadratic formula, we get
x = ± + ±
=2– 48 48 1296 – 48 3600
2 2
= ±– 48 60
2 = 6 or –54
So, the volume of the water to be emptied
= ×1 992 14
m3
= 9928 × 1000 litres =
9900028
litres
Since, 257
litres of water is emptied in 1
second.
So,99000
28 litres of water will be emptied
in 9900028
×725
seconds, i.e., in 16.5 minutes.
ORVolume of the wall= 24 m × 0.4 m × 6 m
= 2400 cm × 40 cm × 600 cm= 2400 × 40 × 600 cm3
Volume of the mortar
= 1
10× Volume of the wall
= 240 × 40 × 600 cm3
Let the number of bricks used in theconstruction be N.Now, Volume of the wall = Volume of themortar + Volume of N bricks⇒ 2400 × 40 × 600 = 240 × 40 × 600 + N × 25 × 16 × 10
⇒ N = × × ×
× ×240 40 600 (10 – 1)
25 16 10= 1440 × 9
⇒ N = 12960Hence, 12960 bricks are used.
SECTION-D
29. Given: In an A.P., t1 = a, t2 = b and tn = c.
We have to show that Sn =( + )( + – 2 )
2( – )a c b c a
b aCommon difference d = t2 – t1 = b – a.
305R SIT C E P A EP A CR P
Since x is the speed of the stream, itcannot be negative. So, we ignore the rootx = – 54. Therefore, x = 6 which gives thespeed of the stream as 6 km/h.
(x + 13) (x – 11) = 0⇒ x = 11∴ Numbers are 11, 13.
31. Proof: We are given a circle with centreO, a point P lying outside the circle andtwo tangents PQ and PR on the circlefrom P(see figure). We are required toprove that PQ = PR.
For this, we join OP, OQ and OR. Then∠OQP and ∠ORP are right angles,because these angles are between the radiiand tangents. Now in right triangles OQPand ORP,
OQ = OR(Radii of the same circle)
OP = OP (Common)Therefore, ∆OQP ≅ ∆ORP (RHS)This gives PQ = PR (CPCT)
Consider: AB + CD= (AS + BS) + (DQ + CQ)
= (AP + BR) + (DP + CR)(Using above theorem)
= (AP + DP) + (BR + CR)∴ AB + CD = AD + BC. Hence proved.
32. Let r1 = radius of top of frustum= radius of base of cone = 3 m
r2 = radius of base of frustum = 10 mFG = h = height of frustum = 24 mAF = H = height of cone = 28 – 24 = 4 m
∴ L = slant height of cone
= + = + =2 2 2 21 H 3 4 5 mr
l = slant height of frustum
= − +2 22 1( )r r h
= + = +2 27 24 49 576 = 25 m
Quantity of canvas required= C.S.A. of cone + C.S.A. of frustum= πr1 L + π(r1 + r2) l= π(3 × 5 + 13 × 25)
= 22 22
× (15 325) = × 3407 7
+
= 1068.57 m2 (approx.)
33. Let AB = 40 m = Height of towerCD = h = Height of lighthouse
∠CAE = 30°; ∠CBD = 60°
In ∆CBD, tan 60° =CDDB
⇒ 3 =CE + DE
DB
306 AM T H E M A T C SI X–
⇒ 3 =CE + 40
DB
(∵AB = DE = 40 m)
3 = +CE 40DB DB
...(i)
Also in ∆AEC,
tan 30° =CEAE
⇒ 13
=CEDB
...(ii) (∵ AE = DB)
Using (ii) in (i), we have
3 = +1 40DB3
⇒40DB
= − =1 23
3 3
⇒ DB = 20 3 m
From equation (ii),
13
=CE
20 3
⇒ CE = 20 m∴ h = CE + DE
= 20 + 40⇒ h = 60 m.Also distance of foot of tower from top oflighthouse = CB.
∴ sin 60° =CDCB
⇒3
2=
60CB
⇒ CB = ×120 33 3
⇒ CB = 40 3 m.
ORIn figure, O is the centre of the balloon.OP = R, ∠PAQ = θ, ∠OAB = φ.
Let the height of the centre of the balloonbe h. Thus OB = hIn right-angled triangle AOP,
sin ∠OAP = OPAO
⇒ sin2θ
= R
AO
⇒ AO = R cosec 2θ ...(i)
Also, in right-angled triangle AOB,
sin φ = OBAO
⇒ sin φ = AO
h
⇒ AO = h cosec φ ...(ii)From equation (i) and (ii), we get
h cosec φ = R cosec 2θ
∴ h = R sin φ cosec 2θ
Hence proved.34. Since the inner diameter of the glass = 5 cm
and height = 10 cmSo, the apparent capacity of the glass
= πr2h= 3.14 × 2.5 × 2.5 × 10 cm3
= 196.25 cm3.
307R SIT C E P A EP A CR P
But the actual capacity of theglass is less by the volume ofthe hemisphere at the base ofthe glass.
i.e., it is less by 23
πr3
= 23
× 3.14 × 2.5 × 2.5 × 2.5 cm3
= 32.71 cm3
So, the actual capacity of the glass
= apparent capacity of glass
– volume of the hemisphere
= (196.25 – 32.71) cm3
= 163.54 cm3.
Practice Paper–5
SECTION-A
1. (A) +
21 1 5–
2 2 4k = 0
⇒ 12
k = 5 1
–4 4
⇒ k = 2.
2. (B) The given A.P. is – 11, – 8, – 5, ........, 49.To find 4th term from last, a = 49, d = – 3.∴ Required term = 49 + (4 – 1) × (– 3)
= 49 – 9 = 40.
3. (A) In ∆OPQ,PQ2 = 132 – 52
⇒ PQ = 12 cm
∴ ar( PQOR) = 2 × ar(∆POQ)
= 2 × 12
× 5 × 12
= 60 cm2.
4. (A) Area of a triangledrawn in a semicircleis directly proportionalto its height.
∴ar(largest ∆ABC) = 12
× 2r × r
= r2 sq. units.
5. (D) 1
2
VV
= 6427
⇒
31
32
4343
r
r
π
π=
3
343
⇒ 1
2
rr
= 43
⇒2
122
rr
= 169
∴ 1
2
SS
=2
122
44
rr
ππ
⇒2
122
rr
= 169
i.e., S1 : S2 = 16 : 9.
6. (D) Let O be the centre of the circle and PAand PB are two tangents inclined at 60°.In right-angled ∆OAP,
∠OPA = 30°
APBOPA = OPB =
2∠ ∠ ∠
∵
∴ tan 30° = OAAP
⇒13 =
3AP
⇒ AP = 3 3 cm.
7. (D) Each pair of opposite sides of aquadrilateral circumscribing a circlesubtends complementary angles at thecentre of the circle.∴ ∠AOB + ∠COD = 180°⇒ ∠COD = 180° – 125°
= 55°.
8. (C) Let the girl bought n tickets.
∴6000
n= 0.08
⇒ n = 480.
308 AM T H E M A T C SI X–
9. (B) Let the sun’s elevation = θ.
⇒ tan θ = ABBC
=6
2 3 = 3
= tan 60°
⇒ θ = 60°.
10. (D) As XY ⊥ AB and CD || XY so CD ⊥ AB.
⇒ CD is bisected by AB at M.
⇒ CD = 2 DM.
In ∆OMD,
DM = 2 2OD – OM
= 2 25 – 3
(... r = 5 cm, OM = AM – AO = 8 – 5 = 3 cm)∴ DM = 4 cmTherefore, CD = 2 × 4 = 8 cm.
SECTION-B
11. Side of square = a = 7.5 × 2 = 15 cm
Area of shaded part = Area of square – Area of circle= a2 – πr2
= 152 – 3.14 × (7.5)2
= 225 – 176.625= 48.375 cm2.
12. Let first term = a1 and commondifference = d.
⇒ 7a7 = 11a11
⇒ 7(a1 + 6d) = 11(a1 + 10d)
⇒ 7a1 – 11a1 = 110d – 42d
⇒ a1 = – 17d
⇒ a1 + 17d = 0
∴ a18 = 0.
13. 6x2 – 2x – 2 = 0Comparing the coefficients of like powersof the given equation with ax2 + bx + c = 0,
Favourable outcomes are: (1, 3), (2, 2),(2, 6), (3, 1), (3, 5), (4, 4), (5, 3), (6, 2),(6, 6). These are 9 in counting.
Hence, the required probability
= 9
36 =
14
.
17. Volume of the remaining solid
= Volume of the cube
– Volume of the cone.
= l3 –13
πr2h
= 73 –13
×227
× 32 × 7
= 343 – 66 = 277 cm3.
18.
Extend AB and CD to meet at P.Since the pair of tangents drawn from anexternal point to a circle are equal,i.e., BP = DP and AP = CP.So, AP – BP = CP – DPi.e., AB = CD. Hence proved.
SECTION-C
19. Number of cards with numbers from 2to 101
= 101 – 2 + 1 = 100∴ n(S) = 100(i) Let E1 be the event of drawing a card
having an even number.Even numbers are : 2, 4, 6, ....., 100,which are 50.∴ n(E1) = 50
Now, P(E1) = 1(E ) 50 1= =
(S) 100 2nn
.
(ii) Let E2 be the event of drawing a cardhaving a square number. Square numbersare: 4, 9, 16, 25, 36, 49, 64, 81, 100.
∴ n(E2) = 9 ∴ P(E2) = 2(E ) 9(S) 100
nn
= .
OR
Number of non-defective bulbs= 24 – 6 = 18
P(First bulbs is not defective)
= Number of non-defective bulbs
Total number of bulbs
=1824
= 34
If the first selected bulb is defective, thenthe number of remaining defective bulbs= 6 – 1 = 5 and the remaing total numberof bulbs = 24 – 1 = 23.
24. Circumcentre of a triangle is equidistantfrom the vertices of the triangleLet O (x, y), be the circumcentre of the given∆ABC.∴∴∴∴∴ OA = OB = OCi.e., OA2 = OB2 = OC2
Taking OA2 = OC2, we get
(x – 8)2 + (y – 6)2 = (x – 2)2 + (y + 2)2
⇒ x2 – 16x + 64 + y2 – 12y + 36
= x2 – 4x + 4 + y2 + 4y + 4
⇒ – 12x – 16y = 8 – 100
⇒ 12x + 16y = 92
⇒ 3x + 4y = 23 ...(i)
Also taking OB2 = OC2,
(x – 8)2 + (y + 2)2 = (x – 2)2 + (y + 2)2
⇒ (x – 8)2 = (x – 2)2
⇒ x – 8 = ± (x – 2)
But x – 8 = (x – 2) as – 8 ≠ – 2
∴ x – 8 = – (x – 2)
i.e., 2x = 10
i.e., x = 5
Substituting x = 5 in equation (i), we get
3 × 5 + 4y = 23
i.e., 4y = 23 – 15 = 8
i.e., y = 2Circumradius = OA
= ( ) ( )+2 2– 8 – 8x y
= ( ) ( )+2 25 – 8 2 – 6
= +9 16 = 5
Hence, the circumcentre is (5, 2) and thecircumradius is 5 units.
25. Given chord is AC. Exterior angleAOC = 270°. ACDA is the major segment.∴ ar(segment ACDA)
= ar(sector AOCDA) + ar(∆AOC)
= 270°360°
× π × (20)2 + 12
× 20 × 20
= 34
× 3.14 × 400 + 10 × 20
= 942 + 200
= 1142 cm2.
26. Area of square = (side)2 = (4)2
= 16 cm2
Area of each quadrant = 14
πr2
= 14
× 3.14 × (1)2
= 3.144
cm2
∴ Area of 4 quadrants = 4 × 3.144
= 3.14 cm2
Also area of the circle drawn in middle ofthe square = πr2
= 3.14 ×22
2
= 3.14 cm2
312 AM T H E M A T C SI X–
27. In the figure drawn here, let AB be a wallof height h m and AC be a ladder 15 mlong that makes an angle of 60° with thewall.
In right-angled ∆ABC,
cos 60° =ABAC
⇒ 12
= 15h
⇒ 2h = 15
∴ h = 152
= 7.5 m.
28. Let the required ratio be λ : 1
We will use section formula,
– 3 = – 2 – 5
+1λ
λ ; p =
– 4 + 3+1
λλ
i.e., – 3λ – 3 = – 2λ – 5 ; p = – 4 + 3
+1λ
λ
i.e., λ = 2 ; p = – 4 + 6
3=
23
Required ratio is 2 : 1 and p = 23
.
SECTION-D
29. r = AO = BO = CO = 3.5 cm
CD = 15.5 cm
OD = CD – CO
= 15.5 – 3.5 = 12 cm
Now, area of shaded part = Area of square– Sum of areas of 4 quadrants – Area ofcircle drawn in centre.
= 16 – 3.14 – 3.14= 16 – 6.28 = 9.72 cm2
OR
Side of ∆ABC = a = 10 cm
Radius of each circle = r = 102
= 5 cm
ar(ABC) = 3
4a2 =
34
× 102
= 25 3 = 25 × 1.732
= 43.30 cm2
Area of each sector situated in ∆ABC
= 260°×
360°rπ =
16
× 3.14 × 52 cm2.
Area of the shaded region
= ar(ABC) – 3 × area of any onesector
= 43.30 – 3 × 16
× 3.14 × 25
= 43.30 – 39.25 = 4.05 cm2.
313R SIT C E P A EP A CR P
l = AD = 2 2OD AO+
= +144 12.25
= 12.5 cm.
Now, total surface area
= C.S.A. of the cone ABD + C.S.A. of the hemisphere ACB
= πrl + 2πr2 = πr(l + 2r)
=227
× 3.5 × (12.5 + 7)
= 22 × 0.5 × 19.5 = 214.5 cm2.
OR
Given: r1 = 20 cm, r2 = 8 cm and h = 16 cm
Volume of the container
V = ( )2 21 2 1 2
13
h r r r rπ + +
= ( )2 21× 3.14 × 16 20 8 20 8
3+ + ×
= ( )1× 50.24 × 400 64 160
3+ +
= 13
× 50.24 × 624 = 10450 cm3
(approx) ∴ Capacity of container
= 104501000
l = 10.45 l
Cost of milk = Volume of the container × Rate per litre
= 10.45 × ` 15= ` 156.75.
30. Let the tap of larger diameter takes t hoursto fill the tank. So the smaller one will take(10 + t) hours on the same work.∴ The part of the tank filled by larger tap in
1 hour = 1t
And the part of the tank filled by the smaller
tap in 1 hour = +
110 t
So, the part of the tank filled by both the taps
simultaneously in 1 hour = 1t
+ +
110 t
...(i)
But it is given that the taps together fill the
tank in 3
98
=758
hrs.
So, the part of the tank filled by both the taps
simultaneously in 1 hour = 1
758
= 8
75hrs
...(ii)From the results (i) and (ii), we have
++
1 110t t
= 875
10( 10)
t tt t
+ ++
= 8
75
⇒ 750 + 150t = 8t2 + 80t
⇒ 8t2 – 70t – 750 = 0,
314 AM T H E M A T C SI X–
i.e., 4t2 – 35t – 375 = 0
D = (– 35t)2 – 4 × 4 × (– 375) = 1225 + 6000
∴ D = 7225± = ± 85
∴ t = 35 85
2 4±×
= 120
8 or
– 508
∴ t = 1208
= 15 hrs
(Time cannot be negative so t = – 50
8is rejected.)
And t + 10 = 25 hrs.Thus, time taken by the larger tap = 15 hrs.and time taken by the smaller tap = 25 hrs.
31. DA and DC are two tangents from D to thecircle with centre F.∴ ∠ADF = ∠CDF ...(i)Similarly,∴ ∠BEF = ∠CEF ...(ii)As l || m and DE is transversal,∴ ∠ADE + ∠BED = 180°
Here, 2nd term – 1st term = 3rd term – 2nd term= 3. So, this is the sum of A.P. with commondifference d = 3 and the first term a = – 4.
Let this A.P. contains n terms.
Then, using an = a + (n – 1)d, we get
x = – 4 + (n – 1) 3
⇒ n – 1 = + 43
x
⇒ n = + 73
x
Now, LHS = + 73
x(– 4 + x)
[Using Sn = 2n
(a + l)]
= 2– 4 – 28 76
x x x+ +
= +2 3 – 28
6x x
Hence, the given equation becomes
+2 3 – 286
x x= 437
⇒ x2 + 3x – 2650 = 0
x2 + 53x – 50x – 2650 = 0
⇒ (x + 53) (x – 50) = 0
⇒ x = – 53 or x = 50
But x = – 53 is not possible because the A.P.has no negative term after the second term.
Therefore, x = 50.
OR
Ruchi has 13 flags to be fixed at one sideof her, each at R1, R2,......., R3 (see figure)and the same no. on the opposite side ofit such that the flags are on the straightpassage.
315R SIT C E P A EP A CR P
Let initially Ruchi is at the point R withher books and 27 flags.
Ruchi fixed one flag at the point R also
RR1 = R1R2 = ..... = R12R13 = 2 m
Fixing one flag at R1, distance covered byRuchi
= RR1 + R1R
= 2 + 2 = 4 m
Fixing one flag at R2, distance covered byRuchi = RR2 + R2R
= 4 + 4 = 8 m and so on.
Therefore distance covered by Ruchi to fixall flags on one side of R.
= 4 + 8 + 12 + ....... to 13 term
= 4(1 + 2 + 3 +.......to 13 terms)
= 4 × 13 × 14
2 = 364 m
Similarly, distance covered by Ruchi to fixall flags on other side of R = 364 m.
Hence, total distance covered
= 364 + 364 = 728 m
Maximum distance travelled by Ruchicarrying a flag = RR3