Set-1 Sec-c

CBSE X Mathematics 2009 Solution (SET 1)

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

16.

Prove that 3

2 is an irrational number.

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

17.

2 should be

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 These equations can also be written as:

CBSE X Mathematics 2009 Solution (SET 1)

a b x y ( a b) 0 b a ax by 2ab 0

...(i) ...(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 b 2 ab a 2 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

CBSE X Mathematics 2009 Solution (SET 1) 1 1 8 x 8 x 15 (8 x x) 8 x(8 x) 15 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.

18.

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 1) 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.

19.

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 1; cot tan 1] 3 3 2 5 3 3 3 3 1

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

CBSE X Mathematics 2009 Solution (SET 1) 20.

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) (5) (6)

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 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.

CBSE X Mathematics 2009 Solution (SET 1) 21.

In Figure, 3, AD

BC and BD

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 AC2 – AB2 = DC2 – BD2 … (3) It is given that BD

1 CD 3

CBSE X Mathematics 2009 Solution (SET 1) 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(AC2 – AB2) = BC2 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]

Comparing BLC and ELA: LBC = AEL [Alternate interior angles as AE||BC] BLC = ELA [Vertically opposite angels]

CBSE X Mathematics 2009 Solution (SET 1) 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.

22.

Find the ratio in which the point (2, y) divides the line segment joining the points A (–2, 2) and B (3, 7). Also find the value of y.

Solution: Let the point P(2, y) divide the line segment joining the points A(–2, 2) and B (3, 7), in the ratio 1: internally. By the section formula, we have: 1(3) 1

P 2, y

( 2) 1(7) , 1

(2)

3 2 7 2 , 1 1 3 2 7 2 2 and y 1 1 2(1 ) 3 2

2 2 4

3 2

3 2 1 4

Also, y

7 2 1

7 2 1 1 4

1 4

28 2 4 4 1 4

30 5

6

CBSE X Mathematics 2009 Solution (SET 1) Thus, the point (2, y) divides the line segment joining points (–2, 2) and (3, 7) in the ratio 4:1 and the value of y is 6.

23.

Find the area of the quadrilateral ABCD whose vertices are A(–4, –2), B(–3, – 5), C (3, –2) and D (2, 3).

Solution: Join AC.

Then, Area of ABCD = Area of ∆ABC + Area of ∆ADC 1 Area of ABC [(–4){–5 – (–2)} (–3){–2 – (–2)} 3{ – 2 – (–5)}] 2 1 [(–4)(–3) (–3){–2 2} 3{ – 2 5}] 2 1 [12 (–3){0} 3{3}] 2 1 21 [12 0 9] square units 2 2

Area of ACD

1 [(–4){–2 – 3} 3{3 – (–2)} 2{( – 2) – (–2)}] 2 1 [(–4)(–2 3) 3{3 2} 2{ – 2 2}] 2 1 [20 15 2{0}] 2 1 35 [35] square units 2 2

Then, Area of ABCD = Area of ∆ABC + Area of ∆ADC 21 35 square units + square units = 2 2 = 28 square units

CBSE X Mathematics 2009 Solution (SET 1) 24.

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:

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=

14 cm = 7 cm 2

C = 60°

[ ∆ABC is equilateral]

CBSE X Mathematics 2009 Solution (SET 1) Area of sector AEF

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 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.

25.

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:

CBSE X Mathematics 2009 Solution (SET 1)

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)

(iii)

P (5 will come up on both dice)

11 36

1 36

11 36

25 36

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