2012 class 10 set-3 section-b

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CBSE X Mathematics 2012 Solution (SET 3)

Section B

Q11. If a point A (0, 2) is equidistant from the points B (3, p) and C ( p, 5), then find the value of

p.

Solution:

The given points are A (0, 2), B, (3, P) and C (P, 5).

According to the question, A is equidistant from points B and C.

AB = AC

2 2 2 2

2 2 2 2

2 2

2 2

3 0 2 0 5 2

3 2 3

9 4 4 9

4 13 9

p p

p p

p p p

p p p

On squaring both sides, we obtain:2 24 13 9

4 4

1

p p p

p

p

Q12. A number is selected at random from first 50 natural numbers. Find the probability that it isa multiple of 3 and 4.

Solution:

Total number of outcomes = 50Multiples of 3 and 4 which are less than or equal to 50 are:

12, 24, 36, 48

Favorable number of outcomes = 4Probability of the number being a multiple of 3 and 4

Number of favourable outcomes

Totalnumberof outcomes

4

502

25

Q13. The volume of a hemisphere is1

24252

cm3. Find its curved surface area.

22Use

7




Given that: OA = 8 cm, OB = 5 cm and AP = 15 cm

To find: BPConstruction: Join OP.

Tangent to a circle is perpendicular to the Now, OA AP and OB BPradius through the point of contact

OAP OBP 90

On applying Pythagoras theorem in OAP, we obtain:

(OP)2 = (OA)

2 + (AP)

2

(OP)2 = (8)

2 + (15)

2

(OP)2 = 64 + 225

(OP)2 = 289

OP = 289

OP = 17

Thus, the length of OP is 17 cm.On applying Pythagoras theorem in OBP, we obtain:

(OP)2

= (OB)2 + (BP)

2

(17)2

= (5)2 + (BP)

2

289 = 25 + (BP)2

(BP)2 = 289 – 25

(BP)2 = 264

BP = 16.25 cm (approx.)

Hence, the length of BP is 16.25 cm.

Q15. In Fig. 4, an isosceles triangle ABC, with AB = AC, circumscribes a circle. Prove that the point of contact P bisects the base BC.




Solution:

Given: An isosceles ABC with AB = AC, circumscribing a circle.To prove: P bisects BCProof: AR and AQ are the tangents drawn from an external point A to the circle.

AR = AQ (Tangents drawn from an external point to the circle are equal)Similarly, BR = BP and CP = CQ.

It is given that in ABC, AB = AC. AR + RB = AQ + QC

BR = QC (As AR = AQ)

BP = CP (As BR = BP and CP = CQ)

P bisects BC

Hence, the result is proved.

OR

In Fig. 5, the chord AB of the larger of the two concentric circles, with centre O, touches the

smaller circle at C. Prove that AC = CB.

Solution:




Given: Two concentric circles C1 and C2 with centre O, and AB is the chord of C 1

touching C2 at C.To prove: AC = CBConstruction: Join OC.Proof : AB is the chord of C1 touching C2 at C, then AB is the tangent to C2 at C with OCas its radius.

We know that the tangent at any point of a circle is perpendicular to the radius through

the point of contact.

OC AB

Considering, AB as the chord of the circle C1. So, OC AB.

OC is the bisector of the chord AB.

Hence, AC = CB (Perpendicular from the centre to the chord bisects the chord).

Q16. In Fig. 6, OABC is a square of side 7 cm. If OAPC is a quadrant of a circle with centre O,

then find the area of the shaded region.22

Use7

Solution:

It is given that OABC is a square of side 7 cm

Area of square OABC = (7)2 cm

2 = 49 cm

2

Also, it is given that OAPC is a quadrant of circle with centre O.

Radius of the quadrant of the circle = OA = 7 cm

Area of the quadrant of circle 21π

4r




2 2

2

2

2

1π 7 cm

4

49π cm

449 22

cm4 7

77 cm

2

Area of the shaded region = Area of Square – Area of Quadrant of circle.

2

2

2

7749 cm

2

98 77cm2

21 cm

2

= 10.5 cm2

Thus, the area of the shaded region is 10.5 cm2.

Q17. Find the sum of all three digit natural numbers, which are multiples of 7.

Solution:

Three digit natural numbers which are multiples of 7 are 105, 112, 119,…, 994. 105, 112, 119, … 994 are in A.P.

First term (a) = 105Common difference (d ) = 7

Let 994 be the nth

term of A.P.

an = 994

105 + (n – 1) × 7 = 994 [an = a + (n – 1)d ]

7 (n – 1) = 994 – 105

7 (n – 1) = 889

n – 1 = 127 n = 128

Sum of all the terms of A.P. 128

105 994 ( ), being last term2 2

n

nS a l l

= 64 × 1099

= 70336




Thus, the sum of all three digit natural numbers which are multiples of 7 is 70336.

Q18. Find the value(s) of k so that the quadratic equation 3 x2 – 2kx + 12 = 0 has equal roots.

Solution:

The given quadratic equation is 3 x2 – 2k x + 12 = 0.

On comparing it with the general quadratic equation ax2 + b x + c = 0, we obtain

a = 3, b = – 2k and c = 12

Discriminant, ‘D’ of the given quadratic equation is given by

2

2

2

D 4

2 4 3 12

4 144

b ac

k

k

For equal roots of the given quadratic equations, Discriminant will be equal to 0.i.e., D = 0

2

2

2

4 144 0

4 36 0

36

6

k

k

k

k

Thus, the values of k for which the quadratic equation 3 x2 – 2k x + 12 = 0 will have equal roots

are 6 and – 6.

2012 class 10 set-3 section-b

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