2012 Class 10 Set-1 Section-b

8/12/2019 2012 Class 10 Set-1 Section-b

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CBSE X Mathematics 2012 Solution (SET 1)

Section B

Q11.Find the value(s) of kso that the quadratic equation x24kx+ k= 0 has equal roots.

Solution:

Given equation is 2 4 0x kx k For the given equation to have equal roots, D = 0.

2

2

2

4 0

4 4 1 0

16 4 0

4 4 1 0

4 0 or 4 1 0

10 or4

b ac

k k

k k

k k

k k

k k

Hence, for k= 0 or1

4, the given equation will have equal roots.

Q12.Find the sum of all three digit natural numbers, which are multiples of 11.

Solution:

The smallest and the largest three digit natural numbers, which are divisible by 11 are 110 and

990 respectively.

So, the sequence of three digit numbers which are divisible by 11 are 110, 121, 132, , 990.Clearly, it is an A.P. with first term, a= 110 and common difference, d= 11.

Let there be nterms in the sequence.

So, an= 990

1 990

110 1 11 990

110 11 11 990

11 99 990

11 990 99

11 891

81

a n d

n

n

n

n

n

n

Now, required sum 2 12

na n d


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812 110 81 1 11

2

81220 880

281

11002

44550

Hence, the sum of all three digit numbers which are multiples of 11 is 44550.

Q13.Tangents PA and PB are drawn from an external point P to two concentric circles with

centre O and radii 8 cm and 5 cm respectively, as shown in Fig. 3. If AP = 15 cm, thenfind the length of BP.

Solution:

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

To find:BPConstruction:Join OP.

Tangent to a circle is perpendicular to theNow, OA AP and OB BP

radius 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


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(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= 28925

(BP)2= 264

BP = 16.25 cm (approx.)

Hence, the length of BP is 16.25 cm.

Q14.In Fig. 4, an isosceles triangle ABC, with AB = AC, circumscribes a circle. Prove that thepoint 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 BCHence, 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.


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Solution:

Given: Two concentric circles C1and C2with centre O, and AB is the chord of C 1touching C2at

C.To prove: AC = CBConstruction: Join OC.

Proof: AB is the chord of C1touching C2at C, then AB is the tangent to C2at C with OC as its

radius.

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

Q15.The volume of a hemisphere is1

24252

cm3. Find its curved surface area.

22Use

7

Solution:3 31 4851Given, volume of hemisphere 2425 cm cm

2 2

Let the radius of the hemisphere be r cm.


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3

3

3

3

3

3

3

2Volume of hemisphere

3

2 4851

3 22 22 4851

3 7 2

4851 3 7

2 2 22

441 21

2 2 2

21 21 21

2 2 2

21 21 212 2 2

21cm ... 1

2

r

r

r

r

r

r

r

r

Curved surface area of hemisphere = 2r2

2

22 212 using 1

7 2

22 21 212

7 2 2

= 693 cm2

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)2cm

2= 49 cm

2


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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 2

1

4 r

2 2

2

2

2

1 7 cm

4

49cm

4

49 22cm

4 7

77cm

2

Area of the shaded region = Area of SquareArea of Quadrant of circle.

2

2

2

7749 cm

2

98 77cm

2

21cm

2

= 10.5 cm2

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

Q17.If a point A (0, 2) is equidistant from the points B (3, p) and C (p, 5), then find the value ofp.

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


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

4 4

1

p p p

p

p

Q18.A number is selected at random from first 50 natural numbers. Find the probability that it is

a multiple of 3 and 4.

Solution:

Total number of outcomes = 50

Multiples of 3 and 4 which are less than or equal to 50 are:12, 24, 36, 48

Favorable number of outcomes = 4

Probability of the number being a multiple of 3 and 4Number of favourable outcomes

Totalnumberof outcomes

4

50

2

25

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2012 Class 10 Set-1 Section-b