CBSE X Mathematics All India 2012 Solution (SET 1)

Section C

Q19. Solve the following quadratic equation for x:

x2 – 4ax – b

2 + 4a

2 = 0

Solution:

The given quadratic equation is 2 2 24 4 0x ax b a .

2 2 2 2 2

2 2 22

2 2 2

2

4 4 0 1, 4 , 4

4 4 4 1 4 4

2 2

4 16 16 4

2

4 4

2

4 2

2

2

2 or 2

x a x a b A B a C a b

a a a b B B ACx x

A

a a a bx

a bx

a bx

x a b

x a b x a b

Thus, the solution of the given quadratic equation is x = 2a + b or x = 2a – b.

OR

If the sum of two natural numbers is 8 and their product is 15, find the numbers.

Solution:

Let the two natural numbers be a and b.

It is given that, sum of two numbers = 8

a + b = 8

CBSE X Mathematics All India 2012 Solution (SET 1)

2

2

2

8 ... 1

It is given that, product of two numbers 15

15

8 15 Using 1

8 15

8 15 0

5 3 15 0

5 3 5 0

3 5 0

3 or 5

When 3, we have

8 8 3 5

When 5, we have

8 8

a b

ab

b b

b b

b b

b b b

b b b

b b

b b

b

a b

b

a b

5 3

Thus, the required natural numbers are 3 and 5.

Q20. Find the sum of all multiples of 7 lying between 500 and 900.

Solution:

The multiples of 7 lying between 500 and 900 are 504, 511, 518, … , 896.

This is an A.P.

First term, a = 504

Common difference, d = 511 – 504 = 7

Last term = 896

Let there be n terms in the A.P.

an = 896

a + (n – 1) d = 896

504 + (n – 1) × 7 = 896

(n – 1) × 7 = 896 – 504

392

17

n

n – 1 = 56

n = 57

Weknow that, S2

n

na l

Sum of all the multiples of 7 lying between 500 and 900

CBSE X Mathematics All India 2012 Solution (SET 1)

57

504 8962

571400

2

57 700

39900

Thus, the sum of all the multiples of 7 lying between 500 and 900 is 39900.

Q21. Draw a triangle ABC with BC = 7 cm, ∠B = 45° and ∠C = 60°. Then construct another

triangle, whose sides are 3

5 times the corresponding sides of ∆ABC.

Solution:

1. Draw a line BC = 7 cm.

2. Draw a ray CN making an angle of 60° at C.

3. Draw a ray BM making an angle of 45° at B.

4. Locate the point of intersection of rays CN and BM and name it as A.

5. ABC is the triangle whose similar triangle is to be drawn.

6. Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.

CBSE X Mathematics All India 2012 Solution (SET 1)

7. Locate 5 (Greater of 3 and 5 in3

5) points X1, X2, X3, X4 and X5 on BX so that BX1 =

X1X2 = X2X3 = X3X4 = X4X5.

8. Join X5C and draw a line through X3 (Smaller of 3 and 5 in3

5) parallel to X5C to

intersect BC at C.

9. Draw a line through Cparallel to the line CA to intersect BA at A.

10. ABCis the required similar triangle whose sides are3

5times the corresponding sides

of ABC.

Q22. In Figure 5, a circle is inscribed in a triangle PQR with PQ = 10 cm, QR = 8 cm and PR =

12 cm. Find the lengths of QM, RN and PL.

Solution:

Given: In PQR, PQ = 10, QR = 8 cm and PR = 12 cm.

We know that, the lengths of tangents drawn from an external point to a circle are equal.

QM = OL = x (Say)

RM = RN = y (Say)

PL = PN = z (Say)

CBSE X Mathematics All India 2012 Solution (SET 1)

QR QM + MR 8cm ... 1

PQ PL + LQ 10cm ... 2

PR PN + NR 12cm ... 3

Adding 1 , 2 and 3 , we have

(8 10 12) 30 cm

2 30 cm

15 cm ... 4

x y

z x

z y

x y z x z y

x y z

x y z

From (2) and (4), we have

10 15

5

From 3 and 4 , we have

12 15

3

From 1 and 4 , we have

8 15

7

QM 3 cm

RN 5 cm

PL 7 cm

y

y

x

x

z

z

x

y

z

Q23. In Figure 6, O is the centre of the circle with AC = 24 cm, AB = 7 cm and ∠BOD = 90°.

Find the area of the shaded region. [Use π = 3.14]

Solution:

O is the centre of circle.

Given: AC = 24 cm, AB = 7 cm and BOD = 90°.

The angle in a semicircle is 90°.

CBSE X Mathematics All India 2012 Solution (SET 1)

BAC = 90°

So, ABC is a right-angled triangle.

Area of ABC, A11

Base×Height2

217 24 7 12 84cm

2

Using Pythagoras Theorem in ABC, we have

(AC)2 + (AB)

2 = (BC)

2

2 2 2

2 2

2 2

24 7 BC

BC (576 49)cm

BC 625cm

BC 25cm

BC is diameter of circle.

OC = Radius of circle 25

cm2

Area of the sector COD, A2

2

2

θ

360

90 22 25 25

360 7 2 2

1 22 25 25

4 7 2 2

11 25 25

56

122.77 cm

r

Area of circle, A3 = πr2

2

22 25 25

7 2 2

11 25 25

14

491.07 cm

Area of the shaded region

= Area of circle – (Area of ABC + Area of sector COD)

= A3 – (A1 + A2)

= 491.07 cm2

– (84 + 122.77) cm2

= 491.07 cm2 – 206.77 cm

2

= 284.30 cm2

Thus, the area of shaded region is 284.30 cm2.

CBSE X Mathematics All India 2012 Solution (SET 1)

OR

In Figure 7, find the area of the shaded region, if ABCD is a square of side 14 cm and

APD and BPC are semicircles.

Solution:

Given: Side of square = 14 cm. Semicircle APD and BPC.

Area of square = (14 cm)2 = 196 cm

2

Diameter of semicircles APD and BPC = 14 cm

Radius of semicircles APD and BPC = 14

cm2

= 7 cm

Area of semicircle APD 21π

2r

2

1 227 7

2 7

77cm

Since the radius of the semicircles APD and BPC is same, their area will be same.

CBSE X Mathematics All India 2012 Solution (SET 1)

Area of semicircle BPC = 77 cm2

Area of shaded region

= Area of square – (Area of semicircle APD + Area of semicircle BPC)

= 196 cm2

– (77 + 77) cm2

= 196 cm2

– 154 cm2

= 42 cm2

Thus, the area of the shaded region is 42 cm2.

Q24. A hemispherical bowl of internal radius 9 cm is full of water. Its contents are emptied in a

cylindrical vessel of internal radius 6 cm. Find the height of water in the cylindrical

vessel.

Solution:

Let the height of water in the cylindrical vessel be h cm.

Given: Radius of the hemispherical bowl, r = 9 cm

Volume of the water in hemispherical bowl, V132

3r

3

1

2V 9cm

3

Given: Radius of the cylinder, R = 6 cm,

Volume of water in the cylindrical vessel, V2 = 2r h

2

2V 6cm h

Since water is emptied from the hemispherical bowl into the cylindrical vessel,

Volume of water in cylindrical vessel = Volume of the water in hemispherical bowl

2 3

3

2

26 9

3

2 9

3 6

27

2

13.5

h

h

h

h

Thus, the height of water in the cylindrical vessel is 13.5 cm.

Q25. The angles of depression of the top and bottom of a tower as seen from the top of a 60 3

m high cliff are 45° and 60° respectively. Find the height of the tower.

Solution:

CBSE X Mathematics All India 2012 Solution (SET 1)

Let AD be the tower and BC be the cliff.

Also, let h be the height (AD) of the tower and x be the distance (AB) of the foot of the

tower from the foot of the cliff.

Angles of depression of the top D and bottom A of the tower from top C of the cliff are

45° and 60° respectively.

CDE = 45° and CAB = 60°

BC 60 3 m (Height of the cliff)

CE = BC – BE

CE = (60 3 – h) m (BE = AD = h)

In DEC,

CEtan 45

DE

60 31 (DE = AB = )

60 3 ... 1

hx

x

x h

In CBA,

CBSE X Mathematics All India 2012 Solution (SET 1)

BCtan 60

AB

60 33

60 3

3

60 ... 2

x

x

x

On equating (1) and (2), we get

60 60 3

60 3 60

60 3 1

Hence, the height of the tower is 60 3 1 m.

h

h

h

Q26. Find the coordinates of a point P, which lies on the line segment joining the points A (–2,

–2), and B (2, –4), such that 3

AP =7

AB.

Solution:

3It is given that,AP AB, where A, P and B are three points on line segment AB.

7

AB 7

AP 3

AB 71 1

AP 3

AB–AP 7 3

AP 3

PB 4

AP 3

Thus, AP : PB = 3 : 4

It is given that, the coordinates of points A and B are (–2, –2) and (2, –4).

Using section formula,

3 2 4 2 3 4 4 2Coordinates of P are ,

3 4 3 4

6 8 12 8 2 20, ,

7 7 7 7

CBSE X Mathematics All India 2012 Solution (SET 1)

Hence, the coordinates of point P are2 20

,7 7

.

OR

Find the area of the quadrilateral ABCD whose vertices are A (–3, –1), B (–2, –4), C (4, –

1) and D (3, 4).

Solution:

Given: Vertices of quadrilateral ABCD such as A (3, 1), B (2, 4), C (4, 1) and D

(3, 4).

If P (x1, y1), Q (x2, y2), R (x3, y3) and are vertices of the triangle then area of PQR

1 2 3 2 3 1 3 1 2

1

2x y y x y y x y y

2

1Area of ABC 3 ( 4 1 2 1 ( 1) 4 1 ( 4)

2

1( 3)( 3) ( 2)(0) (4)(3)

2

19 12

2

121

2

21 unit

2

CBSE X Mathematics All India 2012 Solution (SET 1)

2

1Area of ACD ( 3)[ 1 4] (4) 4 1 3 1 1

2

1( 3)( 5) (4) (5) 3 0

2

115 20

2

135

2

35 unit

2

Area of quadrilateral ABCD = Area (∆ABC) + Area (∆ACD)

2

2

2

21 35 unit

2 2

56unit

2

28 unit

Thus, the area of the quadrilateral ABCD is 28 unit2.

Q27. If the points A (x, y), B (3, 6) and C (–3, 4) are collinear, show that x – 3y + 15 = 0.

Solution:

If the given points A (x, y), B (3, 6) and C (–3, 4) are collinear, then

Area of the triangle ABC = 0

1 2 3 2 3 1 3 1 2

10

2

6 4 3 4 3 6 0

2 12 3 3 18 0

2 6 30 0

2 3 15 0

3 15 0

x y y x y y x y y

x y y

x y y

x y

x y

x y

Q28. All kings, queens and aces are removed from a pack of 52 cards. The remaining cards are

well shuffled and then a card is drawn from it. Find the probability that the drawn card is

(i) a black face card.

(ii) a red card.

Solution:

Total number of cards in a pack = 52

CBSE X Mathematics All India 2012 Solution (SET 1)

A pack contains of 4 kings, 4 queens and 4 aces.

Number of cards removed = 4 + 4 + 4 = 12

Remaining number of cards in the pack = 52 – 12 = 40

(i) Number of black face cards = 2 (jacks of spade and club)

Number of black face cards

P a black face cardReamining number of cards in the pack

2

40

1

20

Thus, the probability of getting a black face card is1

20.

(ii) Remaining number of red cards = 26 (2 + 2 + 2) = 26 6 = 20

Remaining number of red cards

P red cardRemaining number of cards in the pack

20

40

1

2

Thus, the probability of getting a red card is1

2.