# In a right triang

Given:

BC = 12 cm

AB = 5 cm

Property 1: If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

Property 2: The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

Property 3: Sum of all angles of a quadrilateral = 360°.

By property 1,

AP = AQ (Tangent from A)

BP = BR (Tangent from B)

CR = CQ (Tangent from C)

ABC is a right-angled triangle,
by Pythagoras Theorem

AC2 = AB2 + BC2

AC2 = 52 + 122

AC2 = 25 + 144

AC2 = 169

AC = 169

AC = 13 cm

Clearly,

AQ + QC = AC
= 13 cm

AP + RC = 13 cm [ AQ = AP and QC = RC]

Also,

AB + BC = 5 cm + 12 cm
= 17 cm

AP + PB + BR + RC = 17 cm
[
AB = AP + PB and BC = BR + RC]

AP + RC + PB + BR = 17 cm

13 cm + BR + BR = 17 cm
[
AP + RC = 10 cm and PB = BR]

13 cm + 2BR = 17 cm

2BR = 17 cm 13 cm
= 4 cm

BR = 2 cm

Now,

BPO = 90° [By property 2]

BRO = 90° [By property 2]

PBM = 90° [Given]

Now by property 3,

BPO + BRO + PBM + ROP = 360°

ROP = 360° - (BPO + BRO + PBM)

ROP = 360° - (90° + 90° + 90°)

ROP = 360° - 270°

ROP = 90°

Now, ROP = 90° and BP = BR which are adjacent sides

PO = BR = 2 cm