10 Mathematics Problems Chapters 1 to 6

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SSLC Mathematics

Unit – Ari thmet ic Sequences

C L A S S X

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BRAINS MOOZHIKKAL, KOZHIKKODE

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SSLC

Unit 1

Arithmetic Sequences

Prepared By Fassal Peringolam

(Plus Two Maths Science Teacher)

BRAINS MOOZHIKKAL KOZHIKKODE

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Arithmetic SequencesEquations:

First term: f

Common difference: d

Nth term:

1

(a-common difference; a+b first term)

Common difference

Sum of sequence: 2 2 1

2

2

2

2

etc.

;



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Sample Questions

PROBLEM

1. For an arithm etic sequence , the 8th term is 35 and the 11th

term is 47

a) Find 14th term

b) Find 5th term

c) To get the 17th term, how much is to be added to the 8th

term?

Answer

Common difference =d

8th term = 35

11th term = 47

8th term + 3d = 11th term

3d=11th term - 8th term = 47- 35= 12

a) 14th term = 11th term + 3d =47 + 12 = 59

b) 5th term = 8th term – 3d =35-12 = 23

c) 17th term = 8th term + 9d

17th term = 8th term + 3 x12 = 8th term + 36

PROBLEM

2. For the arithm etic sequence 22, 26, 30...

a) W hat is the comm on difference?

b) Find the 7th term

c) W ill 50 be a term of this sequence? W hy?

d) Can the difference between any two term s of this

sequence be 50? Justify your answer.

Answer

a)

Common difference d = 26 – 22 = 4

b) 10 th term is 22 7 14 = 46



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c) If any term of this sequence is divided by the common difference 4, the

remainder is 2. 50 divided by 4 gives remainder 2. So, 50 is a term of this

sequence

d) The difference between any two terms of an arithmetic sequence will

be the multiple of its common difference. Here, 50 is not a multiple of 4.

So, for this sequence, 50 cannot be the difference of two terms.

PROBLEM

3.

Con sider the multiples of 7 in between 100 an d 500.

a) W hat are the first and last numbers?

b) How man y terms are ther in this sequence?

Answer

To get the first term, first find the remainder when 100 is divided by 7,

remainder is 2.

Then subtract 2 from 100 and add 7

So, 1st term 100-2+7 105

To get the last term, find the remainder when 500 is divided by 7,

remainder is 3.

Subtract 3 from 500

So, last term 500-3 497

b) Common difference,

49771057 n (497-98)/7 57

So, number of term 57

PROBLEM

4.

First term o f an arithmetic sequence is 6 and sum of first two

terms is 10.

a)

Find Com mon difference

b) Find the third term.



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Answer

First term a 6

Sum of first two terms 10

a+a+d 10

d 10-12 -2Third term a 3 a+2d 2

PROBLEM

5.

Exam ine whether 685 is a term of the arithm etic sequence 7,

29, 51 … ? Give reason?

Answer

First term a = 7

Common difference d = 29 – 7 = 22

nth term = 685

a + (n-1)d = 685

7 + 22n - 22 = 685

n = 31.82

Thus 685 is not a term of the arithmetic sequence 7, 29, 51 …

Because n is not a whole number

PROBLEM

6.

W hich term o f the sequence, 205,199 … is the first negative

term?

Answer

First term, f =205 and

Common difference d = 6

Let the nth term of the given AP be the first negative term.

Then, x n < 0

f + (n-1) d < 0

205+ (n-1) x 6 < 0

205 + 6n - 6 < 0



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6n > 199

n > 33.17

i.e., n ≥ 34

Thus, the 34 th term of the given sequence is the first negative term.

PROBLEM

7. The 5th term of an arithmetic sequence is 34 and the

15th term is 9.

a) Find the comm on difference

b)

Find the r

th

term of the arithmetic sequence.

Answer

According to the given information,

x 5 = f + 4d = 34 and

x 15 = f + 14d = 9

Solving the two equations, we get

f = 44

d = -2.5

Therefore, rth term = f + (r-1)d

= 44 -(r-1)2.5

= 46.5 - 2.5 r

PROBLEM

8.

Write the sequence got by adding 2 to the numbers got by

mu ltiplying 5 to the natural num bers starting from 1.

a) Check w hether it is an Arithm etic sequence.

b) W ill 100 be a term o f this sequence?

Answer

Natural numbers starting from 1 are 1, 2, 3...

Multiplying with 5;

5, 10, 15...

Adding 2;



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7, 12, 17...

a) This is an Arithmetic sequence with common difference 5.

b) If any term of this sequence is divided by the common difference 5,

the remainder is 2.

100 divided by 5 gives remainder 0, not 2. So, 100 is not a term of this

sequence.

PROBLEM

9.

a) Write the arithmetic sequence with 1st term 7 and

comm on difference -2

b) W rite the algebraic expression of this sequence.

c) W ill -30 be a term of this sequence? Justify your answer.

Answer

a) 7, 5, 3...

b) x n = f + (n-1) d

= 7+ (n-1) × -2 = -2n +9

c) Take -30 as the nth term. If we get n as a natural number, it will be a

term of the sequence.

-2n + 9 = -30n = -39 / -2

n = 19.5

Here, n is not a natural number. So, -30 is not a term of this sequence.

PROBLEM

10. For an arithm etic sequence , the 8th term is 67 and the 18th

term is 147

a) Find the com mon difference

b) Find the sum of 1st term and the 25th term

c) Find 13th term

d) Find the sum of the first 25 terms.



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Answer

a) Common difference,

188

(147-67)/(18-8)

80/10

8

b) 1st term + 25th term= 8th term + 18th term =67+147 = 214

c) 13th term = (12th term + 14th term)/2

= (1st term + 25th term)/2= 214 / 2

= 107

d) Sum of the terms = Number of terms x Middle term

Sum of first 25 terms = 25 x 13th term

= 25 x 107

= 2675

PROBLEM

11.

For what value of n, the nth terms of the arithmetic sequence 63,65, 67… and 3, 10, 17… are equal?

Answer

The first AP is 63, 65, 67…

Here, f=63, d=2

Therefore x n=63 + (n-1)2

The second AP is 3, 10, 17 …

Here, f=3, d=7Therefore, a n= 3 + (n-1)7

Therfore,

63 + (n-1)2 = 3 + (n-1)7

60 = 5(n-1)



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n -1 = 12

n = 12+1 = 13

PROBLEM

12.

Which term of the arithmetic sequence 5, 15, 25 ... will be

130 mo re than the 31

st

term?

Answer

Here, f = 5 and d = 10

x 31= f + 30d = 5 + 30 x 10 = 305

Let the required term be the nth term. Then,

x n = 130 + x 31

f + (n-1)d = 130+305

5 + (n-1)10 = 435(n-1)10 = 430

n-1 = 43

n = 44

Hence, the 44th term of the given AP is 130 more than its 31st term.

PROBLEM

13.

In an arithm etic sequence the first term is -4, the last term is

29 and the sum of all its term is 150. Find its common

difference.

Answer

f = -4, x n = 29, Sn = 150

Given that the last term is 29.

29 -4 + (n - 1)d

33 (n - 1)d … (1) Also, given that the sum of all its term is 150.

150 24 1

300 n [-8 + 33] [Using (1)]



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300 n (25)

n 12

So, from (1), we get: d = 3

PROBLEM

14. Find the sum of first 24 terms of the list of numbers whose

nth term is given by a

n

= 3+2n.

Answer

x n= 3+2n

Now, put n=1,2,3

x 1= 3+2(1) = 5

x 2= 3+2(2) = 7x 3= 3+2(3) = 9

Thus, the terms of the AP are 5,7,9

Here, f = 5 and d = 2

2 25 2412 12[10+46]

12 56

672

PROBLEM

15.

Find the sum: -5+ -8) + -11) +... .. . . .. . . .. .+ -230)

Answer

-5, -8, -11, … -230 forms an AP with f=-5, d=-8-(-5) =-3

Let -230 = x n = f+(n-1)d

= -5+(n-1)(-3)

-230 + 5 = (n-1)(-3)n -1 = 75

n = 76

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[(-5) + (-230)]

38 (-235)

- 8930

PROBLEM

16.

Find the sum of all two-digit odd positive numbers.

Answer

Two digit odd positive numbers are 11, 13, 15, ...., 99.

First term, f = 11

Common difference, d = 13 - 11 = 2

Last term, x n = l =99

Now, x n f + (n - 1)d

99 11 + (n - 1)2

99 11 + 2n - 2

2n 90

n 45

2 452 1199

2475Thus, the sum of all two-digit odd positive numbers is 2475.

PR CTICE EXERCISE

1) Find the 8th term of an arithmetic sequence whose 15th term is

47 and the common difference is 4.

[19]

2) Find the 31st term of an arithmetic sequence whose 11th term is

38 and the 16th term is 73.



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[178]

3) The 10th term of an arithmetic sequence is 52 and 16th term is

82.

(a) Find the 32nd term

(b)

Obtain the general term.[x 32 = 162, x n = 5n + 2]

4) Which term of the arithmetic sequence 3, 8, 13, 18 ... is 248?

[50]

5) The 7th term of an arithmetic sequence is – 4 and its 13th term is

–16.

(a) What is common difference?

(b) Find the arithmetic sequence.

[-2; 8, 6, 4, ..... ]6) The 17th term of an arithmetic sequence exceeds its 10th term by

7. Find the common difference.

[1]

7) If the 10th term of an arithmetic sequence is 52 and 17th term is

20 more than the 13th term, find the arithmetic sequence.

[7, 12, 17, 22.......]

8) The 9th term of an arithmetic sequence is equal to 7 times the

2nd term and 12th term exceeds 5 times the 3rd term by 2.(a) Find the first term

(b) Find the common difference.

[a = 1, d = 6]

9) Find the middle term of an arithmetic sequence with 17 terms

whose 5th term is 23 and the common difference is –2.

[15]

10) Find 20th and 25th terms of an arithmetic sequence 2, 5,8,11…

[59, 74]11) For what value of n the nth term of the arithmetic sequence 23,

25, 27, 29 ….and -17,-10,-3, 4 ….are equal?

[9]



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12) The sixth term of an AP is -10 and tenth term is -26.Find its 15th

term. [-46]

13) The angles of a triangle are in arithmetic sequence. If the greatest

angle equals the sum of the other two, find the angles.

[30°, 60°, 90°]14) Three numbers are in arithmetic sequence. If the sum of these

numbers is 27 and the product is 648, find the numbers.

[6, 9, 12]

15) Find sum of 1 + 3 + 5 + ....... to 50 terms.

[2500]

16) Find the sum of first 30 even natural numbers.

[930]

17)

The 10th term of an arithmetic sequence is 29 and and sum of thefirst 20 terms is 610. Find the sum of the first 30 terms.

[1365]

18) How many terms of the arithmetic sequence 3, 5, 7, 9 … must be

added to get the sum 120?

[10]

19) The 3rd term of arithmetic sequence is –40 and 13th term is 0.

Find the common difference.

[4]20) The sum of three numbers of an AP is 15. Find its first term.

[5]




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SSLC Mathematics

Un it – Circ les

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SSLC

Unit 2

Circles





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CirclesPoints:

The ends of a diameter of a circle are joined to some point and

we get a right angle at that point. At points within the circle

we get an angle larger than a right angle. For points outsidethe circle, we get an angle smaller than aright angle.

Any diameter of the circle divides it into two equal arcs; and

we get a pair of right angles by joining points on each to the

ends of the diameter.

Two points on a circle divide it into a pair of arcs. The angle

got by joining these two points to a point on one of these arcs

is equal to half the central angle of the alternate arc.

The angle made by an arc at any point on the alternate arc is

equal to half the angle made at the centre. Every chord of a circle divides it into two pruts. Such a part is

called a segment of a circle.

Angles in the same segment of a circle are equal.

A chord divides a circle into a pair of segments.

Angles in alternate segments are supplementary.

If all vertices of a quadrilateral are on a circle, then its opposite

angles are supplementary.

Suppose a circle is drawn through three vertices of a

quadrilateral. If the fourth vertex is outside this circle, then the

sum of the angles at this vertex and the opposite vertex is less

than 180°; if the fourth vertex is inside the circle, then this sum

is greater than 180°.

A quadrilateral for which a circle can be draVvn through all

the four vertices is called a cyclic quadrilateral.

All rectangles are cyclic quadrilaterals. Isosceles trapeziums

are also cyclic quadrilaterals.

If the opposite angles of a quadrilateral are supplementary,then we can draw a circle through all four of its vertices.

The exterior angle at a vertex of a cyclic quadrilateral is equal

to the interior opposite angle.



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Sample Questions

PROBLEM

1. In the given figure, O is the centre of the circle. If OAC = 35

o

and OBC = 40

o

, find the value of x.



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Answer

Join OPSince OA = OP ACO = ∠OAP = 35o

Similarly, OB = OP and ∠OPB = ∠OBP = 40o APB = 35o + 40o = 75o ∠AOB2x75 o 150 o

PROBLEM

2.

In figure, O is the centre of circle. If

∠

PAO=35 and

PCO=45,then

Calculate ∠ APC and ∠AOC.

Answer

Join OP.

In ∆OQP,

OQ = OP = r

∠OQP = ∠OPQ = 35o

In ∆OPR,

OR = OP = r

∠ORP = ∠OPR = 45o

∠QPR = 35o + 45o = 80o



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∠QOR = 2∠QPR = 2 80o QOR = 160o

PROBLEM

3.

In the figure find PQ B, O is the centre.

Answer

PQ B

= [180 - (2 x 42)]/2= 48

PROBLEM

4.

In the figure PQ is the diameter of the circle,then

a) Find the value of R

b) Find the value of Q

c)

If QR=6cm,then f ind PR



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Answer

∠R=90 ∠Q=90-30=60 The ratio of sides of triangle is 1:√3:2Thus PR=6√3PROBLEM

5 In the f igure f ind APB and AQ B where O is the centre of

the circ le and OA P = 32 and OBP = 47 .

Answer

Join OP.

In ∆OAP, OA = OP = radius

∠OAP = ∠OPA = 32o

In ∆OPR, OB = OP = radius

∠OBP = ∠OPB = 47o

∠ APB = 32o + 47o = 79o

∠

AQB = 180

o

-79

o

=10

o

PROBLEM

6.

O is the centre of the circle as shown in the figure. Find

∠

CBD



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Answer

Take a point E on the circle, join AE and CE.∠ AEC=100/2=50o ∠ AEC + ∠ ABC = 180o (Opposite Angles of a cyclic quadrilaterals)∠ ABC = 130o ABC + ∠CBD = 180o (linear pair)

130o + ∠CBD = 180o ∠ CBD = 50o

PROBLEM

7. Draw a squre of area 12cm

2



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Answer

PROBLEM

8. Using the follow ing fiqure,

a)

∠

B+

∠

E=......?

b) ∠ D ∠E......?

Answer

(a) 180o

(b) ∠EAD



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PR CTICE EXERCISE

1)

ABCD is a cyclic quadrilateral. If BCD = 100, ABD = 70,find ADB.

[30]

2) ABCD is a cyclic quadrilateral with DBC = 80, BAC = 40,find BCD.

[60] 3) In cyclic quadrilateral ABCD, ADBC. B = 70, find

remaining angles.[110, 70, 110]

4) Draw a rectangle of sides 5cm and 3cm and draw a square of thesame area.

5) 2 chords AB and CD intersect at a point O. If AO = 3.5cm, CO =5cm, DO = 7cm, find OB.

[10cm]

6) 2 chords AB and CD intersect at O. If AO = 8cm, CO = 6cm, OD= 4cm, find OB.

[3cm]7) Two circles intersect at two points A and B. AD and ACrespectively are diameters to the two circles. Prove that B lies onthe line segment DC.

8) How do we draw a 22½ o angle?


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SSLC Mathematics

Unit – Second Degree Equat ions

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SSLC

Unit 3

Second Degree Equations





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Second Degree EquationsPoints:

A quadratic equation in the variable x is of the form ax2 + bx

+ c = 0, where a, b, c are real numbers and a ≠ 0.

Different methods to findout the solutions of second degree

equations

Completing the square method

Quadratic Formula (Shreedharacharya’s rule)method

√ 42

is the solutions of second degree equation ax2+bx+c=0,a≠0

b2

4ac is the discriminant of ax2+bx+c=0

If b2

4ac=0, then the equation has only one solution and the

solution is –b/2a.

If b2‐4ac <0 (a negative number),then the equation has no

solution

If b2‐4ac >0 (a positive number) then the equation has two

different solutions



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Sample Questions

PROBLEM

1.

Find the roots of the second degree equation 2 x

2

- 7x 3 = 0

by the method o f completing the square.

Answer

We have

2 7 3 0

72 3

2 0

2 74 3

2

2 74 74

32 74

74

49

16 32

74

49

16 32

74

49

16 2416 25

16

74 54

3 12

PROBLEM

2.

Sum of the area of two squares is 500 m

2

. If the difference of

their perimeters is 40 m, find the sides of the two squares.

Answer

Let the side of the squares be x and y meters.

According to the condition,

x 2 + y 2 = 500 ….(1)



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4x - 4y = 40

(x – y) = 10

y = x - 10

Substituting the value of y in (1), we get,

x 2 + (x - 10)2 = 500

2x 2 - 20x - 400 = 0

x 2 - 10x - 200 = 0

x = 20 or x = -10

As the side cannot be negative, x = 20

Hence, side of the first square, x = 20m

Side of the second square, y = (20 - 10) = 10 m

PROBLEM

3.

Three consecutive positive integers are taken such that the

sum of the square of the first and the product of the other

two is 232. Find the integers.

Answer

Let the three consecutive positive integers be x, x + 1, x + 2.

x 2 + (x + 1) (x + 2) = 232

x 2 + (x 2 + 3x + 2) = 232

2x 2 + 3x - 230 = 0

√ 42

3 √ 1849

4

x = 10 or -11.5

But, x is a positive integer, so, x = 10.

Thus, the numbers are 10, 11, and 12.



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PROBLEM

4.

The sum of the squares of two consecutive even numbers is

164. Find the numbers.

AnswerLet the consecutive numbers be x, x 2.x2 x 22 164x2 x2 4x 4 1642x2 4x ‐ 160 0x2 2x ‐ 80 0x 8 or x ‐10

Neglecting the negative value, we get, x 8.The numbers are 8 and 10.PROBLEM

5.

250 Rupees is divided equally among a certain number of

children. If there were 25 children more, each would have

received 50 paise less. Find the num ber of children.

AnswerLet the number of children be x.It is given that Rs 250 is divided amongst x children.So, money received by each child 250/xIf there were 25 children more, thenMoney received by each child 250/x25From the given information,

250 2502 5 50100

2 5 1 2 5 0 0 0

= -125 or 100

Since, the number of children cannot be negative, so, x 100.Hence, the number of children is 100.



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PROBLEM

6.

By increasing the speed of a bus by 10 km /hr, it takes one and

half hours less to cover a journey o f 450 km . Find the original

speed of the bus.

Answer

Let speed of the bus be x km/hr

Time t = 450/x

If speed is x + 10, then time T = 450/(x + 10)

By question,

450/x - 450/(x + 10) =3/2450 (x + 10) -450 x = (3/2) (x 2 + 10x)

4500 2 = 3 x 2 + 30x

x 2 + 10x - 3000 = 0

x = 50 or x = -60

x = 50

PROBLEM

7.

A person has a rectangular garden w hose area is 100 sq m. He

fences three sides of the garden with 30 m barbed wire. On

the fourth side, the wall of his of his house is constructed;

find the dimensions of the garden.

Answer

Let the length and breadth of garden be x m and y m respectively.

Area of the garden = 100 sq m

xy = 100 m2 or y =100/x

Suppose the person builds his house along the breadth of the garden.

Then, we have:

2x + y = 30



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2x + (100/x) =30

2x 2 - 30x + 100 = 0

x = 10, x = 5

When x = 10 m, we have: y = 10 m

When x = 5 m, we have: y = 20 m

Thus, the dimensions of the garden are 10 m × 10 m or 5 m × 20 m.

PROBLEM

8.

The hypotenuse of a right triangle is 20m. If the difference

between th e length of the other sides is 4m. Find the sides.

Answer

x 2 + y 2 = 202

x 2 + y 2 = 400

Also x - y = 4

x = 4 + y

(4 + y)2 + y 2 = 400

2 y 2 + 8 y – 384 = 0

y = 12 y = – 16Sides are 12cm and 16cm

PR CTICE EXERCISE

1) Perimeter of a rectangle is 40cm. If area is 96cm2, find the sides.

[12, 8]

2) If from a number, twice its reciprocal is subtracted we get 1. What

is the number?[2 or -1]

3) The sum of the squares of two consecutive odd positive integers is

290. Find them.

[11, 13]



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4) Divide 16 into two parts such that twice the square of the larger

part exceeds the square of the smaller part by 164.

[10, 6]

5) The speed of a boat in still water is 8 km/hr. It can go 15 km

upstream 22 km downstream in 5 hours. Find the speed of thestream.

[3 km/hr]

6) Find two consecutive numbers whose squares have the sum 85.

[6, 7]

7) Sum of 2 numbers is 12. If the sum of their squares is 90, find the

numbers.

[9, 3]

8)

If the sum of the squares of 2 consecutive even natural numbers is244. Find the numbers.

[10, 12]

9) Square of a number is 60 more than 7times the number. Find the

number.

[12 or -5]

10) The sum of squares of 2 consecutive odd numbers is 74. Which is

the smaller of the numbers?

[5 or 7]




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SSLC Mathematics

Unit – Trigonometry

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SSLC

Unit

Trigonometry


(Plus Two Maths & Science Teacher)



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TrigonometryPoints:

If the angles of a triangle are 60

0

,60

0

,60

0

; then its sides willbe in the ratio 1:1:1

If the angles of a triangle are 450,450,900; then its sides will

be in the ratio 1:1:√2

If the angles of a triangle are 300,600,900; then its sides will

be in the ratio 1:√3:2

In triangle ABC

Sin A=BC/AC

Cos A= AB/AC

Tan A=BC/AB The angle of elevation of an object viewed, is the angle

formed by the line of sight with the horizontal when it is above

the horizontal level

The angle of depression of an object viewed, is the angle

formed by the line of sight with the horizontal when it is below

the horizontal level



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Sample Questions

PROBLEM

1.

One angle of a right triangle is 30

0

and its hypotenuse is

4cm .W hat is its area?

Answer

Triangle side ratio is ratio 1:√3:2

Altitude = hypotenuse × √3/2= 4 × √3/2 = 3.46 cm

PROBLEM

2. On e angle of a triangle 60

0

and the length of its opposite s ide

is 4cm.W hat is its circumradius.

Answer

From figure,



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sin60 = 4/BD0.8660 = BC/BD

BD = 4.62 cm

Therefore radius = 2.31 cmPROBLEM

3. Tw o sides of a triangle are 7 and 6 centim eters and the angle

between them is 120

0

. Find length of third side?

Answer

In triangle ADCCD = AC sin60

= 6×0.8660 = 5.2 AD = AC cos60

= 6×0.5 = 3BD = BA + AD = 3+7=10

In triangle BDCBC2 = BD2+CD2

= 102+5.22 = 100+27.04 = 127.04BC = 11.3

Third side is 11.3cm

PROBLEM

4. In the figure, B = 90◦; a l so, AB = 10 cm and C = 60◦

a) W hat is the measure of A

b) W hat are the lengths of AC and BC



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Answer

(a) ∠ A=300

(b) sin 60 = AB/AC = 10/AC AC = 10/0.8660=11.55 cm

tan 60 = AB/BCBC = 10/1.73 =5.78 cm

PROBLEM

5.

In the figure, BAC = 90◦, AD=6cm, CD=9cm, ACD = x

a) W hat is tan x?

b) How much is BAD ?

c) W hat is the length of BD?

Answer

(a) tan x= AD/DC=6/9=2/3(b) x(c) tan x = BD/AD=BD/6

2/3=BD/6



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BD=(2/3)× 6=4 cm

PROBLEM

6.

In the figure, AQB is an arc of a circle centred at O. Also,

AOB = 120◦, AOQ = 60◦, PQ = 3 cm

W hat is the radius of the circle?

Answer

cos 60 = OP/AO=(r-3)/r0.5 r = r - 3r - 0.5 r = 3r = 3/0.5 = 6cm

PROBLEM

7.

The shadow of a tower s tanding on a level ground is found to

be 45 m longer when the sun’s altitude is 30° than when it

was 60°. Find the height of the tower.



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Answer In ∆ ABD,

AB/BD = tan 30

h /(45+x)=1/√3 x = (√3h - 45) …..(1)In ∆ABC,

AB/BC = tan 60h / x = √3 x = h/√3 …….(2)From equation (1) and (2), we get(√3h - 45) = h/√3 h = 38.97m

PROBLEM

8.

The length of shadow of a tower is 24 m, when the sun is at

an angle of elevation of 55

0

. Find height of tower.

Answer In the figure AB is tower.tan 50 = AB/BC

AB = CB tan 55 AB = 24 × 1.4281



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= 34.2744Height of the tower = 34.3 m

PROBLEM

9.

A 1.5 m tall boy is standing at some distance from a 3 0 m tall

building. The angle of elevation from his eyes to the top of

the building increases from 30° to 60° as he w alks towards the

building. Find the distance he walked toward s the building.

Answer

In figure PR = 28.5In ∆PAR,PR/AR = tan 3028.5/AR = 0.5773

AR = 49.3634In ∆PRB,PR/BR = tan 6028.5/BR = √3



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BR = 16.4545ST = AR – BR

= 49.3634 – 16. 3634

= 32.9089 mPROBLEM

10. Th e angle of elevation of the top of the building from the foot

of the tower is 30° an d the angle of elevation of the top o f the

tower from the foot of the building is 60°. If the tow er is 50

m h igh, find the height of the building

In ∆ABC, AB/BC=tan 60°BC=50/√3 …..(1)In ∆DCB,

DC/BC = tan 30h /BC = 1/√3h = BC/√3 h = 16.67m



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PR CTICE EXERCISE

1) In ∆ABC, B = 90, C = 70, BC = 16cm, find AC.[Sin70 = 0.94; Cos 70 = 0.34; tan70 = 2.75]

[47.05cm] 2) One angle of a triangle is 110 and the side opposite to it is 4cm

long. What is its circumradius?[2.13cm]

3) The angle between the radius and slant height of a cone is 600. Findthe radius of the cone, if its slant height is 14 cm

[7cm]

4) A man standing on the deck of a ship, which is 10 m above water

level. He observes the angle of elevation of the top of a hill as 60°and the angle of depression of the base of the hill as 30°.(a) Calculate the distance of the hill from the ship(b) Find the height of the hill.

[10√3m, 40m]

5) An observer in a lighthouse 100 m above the sea-level is watchingthe ship sailing towards the lighthouse. The angle of depression of

the ship from the observer is 30°. How far is the ship from thelighthouse?[100√3m]

6) A ladder is placed along a wall such that its upper end is touchingthe top of the wall. The foot of the ladder is 2m away from the walland the ladder is making an angle of 60° with the level ground.Find the height of the wall.

[3.46m]

7) The top of a tower is seen at an angle of elevation of 400 from a

point 30m away from the base of the tower. What is the height ofthe tower?[Sin 40 = 0.64; Cos 40 = 0.77; tan 40 = 0.84]

[25.20m]



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8) The angle of elevation of a tower at a point is 45°. After going 40m towards the foot of the tower, its angle of elevation becomes 60°.Find the height of the tower.

[94.64m]9)

Two men are on the opposite sides of a tower. They measure theangles of elevation of the top of the towers as 30° and 45°. If theheight of the tower is 60 m, find the distance between them.

[163.92 m]

10) From the top of a building 60m high the angles of depression ofthe top and bottom are observed to be 30º and 60º.Find the heightof the tower.

[40m]

11)

The horizontal distance between 2 towers is 70m.The angle ofdepression of the top of first tower when seen from the top ofsecond tower is 30º. If the height of the second tower is 120m,find the height of the first tower.

[79.6m]

12) An aero plane when 3000m high passes vertically above anotheraero plane at an instant when the angle of elevation of the two aeroplanes from the same point on the ground are 60 and 45

respectively. Find the vertical distance between the aero planes.[1268m]



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SSLC Mathematics

Unit – So l ids

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SSLC

Unit

Solids





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Solids

Points:Square Pyramid:

Lateral surface area = 4×½(Base edge × Slant height)

L.S.A 2 Total Surface Area = Base Area × Lateral surface area

T.S.A 2 Relations connecting base edge a, lateral edge e, slant height

l , height h and base diagonal d :

14

14

14

Volume = Base area × height

Volume

Cones

The radius of the sector becomes the slant height of the cone;the arc length of the sector becomes the base circumference of

the cone.

Suppose that a cone of base radius r and slant height l , radius

of the sector l and the central angle x ,then

360

L.S.A of Cone T.S.A of Cone

Volume



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Spheres

T.S.A 4

Volume =

Hemispheres

L.S.A 2

T.S.A 3

Volume =

Sample Questions

PROBLEM

1. A toy in the shape of a square pyramid has base edge 16cm

and slant height 10cm.

(a) Find lateral surface area.

(b) Find T otal surface area.

(c) Calculate its volume.

Answer

L.S.A = 2=2×16×10=320 cm2

T.S.A= 2 = 16×16 + 320=576 cm2

=100 – 64

6

Volume = = 256 × 6=1536 cm3

PROBLEM

2. H eight of a cone is 40cm . Slant height is 41cm .

(a)

Find diam eter of its base.

(b) Find Volume

41 40=81

9



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∴ Diameter = 18cm

Volume

9 40

= 3394.3 cm3 Answer

PROBLEM

3. Diam eter of a football is 30cm .

(a) What is the least area of leather required to make 50

such footballs?

(b) Also find volum e of air inside 50 such footballs

Answer

Diameter of a football is 30cm.Surface area of a football 4

4 15 900 The least area of leather required to make 50 footballs

50 900 45000 Volume of air inside 50 such footballs

= 50 50

15

= 225000 cm3

PROBLEM

4. A solid is in the form of a cylinder with hemispherical ends.

The total height of the solid is 19cm and diameter of the

cylinder is 7cm.

(a) Find volume.

(b) Total Surface Area of solid.

Answer

Volume of solid 2

3.5 12 3.5

641.67 cm3



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Total Surface Area of solid 24

2 3 . 5 1 2 4 3 . 5

= 418 cm2 PROBLEM

5.

A circus tent is made of canvas and is in the form of a right

circular cylinder and a right circular cone above it . The

diameter and height of the cylindrical part of the tent are

126m and 5m respectively. The total height of the tent is

21m . Find the total cost of the tent if the canvas used costs

Rs.12 per sq.m

Answer

Diameter and height of the cylindrical part of the tent are 126m and

5m.

Total height of the tent is 21m.

16 63

2563969 = 4225

6 5m

The total surface area 2 2 6 3 5 6 3 6 5

4725

The total cost of the tent 4725 12

56700 178200

PROBLEM

6. A cylindrical jar of radius 6cm co ntains oil . Iron spheres each

of radius 1.5cm are imm ersed in the oil . How many spheres

are necessary to raise the oil by 2cm?

Answer

Volume of cylinder with height 2cm = n

Volume of iron spheres

6 2 43 1.5

72 4.5



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724.5 16

Number of spheres are 16.

PROBLEM

7. A G ulab Jamun, contains sugar syrup up to about 30% of i ts

volume. Find approximate ly how much syrup w ould be found

is 45 gulab jamuns, each shaped like a cylinder with two

hemispherical ends with length 5 cm and diameter 2.8 cm.

Answer

Volume of one gulab jamun = volume of cylindrical part

+ 2 × (volume of hemispherical part)

2 23

1 . 42.2

1.4

Volume of such 45 gulab jamun

7.97

Volume of Syrup 30% Volume of such 45 gulab jamun

338 cm3



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PR CTICE EXERCISE

1) A solid is hemispherical at the bottom and conical above. If the

curved surface area of the two parts are equal, then from the ratio

of the radius and height of the conical part.[1: 3]

2) A toy is in form of a cone mounted on a hemi-sphere of radius 3.5

cm. The total height of the toy is 15.5 cm.

(a) Find slant height.

(b) Find its total surface area.

[12.5 cm, 214.5 cm

2

]

3)

A tent is of the shape of a right circular cylinder upto a height of 3metres and conical above it. The total height of the tent is 13.5

metres above the ground. Calculate the cost of painting the inner

side of the tent at the rate of Rs. 2 per square metre, if the radius

of the base is 14 metres.

[2068]

4) A solid sphere of radius 3cm is melted and then cast into small

spherical balls each of diameter 0.6cm.Find the number of smallballs thus obtained.

[1000]

5) A cylinder of radius 12 cm contains water to a depth of 20cm, a

spherical iron ball is dropped into the cylinder and thus the level

of water is raised to 6.75cm.Find the radius of the ball.

[9cm]

6) A vessel is in the form of a hollow hemisphere mounted by a hollow

cylinder. The diameter of the hemisphere is 14 cm and the total

height of the vessel is 13 cm. Find the inner surface area of the

vessel.

[572 cm

2

]



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7) A glass cylinder with diameter 20 cm has water to the height of 9

cm. A metal cube of 8 cm edge is immersed in it completely. Find

the height by which the water will rise in the cylinder.

[1.6 cm]

8) The diameters of the internal and external surfaces of a hollow

spherical shell are 6 cm and 10 cm respectively. If it is melted and

recast into a solid cylinder of diameter 14 cm, find the height of

the cylinder.

[8/3 cm]

9) If the number of square centimetres on the surface of a sphere is

equal to the number of cubic centimetres in its volume, what is thediameter of the sphere?

[12 cm]

10) The diameter of a metallic solid sphere is 12 cm. It is melted and

drawn into a wire having diameter of the cross-section 0.2 cm. Find

the length of the wire.

[1188 cm

2

]

11)

A semi-circular thin sheet of metal of diameter 28 cm, is bent tomake an open conical cap. Find the capacity of the cap.

[622.38 cm

3

]

12) An ice-cream cone has a hemispherical top. If the height of the

cone is 9 cm and base radius is 2.5 cm, find the volume of ice cream

cone.

[91.7 cm

3

]




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Unit – Coordinates

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SSLC

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Coordinates





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Points

The coordinates of a point are the distances of the point from

x-axis and y-axis.

The coordinates of a point on the x-axis are of the form ( x, 0)

and of a point on the y-axis are of the form (0, y).

The coordinates of the origin is (0, 0)

Sample Questions

PROBLEM

1.

Find the coordinates of the other three vertices of the

rectangle in the figure below.

Answer

T

The coordinates of rectangle are (0,0),(5,0),(5,4) and (0,4)



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PROBLEM

2.

A circle is drawn w ith centre at 0, 0) and radius 6 units in a

coordinate system.

a) What are the coordinates of the points at which it cuts

the x-axis?

b) And the points wh ere it cuts the y-axis?

Answer

(a) (6,0) and (-6,0)

(b) (0,6) and (0,-6)

Answer

PROBLEM

3.

From the points given below, find the pair wh ich are on a l ine

parallel to the x-axis and th e pair wh ich are on a l ine parallel

to the y-axis

A 4, 3), B 3, 5), C −6, 3), D 3,−2), E 5, 4)

Answer

Parallel to the x-axis A (4, 3) and C (−6, 3)

Parallel to the y-axis B (3, 5) and D (3,−2), PROBLEM

4.

W hat is the distance between the points −3, 2) and 4, 2)?

Answer

Distance =|-3-4|=7PROBLEM

5.

Th e coordinates of a point on a l ine parallel to the y-axis are

5, 2).

a)

W hat is the distance betwe en this l ine and the y-axis?

b) Find the coord inates of the point where this l ine meets

the x-axis.

c)

W hat is the distance between these two po ints?

Answer



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(a) 5(b) (5,0)

(c) Distance = |2-0|=2

PROBLEM

6.

On e end of a l ine of length 10 units is at the point -3, 2). If

the y coordinate of the other end is 10, then find the x

coordinate of the other end.

Answer

Let the x coordinate of other end be x.

Given, distance of A(-3,2) from B(x,10) is 10 units, i.e., AB =10.

AB2 = (-3-x)2 + (2-10)2

102 =9 + x 2 +6x + 64or

100 = x 2 + 6x + 73

x 2 + 6x - 27 = 0

x = -9 or 3

Thus, the x coordinate of the other end is -9 or 3.

PR CTICE EXERCISE

1) Find a relation between x and y such that the point (x, y) is

equidistant from the points A (7,1) and B(3,5).

[x - y = 2]

2) Find the point on y – axis which is equidistant from (-5, -2) and

(3, 2).

[-2, 0]

3) Find the point on y axis which is equidistant from (-5,2) and (3,2)

[(0,-2)]

4) Draw x and y axis then mark the following points.



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(4,3),(-4,7),(-4,-6),(5,9),(6,-4)

5) Draw x and y axis and mark the points P (-1, 6) and Q (6, 6) then

join PQ. Test whether the following points are on the line PQ

(3, 4), (-6, 6), (4, 6), (5, 6)[(-6, 6), (4, 6), (5, 6)]

6) Find the distance from x axis

(4, 4), (4, 3), (5, 7),(4,-3)

[4, 4, 5, 4]

7) Draw a circle passing through the points (0,-3),(0,-2),(2,3),(0,1)

8) Mark the following points without drawing axis

[(5, 7),(3,4)]




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10 Mathematics Problems Chapters 1 to 6

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