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# CBSE Class 10 Mathematics Solved Question Paper 2014

1.

The incircle of an isosceles triangle ABC, in which AB = AC, touches the sides BC, CA and AB at D, E and F respectively. Prove that BD = DC.

Given: $∆\mathrm{ABC}$ is an isosceles triangle with a circle inscribed in the triangle.

To prove: BD=DC

Proof:

AF and AE are tangents drawn to the circle from point A.

Since two tangents drwan to a circle from the same exterior point are equal.

AF=AE=a

Similarly  BF=BD=b  and  CD=CE=c

We also know that $∆\mathrm{ABC}$ is an isosceles triangle

Thus AB=AC

a+b=a+c

Thus b=c

Therefore, BD=DC

Hence proved.

#### Multiple Choice Questions

2.

If k, 2k- 1 and 2k + 1 are three consecutive terms of an A.P., the value of k is

• 2

• 3

• -3

• 5

B.

3

3.

Two different dice are tossed together. Find the probability

(i) That the number on each die is even.

(ii) That the sum of numbers appearing on the two dice is 5.

The total number of outcomes when two dice are tossed together is 36.

The sample space is as follows

(i).  Favourable outcomes = { (2,2) (2,4) (2,6) (4,2) (4,4)

(6,4) (6,2) (6,4) (6,6) }

Porbability that the number on each dice is even

=

(ii)  Favourable outcomes = { (1,4) (2,3) (3,2) (4,1) }

Probability that the sum of the number appearing on the two dice is 5

=

#### Multiple Choice Questions

4.

In a right triangle ABC, right-angled at B, BC = 12 cm and AB = 5 cm. The radius of the circle inscribed in the triangle (in cm) is

• 4

• 3

• 2

• 1

C.

2

It is given that AB = 5 and BC = 12

Using pythagoras theorem

We know that  two tangents drwan to a circlefrom the same point that is exterior to

the  circle are of equal iengths.

Thus AM = AQ = a

Similarly  MB = BP = b  and  PC = CQ = c

We know AB = a+b = 5

BC = b+c = 12 and  AC = a+c = 13

Solving simultaneously we get  a = 3,  b =2,  c = 10

We also know that the tangent is perpendicular to the radius.

Thus OMBP is a square with side b

Hence the length of the radius of the circle inscribed in the right angled triangle is 2 cm.

5.

Two circles touch each other externally at P. AB is a common tangent to the circlestouching them at A and B. The value of $\angle$ APB is

• $30°$

• $45°$

• $60°$

• $90°$

D.

$90°$

6.

In a family of 3 children, the probability of having at least one boy is

• $\frac{7}{8}$

• $\frac{1}{8}$

• $\frac{5}{8}$

• $\frac{3}{4}$

A.

$\frac{7}{8}$

There are in all 23 = 8 combinations or outcomes for the gender of the 3 children

The 8 combinations are as follows

BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG

Thus the probability of having at least one boy in a family is $\frac{7}{8}$

7.

A chord of a circle of radius 10 cm subtends a right angleat its centre. The length of the chord (in cm) is

• 5$\sqrt{2}$

• 10$\sqrt{2}$

• $\frac{5}{\sqrt{2}}$

• 10$\sqrt{3}$

B.

10$\sqrt{2}$

8.

The angle of depression of a car parked on the road from the top of a 150 m high tower is 30°. The distance of the car from the tower (in metres) is

• $50\sqrt{3}$

• $150\sqrt{3}$

• $150\sqrt{2}$

• 75

A.

$50\sqrt{3}$

Let AB be the tower and BC be the distance between tower and car. Let $\mathrm{\theta }$ be the angle of depression of the car.

According to the given information,

9.

The probability that a number selected at random from the numbers 1, 2, 3, ..., 15 isa multiple of 4, is

• $\frac{4}{15}$

• $\frac{2}{12}$

• $\frac{1}{5}$

• $\frac{1}{3}$

C.

$\frac{1}{5}$

10.

ABCD is a rectangle whose three vertices are B (4, 0), C(4,3) and D(0, 3). The length of one of its diagonals is

• 5

• 4

• 3

• 25

A.

5

We se that AB= 4 units and BC=3 units

Using pythagoras theorem

AC= AB2 + BC2

= 42 + 32

AC2 = 25

Thus AC= 5 Units

Hence length of the diagonal of the rectangle is 5 units