Set - A

MT EDUCARE LTD.SUMMATIVE ASSESSMENT - 1

2013-14 CBSE - X Set - A

• Please check that this question paper contains 6 printed pages.

• Code number given on the right hand side of the question paper should be

written on the title page of the answer-book by the candidate.

• Please check that this question paper contains 34 questions.

• Please write down the serial number of the question before attempting it.

MATHEMATICS

General Instructions:

i) All questions are compulsory.

ii) The question paper consists of 34 questions divided in four sections: A,B,C and D.

Section A comprise 10 questions of 1 mark each,

Section B comprise 8 questions of 2 marks each,

Section C comprise 10 questions of 3 marks each, and

Section D comprise 6 questions of 4 marks each.

iii) Question numbers 1 to 10 in Section A are multiple choice questions where you

have to select one correct option out of the given four.

iv) There is no overall choice. However, internal choice has been provided in 1

question of two marks, 3 questions of three marks each and 2 questions of four

marks each. You have to attempt only of the alternative in all such questions.

v) Use of calculator is not permitted.

Time allowed : 3 hours Maximum Marks : 80

Series RLH

Roll No. Code No. 31/1

Set - A

SECTION - AQuestion number 1 to 10 carry 1 marks each.

1. If a pair of linear equations in two variables is consistent, then the linesrespresented by two equations are(a) intersecting (b) parallel(c) always coincident (d) intersecting or coincident

2. In fig. the value of x for which DE || AB is

(a) 4 (b) 1(c) 3 (d) 2

3. If tan2 45º – cos230º = x sin 45º cos 45º , then x =

(a) 2 (b) –2 (c) –1

2(d)

1

2

4. If two positive integers a and b are expressible in the form a = pq² and b = p³q;p, q being prime numbers, then LCM (a,b) is

(a) pq (b) p³q³ (c) p³q² (d) p²q²

5. If the mean of a frequency distributio is 8.1 and fixi = 132 + 5k, fi = 20,

then k =

(a) 3 (b) 4 (c) 5 (d) 6

6. If are the zeros of the polynomial p(x) = 4x² + 3x + 7, then 1 1

is equal to

(a)7

3(b) –

7

3(c)

3

7(d) –

3

7

7. 9 sec²A – 9tan²A is equal to(a) 1 (b) 9 (c) 8 (d) 0

8. Which of the following rational numbers have terminating decimal ?

(a)16

225(b)

5

18(c)

2

21(d)

7

250

... 2 ...

A

E

C

D

B

xx+3

3x+19 3x+4

Set - A... 3 ...

9. If x + 2 is a factor of x2 + ax + 2b and a + b = 4, then(a) a = 1, b = 3 (b) a = 3, b = 1 (c) a = -1, b = 5 (d) a = 5, b = -1

10. If n is a natural number, then 92n - 42n is always divisible by

(a) 5 (b) 13

(c) both 5 and 13 (d)None of these

SECTION - BQuestion number 11 to 18 carry 2 marks each.

11. For the following grouped frequency distribution find the mode :

Class 3-6 6-9 9-12 12-15 15-18 18-21 21-24Frequency 2 5 10 23 21 12 3

12. Check whether x³ – 3x + 1 is a factor of x5 – 4x³ + x² + 3x + 1.

13. Solve the each of the following system of equations by using the method of cross-multiplication: x +y = 7 , 5x + 12y – 7 = 0

14. In a right triangle ABC, right angled at C, if tan A = 1, then verify that2sinA cos A = 1.

Or14. Given :

ABC & AMP are two right triangles,

right angled at B & M respectively.

Prove that :

i) ABC ~AMP

ii)CA

PA =

BC

MP

15. Given that HCF (306, 657) = 9, find LCM (306, 657)

16.1

1 sin +

1

1 sin = 2sec²

M

BA

C

P

Set - A... 4 ...

17. Given 15 cot A = 8, find sin A and sec A

18. Solve the following pair of linear equations by the substitution method.0.2x + 0.3y = 1.3 ; 0.4 x + 0.5 y = 2.3

SECTION - C

Question numbers 19 to 28 carry 3 marks each.19. The sum of the digits of a two digit number is 8 and the difference between the

number and that formed by reversing the digits is 18. Find the number.

20. If are the zeros of the polynomial f(x) = 2x² + 5x + k satisfying the relation

² + ² + = 21

4, then find the value of k for this to be possible.

21. Find the values of and for which the following system of linear equationshas infinite number of solutions:2x + 3y = 72x + () y = 28

OR21. Find the zeroes of the following quadratic polynomials and verify t2 – 15.

22. Evaluate the following :

0 2

2 0 2 0

5cos² 60 + 4sec 30º – tan² 45ºsin 30 + cos 30

23. Given : LM || CB

LN || CD

Prove that : AM AN=

AB AD

24. Prove the following identities where the angles involved are acute angles forwhich the expressions are defined.

(sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A

OR

24. In OPQ right angled at P,OP = 7cm, OQ – PQ = 1cm. Determine the values of

sin Q and cosQ.

D

CA

N

L

B

M

Set - A

25. In fig. if POS ROQ, prove that PS || QR.

26. Prove that 3 is an irrational number.

27. Find the LCM and HCF of the following pairs 336 and 54.

28. Solve the following system of linear equations graphically:

x – y = 12x + y = 8Shade the area bounded by these two lines and y-axis.

ORProve the following :

cosA sinA + = cos A + sinA

1- tanA 1- cotA

SECTION - DQuestion numbers 29 to 34 carry 3 marks each.

29. Use Euclid’s division lemma to show that the cube of any positive integeris of the form 9m, 9m + 1 or 9m + 8 for some integer m.

ORUse Euclid’s division lemma to show that the square of any positive integereither of the form 3m or 3m + 1 for some integer m.

30. In fig., DE || BC and AD : DB = 5 : 4 Find

Area( DEF)

Area( CFB)

31. During the medical check-up of 35 students of a class, their weights wererecorded as, follows :

Weight (in kg) Number of students

Less than 38 0Less than 40 3Less than 42 5Less than 44 9Less than 46 14Less than 48 28Less than 50 32Less than 52 35

... 5 ...

P

R

Q

S

3

2

4

56

1

O

Set - A... 6 ...

Draw a less than type ogive for the given data. Hence obtain the median weight

from the graph and verify the result by using the formula.

32. Points A and B are 90 km apart from each other on a highway. A car starts from

A and another from B at the same time. If they go in the same direction they

meet in 9hours and if they go in opposite directions they meet in 9/7hours.

Find their speeds.

33. If a line is drawn parallel to one side of a triangle, to intersect the other two

sides in distinct points, then other two sides are divided in the same ratio.

ORThe ratio of the areas of two similar triangles is equal to the ratio of the squares

of their corresponding sides.

34. Prove the following cos A – sin A +1

cos A + sin A –1 = cosec A + cot A using the identity

cosec² A = 1 + cot² A.

All the Best

Set - A

Any method mathematically correct should be given full credit of marks.SECTION - A

1. (d) intersecting or coincident2. (a) 4

3. (d)1

2

4. (c) p³q²

5. (d) 6

6. (d) –3

7

7. (b) 9

8. (d)7

250

9. b) a = 3, b = 1

10. c) both 5 and 13

SECTION - B11. We observe that the class 12 - 15 has maximum frequency. Therefore, this

the modal classWe have,

l = 12 ,h = 3 , f = 23, f1 = 10 and f

2 = 21

Mode = l + 1 0

1 0 2

f f

2f f f × h

Mode = 12 +

23 10

46 10 21 × 3

Mode = 12 + 13

15× 3 = 12 +

13

5 = 14.6

MT EDUCARE LTD.CBSE X

Date :

SUBJECT : MATHEMATICS

SUMMATIVE ASSESSMENT - 1

MODEL ANSWER PAPER

Set - A

Marks : 80

Time : 3 hrs.

Set - A... 2 ...

12. x² – 1

x x x x x x x3 5 4 3 2– 3 + 1 – 0 – 4 + + 3 + 1

x5 – 3x³ + x² – + –

– x³ + 3x + 1

– x³ + 3x – 1+ – +

2

13. The given system of equation isx + y = 75x + 12y – 7 = 0By cross- multiplication, we get

x1 7

12 7 = =

y1 7

12 7 =

11 1

5 12

1 7 12 7

x =

1 7 5 7

-y =

1

1 12 5 1

7 84

x =

7 35

-y =

1

12 5

77

x =

28

-y =

1

7

x = 77

7 and y =

28

7 x = 11 and y = –4

Hence, the solution of the given system of equations is x = 11, y = –4.

14. In ABC, we havetan A = 1

BC

AC = 1

BC = x and AC = xBy Pythagoras therorem, we have

AB² = AC² + BC² AB² = x² + x²

AB = 2x

B

x

CxA

Set - A... 3 ...

sin A = BC

AC =

x

2x =

1

2 and cosA =

AC

AB =

x

2x =

1

2

2 sin A cos A = 2 × 1

2 ×

1

2 = 1

OR14. Proof :

In ABC & AMP

BAC = MAP ..... (common angle)

ABC = AMP ..... (each is 900)

ABC ~ AMP ..... (by AA similarity)

CAPA

=BCMP

..... (corresponding sides of similar triangles)

15. HCF (306, 657) = 9 Now HCF (306, 657) × LCM (306, 657) = 306 × 657

LCM (306, 657) =9

657306

=9

201042

= 22338

16.1

1 sin +

1

1 sin = 2sec²

We have,

LHS =1

1 sin +

1

1 sin

LHS =1 sin 1 sin

(1 sin )(1 sin )

LHS = 2

2

1 sin

LHS = 2

2

cos [1 – sin² = cos²]

LHS = 2sec² = RHS1

seccos

M

BA

C

P

Set - A... 4 ...

17. Given : 15 cot A = 8

cot A =8

15.....(i)

cot A =AB

BC.....(ii)

8

15=

AB

BC.....(iii) ... [From (i) and (ii)]

Let AB = 8k, BC = 15k ... [where k is positive number]

In right ABC, using Pythagoras theorem,

(AC)2 = (AB)2 + (BC)2

(AC)2 = (8k)2 + (15k)2

(AC)2 = 64k2 + 225k2

AC2 = 289k²

AC = 17k

sin A = BC

AC=

15

17

k

k

sin A =15

17

sec A = AC

AB=

17

8

k

k

sec A =178

18. x = 1.3 0.3

0.2

y... (i)

0.4 x + 0.5 y = 2.3 ... (ii)

Substituting eqn (i) in eqn (ii) Substituting y = 3 in eqn (i)

0.4 1.3 0.3

0.2

y + 0.5y = 2.3 x =

0.2

0.3(3)1.3

2.6 – 0.6y + 0.5y = 2.3 = 0.2

4.0

0.3 = 0.1y x = 2 y = 3

Solution is x = 2 , y = 3

A

CB

8k

15k

Set - A... 5 ...

SECTION - C19. Let the digit at unit’s place be x and the digit at ten’s place be y. Then,

Number = 10y + xNumber formed by reversing the digits = 10x + yAccording to the given conditions, we have

x + y =8and, (10y + x ) – (10x + y) = 18 9(y –x) = 18 y – x = 2On solving equations (i) and (ii), we get x = 3, y = 5Hence, number = 10y + x = 10 × 5 + 3 = 53

20. Since and are the zeros of the polynomial f(x) = 2x² + 5x + k.

+ = 5

2 and =

2

k

Now,

² + ² + = 21

4

(² + ² + 2) – = 21

4

( + )² – = 21

4

25

4 –

k

2 =

21

4

5 kα+β= - and αβ =

2 2

– k

2 = –1

k = 2

21. The given system of equations will have infinite number of solution, if

2

2 =

3

= 7

28

1

= 3

= 1

4

1

= 1

4 and

3

= 1

4

= 4 and + = 12 = 4 and = 8

Set - A... 6 ...

Hence, the given system of equation will have infinitely many solutions,if = 4 and = 8.

OR

21. t² + 0t – 15

(t)2 – 215

(t + 15 ) (t – 15 ) (using 2 2-a b = (a + b) (a – b))

So, the value of t² – 15 is zero, When (t + 15 ) = 0 or (t – 15 ) = 0,

i.e. when t = – 15 or t = 15 .

Therefore, the zeroes of t2 – 15 are – 15 and 15 .

Now, Sum of zeroes = – 15 + 15 = 0 = (0)

1

= 2

– (coefficient of )

coefficient of

t

t

Product of zeroes = – 15 × 15 = 15

1

= 2

constant term

coefficient of t

22.2 0 2 0 2 0

2 0 2 0

5cos 60 + 4sec 30 – tan 45

sin 30 + cos 30=

2221 2

5 × + 4× –(1)2 3

÷

221 3

+2 2

=1 4

5 × + 4 × – 14 3

÷ 1 3

+4 4

=5 16 1

+ –4 3 1

÷ 1 + 3

4

=15 + 64 – 12

12÷

4

4

=79 – 12

12 × 1

=6712

23. Proof : In ABC,LM || CB ......... (given)

AM

BM=

AL

CL........(I) ......... (by basic proportionality theorem.)

Set - A... 7 ...

In ADC,

LN || CD ......... (given)

AN

DN=

AL

CL......(II) ......... (by basic proportionality theorem)

From (I) and (II)

AM

BM=

AN

DNBy Invertendo (taking reciprocal)

BM

AM=

DN

AN

BM1

AM =

DN1

AN ......... (Adding 1 on both sides)

BM AM

AM

=

DN AN

AN

AB

AM=

AD

ANBy Invertendo (taking reciprocal)

AMAB

=ANAD

24. Proof

L.H.S = (sin A + cosec A)2 + (cos A + sec A)2

= sin2A + 2sinA cosecA + cosec2A + cos2A + 2cosA secA + sec2A

= (sin2A + cos2A) + (cosec2A )+ (sec2A) + 2sinA cosecA + 2 cosA secA

= 1+ (1+ cot2A) + (1 + tan2A) + 2 sinA 1 1

+2cos Asin A cos A

.... [ sin² A + cos² A = 1, cosec² A = 1 + cot² A, sec² A = 1 + tan² A]= 1 + 1 + cot² A + 1 + tan² A + 2 + 2

= 7 + tan2A + cot2A

= R.H.S.

L.H.S. = R.H.S. .... Hence proved.

OR24. In OPQ, we have

OQ² = OP² + PQ² (PQ + 1)² = OP² + PQ² [ OQ – PQ = 1 OQ = 1 + PQ]

PQ² + 2 PQ + 1 = OP² + PQ² 2PQ + 1 = 49 2PQ = 48

D

CA

N

L

B

M

Set - A... 8 ...

PQ - 24cm OQ – PQ = 1cm OQ = (PQ + 1 ) cm = 25cm

Now sinQ = 7

25

and, cosQ = 24

25

25. We have,POS ROQ3 = 4 and 1 = 2Thus, PS and QR are two lines andthe transversal PR cuts them in sucha way that 3 = 4i.e., alternate angles are equal.Hence, PS || QR.

26. Let us assume that 3 is a rational number.

There exist co-prime integers a and b ( 0) such that,

3 = a

b

b3 = a

squaring both sides,3b2 = a2 ... (1)

3 divides a2 3 divides a ... (2)Let a = 3c where c is some integersubstituting this value of a in (1)3b2 = (3c)2

3b2 = 9c2

b2 = 3c2

3 divides b2 3 divides b ... (3)From (2) and (3), we get,a and b both have common factor 3. This contradicts the fact that aand b are co-prime.

Our assumption that 3 is a rational number is wrong.

3 is an irrational number.

Q

O P7cm

P

R

Q

S

3

2

4

56

1

O

Set - A... 9 ...

27. 336 = 24 × 3 × 7 54 = 2 × 33

HCF (336 and 54) = 2 × 3 = 6(Product of common factors raised to least powers)

LCM (336 and 54) = 24 × 33 × 7 = 3024(Product of all the prime factors raised tohighest powers)

Verification :LCM × HCF = 6 × 3024 = 18144Product of the two numbers = 336 × 54 = 18144 LCM × HCF = Product of the two numbers.

28. We have,x – y = 12x + y = 8Graph of the equation x – y = 1:We have,

x – y = 1 y = x – 1 and x = y + 1Putting x = 0, we get y = –1Putting y = 0, we get x = 1Thus, we have the following table for the points on the line x – y = 1:

Plotting points A(O,–1), B(1,0) on the graph paper and drawing a linepassing through them, we obtain the graph of the line reresented by theequation x –y = 1 as shown in

x 0 1y -1 0

Set - A... 10 ...

Graph of the equation 2x + y = 8;We have,

2x + y = 8 y = 8 – 2x and x = 8 y

2Putting x = 0, we get y = 8Putting y = 0, we get x = 4Thus , we have the following table giving two points on the line representedby the equation 2x + y = 8.

Plotting points C (0,8) and D (4,0) on the same graph paper and drawing aline passing though them, we obtain the graph of the line represented bythe equation 2x + y =8 as shown.

OR

x 0 4y 8 0

X

Y'

0

M(0,2)

B(0,1) D(0,4)

P(3,2)

X'X

C(0,8)

x-y =1

2x+y = 8

Set - A... 11 ...

28. we have,

LHS =cosA

1- tanA +

sinA

1- cotA

LHS =

cosAsinA

1-cos A

+

sinAcosA

1-sin A

LHS =

cosAcosA - sinA

cos A +

sinAsinA - cosA

sin A

LHS =cos²A

cosA - sinA +

sin²A

sinA - cosA

LHS =cos²A

cosA - sinA –

sin²A

cosA - sinA

LHS =cos²A - sin²A

cosA - sinA

LHS =(cosA - sinA)(cosA + sinA)

cosA - sinA

LHS = cosA + sinA =RHS

SECTION - D 29. Let x be any positive integer and b = 3

Applying Euclid’s Division Algorithm x = 3q + r where 0 < r < 3 The possible remainders are 0, 1, 2 x = 3q or 3q + 1 or 3q + 2i) If x = 3q x3 = (3q)³ = 27q³ = 9(3q³) = 9m for some integer m,

where m = 3q³

ii) If x = 3q + 1 x³ = (3q + 1)³ = (3q)3 + 3(3q)²(1) + 3(3q) (1)² + (1)³ [ since (a +b) ³ = a³ + 3a²b + 3ab² + b³]

= 27q3 + 27q2 + 9q + 1= 9q(3q² + 3q + 1) + 1= 9m + 1 for some integer m,

where m = q (3q² + 3q + 1)

iii) If x = 3q + 2 x3 = (3q + 2)3

= (3q)3 + 3(3q)² (2) + 3(3q) (2)² + (2)3

[ since (a + b)³ = a³ + 3a²b + 3ab² + b³]= 27q3 + 54q2 + 36q + 8

Set - A

= 9q(3q² + 6q + 4) + 8= 9m + 8 for some integer m,

where m = q (3q² + 6q + 4)

Hence, the cube of any positive integer is either of the form 9m, 9m+1or 9m + 8

OR29. Let x be any positive integer and b = 3

Applying Euclid’s Division Algorithmx = 3q + r where 0 < r < 3 The possible remainders are 0, 1, 2 x = 3q or 3q + 1 or 3q + 2Now,i) If x = 3q x2 = (3q)2

= 9q2

= 3 (3q2)= 3m for some integer m, where m = 3q²

ii) If x = 3q + 1 x2 = (3q + 1)2

= 9q2 + 6q + 1= 3q (3q + 2) + 1= 3m + 1 for some integer m, where m = q (3q + 2)

iii) If x = 3q + 2 x2 = (3q + 2)2

= 9q2 + 12q + 4= 9q² + 12q + 3 + 1= 3 (3q2 + 4q + 1) + 1= 3m + 1 for some integer m,

where m = 3q² + 4q + 1

Hence, the square of any positive integer is either of the form 3m or 3m + 1

30. In ABC, we haveDE BC ADE = ABC and AED = ACB [Corresponding angles]Thus, in triangles ADE and ABC, we have

A = A [Common]ADE = ABC

and, ADE = ACB AED ABC

AD

AB =

DE

BC

We have,AD

AB=

5

4

... 12 ...

Set - A... 13 ...

DB

AD =

4

5

DB

1AD

= 4

5+ 1

DB AD 9

AD 5

AB 9 AD 5

AD 5 AB 9

DE 5

BC 9

In DFE and CFG, we have1 = 3 [Alternate interior anges]2 = 4 [Vertically interior anges]

Therefore, by AA -similarity criterion, we haveDFE ~ CFG

Area(DFE) DE²

Area(CFB) BC²

2Area(DFE) 5

Area(CFB) 9 =

25

81[Using (i)]

31. Weight Number of Weight C.f Points to (in kg) students (in kg) beplotted

36 - 38 0 Less than 38 0 (38, 0)

38 - 40 3 – 0 = 3 Less than 40 3 (40, 3)

40 - 42 5 – 3 = 2 Less than 42 5 (42, 5)

42 - 44 9 – 5 = 4 Less than 44 9 (44, 9)

44 - 46 14 – 9 = 5 Less than 46 14 c.f. (46, 14)

46 - 48 28 – 14 = 14 f Less than 48 28 (48, 28)

48 - 50 32 – 28 = 4 Less than 50 32 (50, 32)

50 - 52 35 – 32 = 3 Less than 52 35 (52, 35)

D E

B C

2

A

F1 54

Set - A

Median from the graph is 46.5 kg

Now, = 35

2 2

n = 17.5 which lies in the class 46 - 48. (See the table)

Median class is 46 - 48.l = 46, h = 2, f = 14, c.f. = 14

Median =– .

2n

c fl h

f

= 214

14–17.546

=7

3.546

= 46 + 0.5

Median = 46.5 kg.

Hence the median is same as obtained from the graph.

... 14 ...

5

10

15

20

25

30

35

Y

3517.5

2 2n

No. of st

uden

ts

Weight in kg

•38 40 42 44 46 48 50 52

••

•

•

••

•

Median = 46.5(38, 0)

(40, 3)(42, 5)

(44, 9)

(46, 14)

(50, 32)(52, 35)

X 0

Y

•

•

•

•

•

•

•

• • • • • • •

(48, 28)

X

SCALE : X - axis, 1cm = 1 kgY - axis, 1cm = 5 students

Set - A... 15...

32. Let X and Y be two cars starting from points A and B respectively. Let thespeed of car X and xkm/ hr and that of car Y be ykm/hr.Case I When two cars move in the same directions.Suppose tow cars meet at a point Q. Then,

Distance travelled by car X = AQ,Distance travelled by carYX = BQ,

It is given that two cars meet in 9 hours. Distance travelled by car X in 9 hours = 9xkm. AQ = 9xDistance travelled by car y 0n 9 hours = 9ykm.

BQ = 9y

Clearly, AQ – BQ = AB [AB =90km]

9x – 9y = 90 ... (i)

x – y = 10

Case II When two cars move in opposite directions :

Suppose two cars meet at point P. Then,

Distance travelled by car X = AP,

Distance travelled by car Y = BP,

In this case, two cars meet in 9/7 hours.

Distance travelled by car X in 9

7hours =

9

7xkm

BP = 9

7y

Clearly, AP + BP = AB

9

7x +

9

7y = 90

9

7(x +y) = 90

(x +y) = 70 ... (ii)

Solving equations (i) and (ii), we get

x = 40 and y = 30.

Hence, speeed of car X is 40km/hr and speed of car Y is 30km/hr.

90km

A BP Q

Set - A

33. Given : A ABC, in which DE || BC such that DE

intersects AB and AC at D and E

respectively.

To Prove :AD AE

=DB EC

Construction : Join BE and CD and draw EF AB, DG AC.

Proof : We have :

ar (ADE) =1

2 × AD × EF .... [Base = AD and height = EF]

andar (DBE) =1

2 × DB × EF .... [Base = DB and height = EF]

ar ADE

ar DBE

=

1 × AD × EF

21

× DB × EF2

= AD

DB.... (i)

Again, ar (ADE) =1

2 × AE × DG .... [Base = AE and height = DG]

and ar (DCE) =1

2 × EC × DG .... [Base = EC and height = DG]

ar ADE

ar DCE

=

1 × AE × DG

21

× EC × DG2

= AE

EC.... (ii)

Now, DBE and DCE are on the same base DE and between the sameparallels DE and BC. ar (DBE) = ar (DCE)Therefore, equation (ii) becomes,

ar ADE

ar DBE

=

AE

EC.... (iii)

From (i) and (iii), we get : AD AE

= DB EC

OR

... 16 ...

A

ED

B C

GF

Set - A

33. Given : ABC ~ DEF

To prove :

ar ABC

ar DEF

=

AB²

DE² =

BC²

EF² =

AC²

DF²

Construction : Draw AM BC and DN EF

Proof : ABC ~ DEF

A = D, B = E, C = F and AB

DE =

BC

EF =

AC

DF.... (i)

Now, ar (ABC) =1

× BC × AM2

and

ar (DEF) =1

× EF × DN2

ar ABC

ar DEF

=

1 × BC × AM

21

× EF × DN2

= BC

EF ×

AM

DN.... (ii)

Now, in AMB and DNE we have :AMB = DNE .... (Each being a right angle)

and B = E .... [Using (i)]

AMB ~ DNE .... (AA similarity)

and so,AM

DN=

AB

DE... (corresponding sides of similar triangles)

AM

DN=

BC

EF.... (iii) ..... [Using (i) we have

AB

DE =

BC

EF]

From (ii) and (iii) we get :

ar ABC

ar DEF

=

BC BC ×

EF EF

= BC²

EF²Similarly, we have :

ar ABC

ar DEF

=

AB²

DE² and

ar ABC

ar DEF

=

AC²

DF²

Hence,

ar ABC

ar DEF

=

AB²DE²

= BC²EF²

= AC²DF²

... 17 ...

B

A

CM E

D

FN

Set - A... 18 ...

34. L.H.S. =cos A –sin A+1

cos A+sin A –1

Dividing Numerator & Denominator by Sin A

=cot A – 1 + cosecA

cot A + 1 – cosecA

=2 2(cot A + cosec A) – (cosec A – cot A)

(1 + cot A – cosec A) .... ( cosec² A – cot² A = 1)

=(cot A + cosec A) – (cosec A + cot A) (cosec A – cot A)

(1+cot A – cosec A)

=(cot A+cosec A) (1-cosec A+cot A)

(1+cot A -cosec A)

= cot A + cosec A

= R.H.S.

L.H.S. = R.H.S. .... Hence proved.