Question

In the figure 35, angle ADC = 90 degrees, BC=38cm, CD=28cm and BP=25cm. Find the radius of the circle.

Answer 1

Given Angle ADC = 90° 

BC = 38cm  

CD = 28cm 

BP = 25cm 

Join OR and OS 

Angle D = 90° 

OS is perpendicular to DA 

= Angle OSD = 90° 

Now OR is perpendicular to CD 

Angle ORD = 90° 

ORDS is square as each angle is 90° 

Now from B:BP = BQ = 25cm 

CQ= BC- BQ 

=38-25=13cm 

From C:CQ = CR = 13cm 

RD = DC – CR 

RD = 28-13 = 15cm 

As ORDS is square has all sides equal 

So RD = RO = OS = DS = 15cm 

Radius = OS = 15cm

Answer 2

Since tangent to a circle is perpendicular to the radius through the point of contact.
 
∴ ∠OSD = ∠ORD = 90°,
OR = OS
⇒ DROS is a square.
Also, BP = BQ [Tangents from an external point are equal]

⇒ BQ = 25 cm                     [BP = 25 cm]
⇒ BC – CQ = 25
⇒ 38 – CQ = 25                  [BC = 38 cm]
⇒ CQ = 38 – 25 = 13 cm
⇒ CR = CQ = 13                 [CQ = 13 cm]
⇒ CD – DR = 13                 [CR = CD – DR]
⇒ 28 – DR = 13                  [CD = 28 cm]
⇒ DR = 28 – 13 = 15 cm
Since DROS is a square,
so OR = DR = 15 cm.
 Hence, radius of the circle = 15 cm.