A backyard is in the shape of a triangle ABC with right angle . . .

Question : 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 AP = x m. 

Based on the above information, answer the following questions : 

(i) Find the length of AR in terms of x. [1 Marks]

(ii) Write the type of quadrilateral BQOR. [1 Marks]

(iii) 

(a) Find the length PC in terms of x and hence find the value of x. [2 Marks]

OR

(b) Find x and hence find the radius r of the circle. [2 Marks]

CBSE Delhi 2024 (30/1/1)

Solution : 

This is a case study based question from circles.

In Δ ABC
∠ABC=90°
AB=7 m
BC=15 m

AP=x m

(i) Find the length of AR in terms of x. 
Ans : 

AP=x m (Given)

AP=AR (Length of tangent drawn from an external point to a circle are equal in length)
∴ AR = x m

(ii) Write the type of quadrilateral BQOR.

Ans :

In BQOR
BR=BQ (Length of tangent drawn from an external point to a circle are equal in length)

OR=OQ (Radii of the same circle)

∠ABC=90° (Given)
∠BRO=∠BQO=90° (Tangent at any point of the circle is perpendicular to the radius through the point of contact)

∠ROQ=90° (By ASP of a quadrilateral)

∵ All sides are equal and all angles are of 90°.

∴ BQOR must be a square.

(iii) 

(a) Find the length PC in terms of x and hence find the value of x.

Ans : 
We know, length of tangent drawn from an external point to a circle are equal in length

AP=AR = x 
BR=BQ = (7-x) cm
PC=QC
= 15-(7-x)
= (15-7+x) cm
= (8+x) cm

∴ PC=(8+x) cm

In Rt. ΔABC

AC²=AB²+BC² (By Pythagoras Theorem)
AC²=(7)²+(15)²
AC²=49+225
AC²=274
AC=√274
AC=√274 cm 
AP+PC=√274
x+8+x=√274
2x+8=√274
2x=(√274)-8
x=[(√274)-8]/2 cm

OR

(b) Find x and hence find the radius r of the circle. 

Ans :
x=[(√274)-8]/2 cm

Radius (r)
= BQ=BR 

= (7-x) cm 
= 7-{(√274)-8]/2} cm