Question Paper Design & Guidelines for Formative Assessment · 2018-10-01 · (vi) In case of Multiple Choice, Fill in the blanks and Matching questions, scratching / rewriting/ marking

Question Paper Design & Guidelines for Formative Assessment(Issued By SSLC Examination 2015)

Dimension-1 : Weightage to Content : Mathematics

Sl. No. Unit Marks

1 Real Numbers 03

2 Sets 03

3* Progressions 08

4* Permutations and Combinations 05

5 Probability 03

6 Statistics 04

7 Surds 04

8* Polynomials 04

9* Quadratic Equations 09

10* Similar Triangles 06

11* Pythagoras Theorem 04

12* Trigonometry 06

13 Co-ordinate Geometry 04

14* Circle-chord Properties and Tangent Properties 10

15* Mensuration 05

16 Network and Polyhedral 02

Total 80

Dimension-2 : Weightage to Objectives

Sl. No. Objectives % Marks

1 Remembering 10%

2 Understanding 55%

3 Applying (Including Analysis) 20%

4 Skill 15%

Total 100%

ii| OSWAAL SSLC Karnataka Question Bank MATHEMATICS, Class – 10

ObjectivesMCQs1 Mark

1 MarkQuestions

S.A.2 Marks

L.A.3 Marks

L.A.4 Marks

Total Marks

Percentage

Remembering 1 × 2 = 2 1 × 4 = 4 2 × 1 = 2 – – 08 10%

Understanding 1 × 6 = 6 1 × 2 = 2 2 × 10 = 20 3 × 4 = 12 4 × 1 = 4 44 55%

Applying (Including Analysis)

– – 2 × 3 = 6 – 4 × 1 = 4 16 20%

Skill – – 2 × 2 = 4 3 × 2 = 6 4 × 2 = 8 12 15%

Total 1 × 8 = 8 1 × 6 = 6 2 × 16 = 32 3 × 6 = 18 4 × 4 = 16 80 100%

Sl. No. Type of Questions No. of Questions Marks

1 M.C.Q. 08 08

2 Short Answer Type (1 Mark) 06 06

3 Short Answer Type (2 Marks) 16 32

4 Long Answer Type (3 Marks) 06 18

5 Long Anwer Type (4 Marks) 04 16

Total 40 80

Eassy 30%

Average 50%

Dif cult 20%

SYLLABUS | iii

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iv| OSWAAL SSLC Karnataka Question Bank MATHEMATICS, Class – 10

Key : * Indicates Internal Choice Question Unit

Note : (i) Numbers outside the bracket indicates Marks.

(ii) Numbers inside the bracket indicates Questions.

(iii) Internal choice in Questions to be given in the following Units, which are comparitively having more contents for 2, 3

and 4 Marks. The Units are 3, 4, 8, 9, 10, 11, 12, 15 and 16.

(iv) In case of Questions on proving theorems, the Choice Questions can be the converse of the theorems OR

Corollary having equal weightage in marks.

(v) In case of Questions on Riders based on theorems, Choice Questions to be the Riders based on the same

theorem OR Converse or Corollary.

CCE RFSOLVED PAPER, APRIL 2015SUBJECT-MATHEMATICS

Code No. : 81–E Time : 9-30 A.M. to 12-30 P.M.Date : 06.04.2015 Max. Marks : 80

General Instructions :(i) The Question-cum-Answer Booklet consists of objective and subjective types of questions having 40 questions.

(ii) Space has been provided against each objective type question. You have to choose the correct choice and write the complete answer along with its letter in the space provided.

(iii) For subjective type questions enough space for each question has been provided. You have to answer the questions in the space.

(iv) Follow the instructions given against both the objective and subjective types of questions.(v) Candidates should not write the answer with pencil. Answers written in pencil will not be evaluated. (Except Graphs,

Diagrams & Maps)(vi) In case of Multiple Choice, Fill in the blanks and Matching questions, scratching / rewriting/ marking is not permitted,

thereby rendering to disqualification for evaluation.(vii) Candidates have extra 15 minutes for reading the question paper.

(viii) Space for Rough Work has been printed and provided at the bottom of each page.(ix) Do not write anything in the space provided in the right side margin.

I. Four alternatives are given for each of the following questions / incomplete statements. Only one of them is correct or most appropriate. Choose the correct alternative and write the complete answer along with its letter in the space provided against each question. 8 × 1 = 8

1. The distance between the points (2, 3) and (6, 6) is :(A) 5 units (B) 7 units(C) 3 units (D) 4 units.

2. The value of nCn is(A) n (B) 0(C) 1 (D) n !

3. The given pie-chart shows annual agricultural yield of certain place. If the total production is 8100 tons, then the yield of Ragi in tons is :

Ragi

100°

80° 140°

40°

OthersRice

Sug

arca

ne

(A) 225 (B) 2250(C) 22.5 (D) 2.250.

4. The value of tan2 60° is :

(A) 3 (B) 2 3

(C)13

(D) 3.

5. The slope of the line joining the points (3, –2) and (4, 5) is :

(A)17

(B) 1

(C) 7 (D)37

.

vi | OSWAAL SSLC Karnataka Question Bank MATHEMATICS, Class-10

6. The harmonic mean between 2 and 4 is :

(A)83

(B)38

(C)86

(D)6

.8

7. The probability of winning a game is 5

.6

Then the probability of losing it is :

(A)56

− (B)56

(C)16

− (D)1

.6

8. Which of the following is the zeroes of the polynomial x2 + 4x + 4 ?(A) 2 (B) –2(C) 4 (D) –4.

II. Answer the following : 6 × 1 = 69. In the figure, BC is the diameter. What is the measure of x ?

A

B CO

x

10. If A = {1, 2, 3, 4} and B = {1}, find A / B.11. If f(x) = x2 –4, find f(4).12. Express 210 as the product of prime factors.13. Mean and standard deviation of runs scored by a cricket player are 80 and 4 respectively. Find coefficient of

variation.

14. In D ABC, XY || BC, 12

AYCY

= and AX = 4. Find BX.

A

X Y

B C

III.15. Two dice are thrown simultaneously. Find the probability of getting

(a) same number on both faces and(b) both faces having multiples of five. 2

16. Find the product of 2 and 3 5. 317. Find the value of a if (x – 5) is a factor of (x3 – 3x2 + ax – 10).

ORWhat must be added to the polynomial

P(x) = x4 + 2x3 – 2x2 + x – 1so that the resulting polynomial is exactly divisible by x2 + 2x – 3 ? 2

SOLVED PAPER-2015 | vii

18. In D ABC, AD is the median and PQ || BC. Prove that PE = EQ. 2A

QP

D CB

E

19. Find the perimeter of a triangle whose vertices have the coordinates (3, 10), (5, 2) and (14, 12). 2

20. Draw a circle of radius 3 cm and construct two tangents to it from an external point 8 cm away from the centre.2

21. In how many ways 4-digit numbers can be formed using the digits 1, 2, 3, 7, 8, 9 without repetitions ? How many of these are even numbers ? 2

22. Find the number of sides of a polygon having 35 diagonals. 2

23. Calculate variance for the following data : 2

X = 2, 4, 6, 8, 10

(Scores)

24. Rationalise the denominator and simplify :

6 3.

6 3+−

2

25. Solve the equation by using formula : x2 – 4x + 2 = 0. 2

26. In the given figure, find the values of cos q and tan q. 2A

5 13

B C�

27. Draw a plan of a level ground using the given information : 2

[Scale : 20 m = 1 cm]

Metre To D

To E 60

1501008040

80 to C

40 to B

From A

28. Verify Euler's formula for a Hexahedron. 2

29. Prove that 3 + 5 is an irrational number. 2

30. In a class of 100 students, 55 students have passed in Physics and 67 students have passed in Mathematics. Find the number of students passed in Physics only.

OR

A and B are the two subsets of Universal Set such that n( ) = 700, n(A) = 200, n(B) = 300 and n(A B)

= 100. Find n(A' B'). 2

viii | OSWAAL SSLC Karnataka Question Bank MATHEMATICS, Class-10

IV.

31. In a harmonic progression, T3 = 17

and T7 =1

.5

Find T15. 3

32. In DABC, ∠A = 90°, AD ⊥ BC and ∠B = 45°. If AB = x, find AD in terms of x.OR

ABC is a right angled triangle with ∠C = 90°, BC = a, AC = b, CD ⊥ AB and CD = P. Show that

21

P= 2 2

1 1.

a b+ 3

33. Prove that 'the tangents drawn from an external point to a circle are equal'. 334. Aniruddha bought some books for Rs. 60. Had he bought 5 more books for the same amount, each book would

have cost him 1 rupee less. Find the number of books bought by Aniruddha.OR

The ages of Kavya and Karthik are 11 years and 14 years respectively. In how many years time will the product of their ages will be 304 ? 3

35. Prove thatsec tansec tan

q − qq + q

= 1 – 2 sec q. tan q + 2 tan2 q.

ORProve that

+ q− q

1 cos1 cos

= cosec q + cot q. 3

36. A solid cylinder has a total surface area of 462 cm2. Its curved surface area is one-third of its total surface area. Find the radius of the cylinder.

ORFrom the top of a cone of base radius 12 cm and height 20 cm, a small cone of base radius 3 cm is to be cut off. How far down the vertex is the cuts should be made ? Find the volume of the frustum so obtained. 3

V. 37. The sum of three consecutive terms in an arithmetic progression is 6 and their product is – 120. Find the three

numbers.OR

The product of three consecutive terms in a geometric progression is 216 and sum of their products in pairs is 156. Find the three terms. 4

38. 'In a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.' Prove. 4

39. Construct direct common tangents to two circles of radii 4 cm and 2 cm whose centres are 9 cm apart from their centres. Measure their length and write the length of direct common tangents. 4

40. Solve the equation graphically :x2 + x – 6 = 0 4

SOLVED PAPER-2015 | ix

SSLC Examination March/April-2015Model Answers

(Issued by KSEEB)

Date : 06.04.2015 Code No. : 81–E Max. Marks : 80

Q.No. Answer Key Value Point Marks Alloted

I. 1. (A) 5 units 1 2. (C) 1 1 3. (B) 2250 1 4. (D) 3 1 5. (C) 7 1

6. (A)

83

1

7. (D)

16

1

8. (B) –2 1II. 9. x = 90° 1 10. A / B = {2, 3, 4} 1 11. f(x) = x2 – 4 f(4) = 42 – 4 ½ = 16 – 4 \ f(4) = 12 ½ 12. 210 = 2 3 5 7× × ×

2 210

3 105

5 35

7 ½ + ½ = 1

13. CV = – 100x

σ ×

½

=

4100

80×

\ CV = 5 ½

14.

AXBX

=

AYCY

½

4BX

=

12

\ BX = 8 ½ 15. S = {(1, 1), (1, 2), .... .... (1, 6) (6, 1), (6, 2), .... .... (6, 6)} \ n(S) = 36. ½ (a) Same number on both faces : Let A be the event. A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)} n(A) = 6

P(A) =

( )( )

n An S

P(A) =

1.

36 ½

x | OSWAAL SSLC Karnataka Question Bank MATHEMATICS, Class-10

(b) Both faces having multiples of five :Let B be the event. B = {(5, 5)}\ n(B) = 1 ½

P(B) = ( )( )

n Bn S

\ P(B) = 1

.36

½

Let B be the event. B = {(5, 5)}

16. L.C.M. of orders = 6 2 =1 3

62 62 2 8= = ½3 5 =

1 263 65 5 25= = ½

\ 32 5× = 6 68 25× ½

= 6 200. ½17. Let P(x) = x3 – 3x2 + ax – 10 ½

By factor theorem,(x – 5) is a factor of P(x) if P(5) = 0 ½

P(5) = (5)3 –3(5)2 + a(5) – 100 = 125 – 75 + 5a – 100 = 5a + 40 ½

5a = – 40a = –8. ½

ORBy division algorithm for polynomials, P(x) = [g(x). q(x)] + r(x)

P(x) – r(x) = g(x). q(x)P(x) + {– r(x)} = g(x). q(x)

x2 + 1 ½x2 + 2x – 3 )x4 + 2x3 – 2x2 + x – 1

x4 + 2x3 – 3x2

– – + ½x2 + x – 1x2 + 2x – 3 ½

– – + – x + 2

\ r(x) = – x + 2 ⇒ {– r(x)} = x – 2 ½Hence, we should add (x – 2) to P(x) so that the resulting polynomial is exactly divisible by g(x).

18.

A

QP

D CB

E

To prove : PE = EQProof : In D ABD,

PE || BD Q PQ || BC

\APAB

= =AQ PEAC BD

...(i) (Q Thale's theorem) ½

Similarly In D ADC,AEAD

=AQ EQAC DC

= ...(ii) ½

From (i) and (ii)PEBD

=EQDC

½

But, BD = DC (Q AD is the median) ½\ PE = EQ.

SOLVED PAPER-2015 | xi

19. Let A = (x1, y1) = (3, 10)B = (x2, y2) = (5, 2)C = (x3, y3) = (14, 12)

AB =2 2

2 1 2 1( ) ( )x x y y− + −

= 2 2(5 3) (2 10)− + −

AB = 4 64 68 units.+ = ½

Similarly BC = 2 2(14 5) (12 2)− + − = 81 100+ = 181 units½

AC = 2 2(14 3) (12 10)− + − = 121 4+ = 125 units½

Perimeter = AB + BC + AC P = ( )68 181 125 units.+ + ½

20. r = 3 cm d = 8 cm.

A

O x P

B

PA and PB are tangents.Drawing circle of radius 3 cm ½Bisecting OP ½Drawing tangents PA, PB. 1

21. Given digits : 1, 2, 3, 7, 8, 9(a) 4-digit number can be formed in 6P4 ways ½

6P4 = 6 × 5 × 4 × 36P4 = 360. ½

(b) Even numbers :

T H Ten U3P1

4P15P1

2P1

Units place can be filled in 2P1 waysTens place can be filled in 5P1 waysHundreds place can be filled in 4P1 waysThousands place can be filled in 3P1 waysTotal number of ways = 2P1 × 5P1 × 4P1 × 3P1 ½

= 2 × 5 × 4 × 3 = 120. ½22. D = 35, n = ?

D = nC2 – n

35 = 2

2!

n P– n ½

35 =( 1)

2n n −

– n

35 =2

2n n−

– n ½

35 =2 3

2n n−

n2 – 3n – 70 = 0 ½

xii | OSWAAL SSLC Karnataka Question Bank MATHEMATICS, Class-10

n2 – 10n + 7n – 70 = 0n(n – 10) + 7(n – 10) = 0

n = 10 or n = – 7 ½Neglecting n = – 7

n = 10⇒ Number of sides = 10.

23. X = 2, 4, 6, 8, 10

X D = X – –x D2

2 –4 16

4 –2 4

6 0 0

8 2 4

10 4 16

n = 5 SD2 = 40 ½–x =

306

5Xn

S= = \

–x = 6 ½

Variance = s2 = 2D

nS

½

=405

Variance = s2 = 8. ½

24.6 3 6 36 3 6 3

+ +×

− +(Multiplying by RF) ½

=( )2

6 3

6 3

+

−=

6 3 2 183

+ +½

=+9 6 2

3=

( )3 3 2 2

3

+

½

= 3 +2 2 . ½25. x2 – 4x + 2 = 0

a = 1b = –4 ½c = 2

x =2 4

2b b ac

a− ± − =

( 4) 16 4(1)(2)2(1)

− − ± −½

=4 16 8

2± −

=4 8

2±

½

=4 2 2

2±

=( )2 2 2

2

±

x = 2± 2. ½

\ Roots are 2 + 2 or 2 – 2.

SOLVED PAPER-2015 | xiii

26. A

5 13

B C�

In D ABC,∠B = 90°

\ AC2 = AB2 + BC2

132 = 52 + BC2 (Pythagoras theorem) ½BC2 = 169 – 25 = 144BC = 12

cos q =Adjacent sideHypotenuse

BCAC

= ½

\ cos q =1213

½

tan q = Opposite sideAdjacent side

ABBC

=

\ tan q =5

.12

½

27. Scale : 20 m = 1 cm40 m = 2 cm60 m = 3 cm80 m = 4 cm

100 m = 5 cm150 m = 7.5 cm

D

C

B

A

E60

80

40

Calculation ½ Field drawing 1½28. F = 6

V = 8E = 12 ½

F + V = E + 2 ½6 + 8 = 12 + 2 ½

14 = 14. ½

29. If possible, let us assume 3 + 5 is a rational number.

3 + 5 =pq

, where p, q ∈ z, q ≠ 0 ½

xiv | OSWAAL SSLC Karnataka Question Bank MATHEMATICS, Class-10

3 –pq

= – 5

3q pq−

– 5 ½

⇒ – 5 is a rational number.

Q3q p

q−

is a rational number. ½

But, – 5 is not a rational number

\ Our supposition 3 + 5 is a rational number is wrong. ½

⇒ 3 + 5 is an irrational number.30. Number of students who have passed in Physics, n(P) = 55.

Number of students who have passed in Mathematics, n(M) = 67.Number of students who have passed in both subjects, n(P ∩ M) = ?Number of students in the classroom, n(P ∪ M) = 100 ½

n(P) + n(M) = n(P ∪ M) + n(P ∩ M) ½55 + 67 = 100 + n(P ∩ M)

n(P ∩ M) = 122 – 100n(P ∩ M) = 22 ½

Number of students who have passed in Physics only= n(P) –n(P ∩ M)= 55 – 22= 33. ½

ORn(U) = 700n(A) = 200n(B) = 300

n(A ∩ B) = 100n(A) + n(B) = n(A ∪ B) + n(A ∩ B) ½

200 + 300 = n(A ∪ B) + 100500 – 100 = n(A ∪ B)n(A ∪ B) = 400 ½

n(A ∪ B)’ = n(A’ ∩ B’)= n[U \ A ∪ B] ½= n(U) \ n(A ∪ B)= 700 – 400

n(A ∪ B)’ = 300 ½or n(A’ ∩ B’) = 300

31. T3 = 71 1

,7 5

T =

\ T3 and T7 of A.P. are 7, 5

d =p qT T

p q

−

−

= 7 3

7 3T T−

−=

5 74−

=2

4−

= – 12

1

\ d = – 12

T3 = 7a + 2d = 7

a + 2 12

−

= 7 a – 1 = 7 \ a = 8 1

Tn = a + (n – 1)d

SOLVED PAPER-2015 | xv

T15 = 8 + ×

114 –

2½

= 8 – 7T15 = 1

\ 15th term of HP = 1 ½Note : For alternate method full marks may be given.

32. C

D

BA45°

x

x ½

In DABC,∠A = 90°, AB = x, ∠B = 45° ⇒ ∠C = 45° ⇒ AB = AC = x

½

BD = CD = 2

BC

BC2 = AC2 + AB2 ..... Pythagoras theorem= x2 + x2

BC2 = 2x2 ½

\ BC = x 2AD2 = CD. BD

= 12

BC.12

BC =2

2BC

=2

22

x

1

=2.24

x

\ AD2 =2

2x

\ AD = 2

x½

ORA

D

BC a

b

P

To prove : 21

P= 2 2

1 1a b

+ ½

Proof : In DABC,CD2 = AD . BD

P2 = AD . BD ...(i) ½

xvi | OSWAAL SSLC Karnataka Question Bank MATHEMATICS, Class-10

CB2 = AB . AD \

a2 = AB . AD \ 21a

=1.AB BD ...(ii) ½

AC2 = AB . AD

b2 = AB . AD \ 21b

=1.AB AD

...(iii) ½

Adding (ii) and (iii)

2 21 1a b

+ = +1 1. .AB BD AB AD =

1 1 1AB BD AD

+

1

= +

1.

AD BDAB BD AD = 2

1.

ABAB P

[from (i)]

\ 2 21 1a b

+ =2

1P

33. P

Ox

B

Q

½

Data : O is the centre of the circle ½B is the external pointBP and BQ are the tangents.To prove : BP = BQ ½

Proof : In D BOP, B P O = 90°Similarly in DBOQ, BQO = 90°(Q radius, tangents at the point of contact are perpendicular)In 2 right angled triangles BOP and BOQHypotenuse BO = Hypotenuse BO.. Common side

OP = OQ ...radii\ DBOP ≅ DBOQ\ BP = BQ. 1½

34. Let the number of books be xTotal cost = Rs. 60

\ Cost of each book = Rs. 60x

½

If the number of books = (x + 5)

Then cost of each book = Rs.60

( 5)x + ½

But,60 60

.( 5)x x

−+

= 1 ½

60( 5) 60( 5)

x xx x

+ −+

= 1

260 300 60

5x x

x x+ −

+= 1

x2 + 5x = 300 ½x2 + 5x – 300 = 0

x2 + 20x – 15x – 300 = 0x(x + 20) –15(x + 20) = 0 ½

x = – 20 or 15

SOLVED PAPER-2015 | xvii

Neglecting x = –20 \ x = 15 ½⇒ No. of books purchased by Aniruddha = 15

ORLet the number of years be x\ Kavya’s and Karthik’s ages are (11 + x) and (14 + x) ½Product of their ages = 304

(11 + x) (14 + x) = 304 ½154 + 11x + 14x + x2 – 304 = 0 ½

x2 + 25x – 150 = 0 ½x2 + 30 x – 5x – 150 = 0

x(x + 30) –5(x + 30) = 0 ½x = 5 or –30 ½

Neglecting x = –30x = 5. ½

i.e. After 5 years, product of their ages will be 304.

35. To prove :sec tansec tan

q − qq + q

= 1 – 2sec q. tan q + 2tan2 q

Proof : LHS = sec tansec tan

q − qq + q

=sec tansec tan

q − qq + q

×sec tansec tan

q − qq − q

½

=2

2 2(sec tan )sec tan

q − qq − q

½

=2 2sec tan 2sec .tan

1q + q − q q

(Q sec2 q – tan2 q = 1)

= sec2 q + tan2 q – 2 sec q. tan q 1= 1 + tan2 q + tan2 q – 2 sec q. tan q (Q sec2 q = 1 + tan2 q)= 1 – 2 sec q. tan q + 2 tan2 q ½ 2= RHS.

OR

LHS = 1 cos1 cos

+ q− q

Multiplying by RF =1 cos 1 cos1 cos 1 cos

+ q + q×

− q + q½

= 2

2(1 cos )1 cos

+ q− q

½

=2

2(1 cos )

sin+ q

q½

=1 cos

sin+ q

q½

=1 cos

sin sinq

+q q

½

= cosec q + cot q = RHS ½

36. CSA of cylinder =13

TSA of cylinder ½

2prh =13

. 2pr (r + h)

=13

× 462

2prh = 154 ½

xviii | OSWAAL SSLC Karnataka Question Bank MATHEMATICS, Class-10

TSA of cylinder = 2pr (r + h) ½462 = 2pr2 + 2prh462 = 2pr2 + 154

462 – 154 = 2pr2

2pr2 = 308 ½

r2 =308

2227

×

=308 7

44×

½

r2 = 49 ½r = 7 cm.

ORr1 = 12 cm, h1 = 20 cm, r2 = 3 cm, h2 = ?We know,

1

2

rr = 1

2

hh ½

123

= 2

20h

12 h2 = 60 \ h2 = 5 cm 1

Volume of frustum = 13

ph (r12 + r2

2 + r1 r2) ½

=1 223 7

× × 15 (122 + 32+ 12 × 3)

=1107

(114 + 9 + 36) ½

=1107

× 189

Volume of frustum = 2970 cubic cm. ½ 37. Let the three consecutive terms of the A.P. be (a – d), a, (a + d)

Sum = 6 ½

a – d + a + a + d = 63a = 6

\ a = 2 1Product = – 120

(a – d). a (a + d) = – 120 ½(a2 – d2)a = – 120(22 – d2)2 = – 120

4 – d2 = – 60 ½–d2 = – 64

d2 = 64 \ d = ± 8 ½If a = 2, d = 8, then the three numbers are –6, 2, 10If a = 2, d = – 8, then the three numbers are 10, 2, – 6 1

OR

Let the three consecutive terms bear

, a, ar

Product = 216ar

.a. ar = 216

a3 = 216 \ a = 6 1

SOLVED PAPER-2015 | xix

Sum of their products in pairs = 156

.a

ar

+ (a. ar) + .a

arr

= 156 ½

2ar

+ a2r + a2 = 156

36r

+ 36r + 36 = 156 ½

36r

+ 36r = 120

236 36rr

+= 120

36r2 – 120r + 36 = 03r2 – 10r + 3 = 0

r = 3 or13

½ + ½

(a) If a = 6, r = 3, the 3 consecutive numbers are 2, 6, 18 ½

(b) If a = 6, r = 13

, the 3 consecutive numbers are 18, 6, 2 ½

38.A

D

CB ½

Data : In DABC, B = 90° ½To prove : AC2 = AB2 + BC2 ½Constn. : Draw BD ⊥ AC ½Proof : In 2 triangles ABC, ADB

A B C = 90°, A D B = 90°

B A D .... common angle\ DABC ~ DADB

⇒ABAD

=ACAB

AB2 = AC. AD. ...(i) ½In 2 triangles ABC and BDC

A B C = B D C = 90°

A C B is common\ DABC ~ DBDC

⇒BCDC

=ACBC

\ BC2 = AC. DC ..(ii) ½

Adding (i) and (ii)AB2+ BC2 = (AC. AD) + (AC. DC)

= AC(AD + DC)AB2 + BC2 = AC2 or AC2

= AB2 + BC2 1

cont inue

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z #

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