Set-3 Sec-c

CBSE X Mathematics 2009 Solution (SET 3)

CBSE X 2009 Mathematics Section C Question Number 16 to 25 carry 3 marks each.

16.

Prove that (3 5 2) is an irrational number. Solution: If possible, suppose (3 5 2) is a rational number. Then, we can find two integers a, b (b ≠ 0) such that a (3 5 2) b a 5 2 3 b 1 a 2 3 5 b 1 a Since a, b, 5 and 3 are integers, 3 is rational and so 2 is rational. 5 b This conclusion contradicts the fact that 2 is an irrational number. Hence, our assumption that (3 5 2) is rational is false. Thus, (3 5 2) is irrational.

17.

Draw a right triangle in which sides (other than hypotenuse) are of lengths 8 3 cm and 6 cm. Then construct another triangle whose sides are times the 4 corresponding sides of the first triangle. Solution: Let us assume that right ABC has base BC = 6 cm, side AB = 8 cm, and 3 = 90 . Let A'BC' have sides that are times those of ABC. 4 Now, ABC and A'BC' can be drawn as follows: (1) (2) (3) (4)

B

Draw a line segment BC = 6 cm. Draw a ray BX making 90 with BC. Draw an arc of 8 cm radius taking B as its centre to intersect BX at A. Join AC. ABC is the required triangle. Draw a ray BY making any acute angle with BC on the other side of line segment BC. Locate 4 points B1, B2, B3, and B4 on ray BY such that BB1 = B1B2 = B2B3 = B3B4

CBSE X Mathematics 2009 Solution (SET 3) (5) (6)

18.

Join B4C. Draw a line through B3 parallel to B4C intersecting BC at C'. Through C', draw a line parallel to AC intersecting AB at A'. A'BC' is the required triangle.

Two dice are thrown simultaneously. What is the probability that (i) 5 will not come up on either of them? (ii) 5 will come up on at least one? (iii) 5 will come up at both dice? Solution: The sample space for the given experiment can be shown as:

1 2 3 4 5 6

1 (1, 1) (2, 1) (3, 1) (4, 1) (5, 1) (6, 1)

2 (1, 2) (2, 2) (3, 2) (4, 2) (5, 2) (6, 2)

3 (1, 3) (2, 3) (3, 3) (4, 3) (5, 3) (6, 3)

4 (1, 4) (2, 4) (3, 4) (4, 4) (5, 4) (6, 4)

5 (1, 5) (2, 5) (3, 5) (4, 5) (5, 5) (6, 5)

Total number of possible outcomes = 6 × 6 = 36 (i)

P (5 will come up on either of them)

11 36

P (5 will not come up on either of them) 1 (ii)

P (5 will come up on at least one)

11 36

11 36

25 36

6 (1, 6) (2, 6) (3, 6) (4, 6) (5, 6) (6, 6)

CBSE X Mathematics 2009 Solution (SET 3) (iii)

19.

P (5 will come up on both dice)

In Figure, 3, AD

BC and BD

1 36

1 CD. Prove that 2CA2 = 2AB2 + BC2. 3

OR In Figure 4, M is mid-point of side CD of a parallelogram ABCD. The line BM is drawn intersecting AC at L and AD produced at E. Prove that EL = 2 BL.

Solution:

On applying Pythagoras theorem to ∆ABD, we obtain AB2 = BD2 + AD2 AD2 = AB2 – BD2 … (1) On applying Pythagoras theorem to ∆ CD, we obtain AC2 = AD2 + DC2 AD2 = AC2 – DC2 … (2) On comparing equations (1) and (2), we obtain AB2 – BD2 = AC2 – DC2

CBSE X Mathematics 2009 Solution (SET 3) AC2 – AB2 = DC2 – BD2 … (3) It is given that BD

1 CD 3

CD = 3BD Now, BD + DC = BC BD + 3BD = BC 1 BD BC 4

… (4)

Also, CD = 3BD 3 DC BC 4

… (5)

[From (4)]

On substituting the values of BD and DC from (4) and (5) in (3), we obtain 2

3 1 AC – AB BC BC 4 4 9 1 BC2 16 16 8 1 BC2 BC2 16 2 2 2 2 2(AC – AB ) = BC 2AC2 – 2AB2 = BC2 2CA2 = 2AB2 + BC2 2

2

2

Hence, proved. OR

Comparing BCM and EDM: CM = MD [M is the mid-point of CD] BMC = DME [Vertically opposite angels] BCM = EDM [Alternate interior angles as AE||BC] BCM EDM BC = ED BC = AD = ED

[ASA congruence criterion] [C.P.C.T.] [Opposite sides of parallelogram are equal]

CBSE X Mathematics 2009 Solution (SET 3) Comparing BLC and ELA: LBC = AEL [Alternate interior angles as AE||BC] BLC = ELA [Vertically opposite angels] BLC

ELA

[AA similarity criterion]

It is known that the corresponding sides of similar triangles are proportional. BL BC EL AE BL BC [AE AD DE] EL AD DE BL DE [BC AD DE] EL DE DE BL DE EL 2DE BL 1 EL 2 EL 2BL Hence, proved.

20.

The sum of first six terms of an arithmetic progression is 42. The ratio of its 10th term to its 30th term is 1 : 3. Calculate the first and the thirteenth term of the A.P. Solution: Let a be the first term and d be the common difference of the given A. P. It is known that the nth term of an A.P. is given by an = a + (n – 1)d and the n [2a (n 1)d ] . sum of the first n terms is given by S n 2 Sum of the first 6 terms = 42 6 [2a (6 1) d ] 42 2 3[2a 5d ] 42 2a + 5d = 14 … (1) 10th term = a10 = a + (10 – 1)d = a + 9d 30th term, a30 = a + (30 – 1)d = a + 29d It is given that

a10 a30

1 3

CBSE X Mathematics 2009 Solution (SET 3) a 9d 1 a 29d 3 3a 27 d a 29d 2a 2d a d

Substituting a = d in equation (1), we get: 2d + 5d = 14 7d = 14 d=2 a=d=2 Now, 13th term = a13 = a + 12d = 2 + 12 × 2 = 2 + 24 = 26 Thus, the first term and the 13th term of the given A.P. are 2 and 26 respectively.

21.

The area of an equilateral triangle is 49 3 cm2. Taking each angular point as centre, circles are drawn with radius equal to half the length of the side of the triangle. Find the area of triangle not included in the circles. [Take 3 = 1.73] OR Figure 5 shows a decorative block which is made of two solids – a cube and a hemisphere. The base of the block is a cube with edge 5 cm and the hemisphere, fixed on the top, has a diameter of 4.2 cm. Find the total surface 22 area of the block. [Take = ] 7

Solution:

CBSE X Mathematics 2009 Solution (SET 3)

Let ABC be the given equilateral triangle whose area is 49 3 cm2. Area of equilateral triangle

3 (Side) 2 4

3 (Side) 2 49 3 cm2 4 4 49 3 (Side)2 cm2 3 Side = 2 × 7 cm = 14 cm

Thus, radius of each circle We know,

A=

B=

Area of sector AEF

14 cm = 7 cm 2

C = 60°

[ ∆ABC is equilateral]

60 22 7 7 360 7 22 7 77 cm 2 6 3

Area of the three sectors

77 cm2 3 = 77 cm2 3

Thus, area of the triangle not included in the circle

49 3 77 cm 2

= (84.77 – 77) cm2 = 7.77 cm2

OR TSA of the block = TSA of the cube – Base area of the hemisphere + CSA of the hemisphere …(1) TSA of the cube = 6 (side)2 = 6 × 52 cm2 = 150 cm2 Base area of the hemisphere = π (radius)2

CBSE X Mathematics 2009 Solution (SET 3) 2

22 4.2 7 2 22 4.2 4.2 7 2 2 11 0.6 2.1 13.86 cm 2 CSA of hemisphere = 2 × π(radius)2 = 2 × 13.86 cm2 From equation (1), we get: T.S.A of the block = (150 – 13.86 + 2 × 13.86) cm2 = 150 + 13.86 cm2 = 163.86 cm2 Thus, the total surface area of the block is 163.86 cm2.

22.

Find the ratio in which the point (x, –1) divides the line segment joining the points (–3, 5) and (2, –5). Also find the value of x. Solution: Let the point (x, –1) divide the line segment joining the points (–3, 5) and (2, –5) internally in the ratio k:1. Then, by using the section formula, we have: k (2) 1( 3) k ( 5) 1(5) ( x, 1) , k 1 k 1 2 k 3 5k 5 , k 1 k 1

( x, 1)

...(1)

Comparing the y coordinates from equation (1), we get: 5k 5 1 k 1 (k 1) 5k 5

k 1 5k k 4k k

5k 5 5 1

6 6 3 4 2

Therefore, the point (x, –1) divides the line segment joining the points (–3, 5) and (2, –5) in the ratio 3:2.

CBSE X Mathematics 2009 Solution (SET 3) Now, comparing the x coordinates of equation (1), we get: 3 2 3 2k 3 3 2 x [k ] 3 k 1 2 1 2 3 3 0 3 1 2 Hence, the value of x is 0.

23.

Find the area of the quadrilateral ABCD whose vertices are A (1, 0), B (5, 3), C (2, 7) and D (–2, 4) Solution: The given quadrilateral ABCD can be drawn as:

We join AC to obtain ABC and ACD. We know that the area of a triangle whose vertices are (x1, y1), (x2, y2) and (x3, y3) is given by the numerical value of the expression: 1 x1 y2 y3 x2 y3 y1 x3 y1 y2 2

Area of ABC

1 [1(3 7) 5(7 0) 2(0 3)] 2 1 25 [ 4 35 6] square units 2 2

1 [1(7 4) 2(4 0) ( 2)(0 7)] 2 1 25 (3 8 14) square units 2 2 Thus, area of ABCD Area of ABC Area of ACD Area of ACD

25 2

25 square units 2

25 square units

CBSE X Mathematics 2009 Solution (SET 3) 24.

Evaluate: 2 cosec 2 58 3

2 cot 58 tan 32 3

5 tan13 tan 37 tan 45 tan 53 tan 77 3

Solution: 2 2 5 cosec 2 58 cot 58 tan 32 tan13 tan 37 tan 45 tan 53 tan 77 3 3 3 2 2 5 cosec 2 58 cot 58 cot(90 32 ) cot(90 13 ) cot(90 37 ) 1 tan 53 tan 77 3 3 3 [tan cot(90 ); tan 45 1]

2 2 5 cosec 2 58 cot 58 cot 58 cot 77 cot 53 tan 53 tan 77 3 3 3 2 5 (cosec 2 58 cot 2 58 ) (cot 77 tan 77 )(cot 53 tan 53 ) 3 3 2 5 1 1 1 [cosec 2 cot 2 3 3 2 5 3 3 3 3 1

1; cot

Thus, the value of the given expression is –1.

25.

Solve for x and y: ax by a b b a ax – by = 2ab OR The sum of two numbers is 8. Determine the numbers if the sum of their 8 reciprocals is . 15 Solution: The given pair of linear equations is: ax by a b b a ax – by = 2ab

tan

1]

CBSE X Mathematics 2009 Solution (SET 3) These equations can also be written as: a b x y ( a b) 0 ...(i) b a ax by 2ab 0 ...(ii) Now, the given equations can be solved by cross-multiplication method as: x y 1 b a a b ( 2ab) ( b){ (a b)} a(a b) ( 2ab) b a a b b a x y 1 2 2 2 2 2b ab b a ab 2a a b x y 1 2 2 b ab a ab b a x y 1 b(b a ) a (a b) b a On taking the first and last terms, we get: x 1 b(b a ) b a b(b a ) x (b a) x b On taking the last two terms, we get: y 1 a a b b a y y

a a b b a a

Thus, the solution to the given pair of equations is x = b and y = – a. OR Let one of the two numbers be x. Then, the other number is 8 – x. 8 It is given that the sum of the reciprocals of the numbers = 15

1 1 8 x 8 x 15 (8 x x) 8 x(8 x) 15

CBSE X Mathematics 2009 Solution (SET 3)

8 8 x(8 x) 15 x(8 x) 15 8x x2

15

x 2 8 x 15 0 x 2 5 x 3x 15 0

[( 5) ( 3)

8, ( 5) ( 3) 15]

x( x 5) 3( x 5) 0 ( x 5)( x 3) 0 x 5 or 3 If we take x = 5, then the other number is 8 – 5 = 3. If we take x = 3, then the other number is 8 – 3 = 5. In either case, the numbers are 3 and 5. Thus, the required two numbers are 3 and 5.