Class 10 Maths circles questions with solutions

Q 1 :

A backyard is in the shape of a triangle ABC with right angle at B. AB = 7 m and BC 15 m. A circular pit was dug inside it such that it touches the walls AC, BC and AB at P, Q and R respectively such that $A P = x m$ .

Based on the above information, answer the question:

Q. Find the length of AR in terms of $x$ .

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Given, AB = 7 m, BC = 15 m and AP = x m

Hence, AP = AR (Tangent drawn from an external point to the circle are equal in length)

$∴$ AR = x m

Q 2 :

A backyard is in the shape of a triangle ABC with right angle at B. AB = 7 m and BC 15 m. A circular pit was dug inside it such that it touches the walls AC, BC and AB at P, Q and R respectively such that $A P = x m$ .

Based on the above information, answer the question:

Q. Write the type of quadrilateral BQOR.

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Since, AR = x m and AB = 7 m

$∴$ RB = (7 – x)m

Also, RB = BQ

(Tangents drawn from an external point to the circle)

OR = OQ (radii of circle)

$∠$ ORB = $∠$ OQB = 90° (Angle between radius and tangent)

Also, $∠$ RBQ = 90° (angle between the walls AB and BC)

Thus, $∠$ ROQ = 90°

Thus, BQOR is square.

Q 3 :

A backyard is in the shape of a triangle ABC with right angle at B. AB = 7 m and BC 15 m. A circular pit was dug inside it such that it touches the walls AC, BC and AB at P, Q and R respectively such that $A P = x m$ .

Based on the above information, answer the question:

Q. Find the length PC in terms of $x$ and hence find the value of $x$

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(4.28)

Here, BC = 15 m

BQ = (7 – x)m

$∴$ QC = 15 – (7 – x)

or, QC = (8 + x)m

Also, QC = PC (Tangents from an external points C to the circle)

i.e., PC = (8 + x)m

In right DABC, using Pythagoras theorem,

$A C^{2} = A B^{2} + B C^{2}$

$\Rightarrow A C^{2} = 7^{2} + 15^{2} = 49 + 225 = 274$

$\Rightarrow A C = 16.55$

$\Rightarrow A P + P C = 16.55$

$\Rightarrow x + 8 + x = 16.55$

$\Rightarrow 2 x = 8.55$

$\Rightarrow x = 4.275 ~ 4.28 m$

Q 4 :

The chain and gears of bicycles or motorcycles, as well as the belt around pulleys, are real-life examples of tangents to a circle. One such example is shown in Fig, where PI and PA represent tangents to the gears of a bicycle.

Based on the above information, answer the following questions:

(i) If PI = 20 cm, then find PA.

20 cm
40 cm
60 cm
30 cm

1

2

3

4

(1)

PI and PA are tangents to the circle.
As tangents from an external point to a circle are equal,
∴ PI = PA = 20 cm

Q 5 :

The chain and gears of bicycles or motorcycles, as well as the belt around pulleys, are real-life examples of tangents to a circle. One such example is shown in Fig, where PI and PA represent tangents to the gears of a bicycle.

Based on the above information, answer the following questions:

(ii) If PO = 25 cm, then find IO.

10 cm
11 cm
14 cm
15 cm

1

2

3

4

(4)

Consider Fig.

In right triangle ΔPIO, we have

$O P^{2} = P I^{2} + I O^{2} (25)^{2} = (20)^{2} + I O^{2}$

$I O^{2} = 625 - 400 = 225 I O = 15 cm$

Q 6 :

The chain and gears of bicycles or motorcycles, as well as the belt around pulleys, are real-life examples of tangents to a circle. One such example is shown in Fig, where PI and PA represent tangents to the gears of a bicycle.

Based on the above information, answer the following questions:

(iii) (a) If arc IZA subtends an angle of 260° at the centre of the circle, then find ∠IPA

$70 °$
$60 °$
$80 °$
$40 °$

1

2

3

4

(3)

Given, reflex ∠IOA = 260°

$∠ I O A = 360 ° - 260 ° = 100 °$

The angle between two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.

$∠ I P A + ∠ I O A = 180 ° ∠ I P A + 100 ° = 180 ° ∠ I P A = 180 ° - 100 ° = 80 °$

Q 7 :

The chain and gears of bicycles or motorcycles, as well as the belt around pulleys, are real-life examples of tangents to a circle. One such example is shown in Fig, where PI and PA represent tangents to the gears of a bicycle.

Based on the above information, answer the following questions:

(iii) (b) If PI = 2x + 8, then find x.

6 cm
5 cm
4 cm
3 cm

1

2

3

4

(1)

Given, PI = 2x + 8

From part (i), PI = 20 cm

$20 = 2 x + 8 2 x = 20 - 8 = 12 x = 6 cm$

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