Question

ABCD is a quadrilateral such that ∠D = 90°. A circle with centre O and radius r, touches the side AB, BC, CD and DA at P, Q, R and S respectively. If BC = 40 cm, CD = 25 cm and BP = 28 cm, find r.

Answer 1

Answer : r=14 cm

Step-by-step explanation:

So, BP = BQ

(Tangents from external point B)

But BP = 27 cm

⇒ BQ = 27 cm

It is given that BC = 38 cm

⇒ BQ + CQ = 38

⇒ 27 + CQ = 38

⇒ CQ = 11 cm

⇒ CQ = CR (Tagents from an external point C)

But CQ = 11 cm

⇒ CR = 11 cm

It is given that : CD = 25 cm

⇒ CR + DR = 25

⇒ 11 + DR = 25

⇒ DR = 14 cm

Since, tangent to a circle is perpendicular to the radius through the point of contact.

∴ ∠ORD = ∠OSD = 90°

It is given that

∠D = 90°

Now, in quadrilateral ORDS,

∠ORD = ∠OSD = ∠RDS = ∠ROS = 90°

and OR = OS [radii of circle]

Therefore, ORDS is a square

So, OR = DR = 14 cm

Hence r = 14 cm.