Question

in the adjoining figure,PQR is a right angle traingle with PQ=12 cm and QR=5cm. A circle with cente O and radius x is inscribed in triangle PQR. Find the value of x.

Answer 1

Here is ur ans

PR^2 = QR^2 + PQ^2
= 144 + 25
= 169
-> PR = 13 cm.

Let the circle touch sides PQ, QR and PR at A, B and C respectively.

Let PA = x,
BQ = y (Radius),
CR = z

OAQB is a square as it has four right angles and has two adjacent sides (Radii OA and OB) equal.

Using tangents from an external point to a circle are equal,
PC = PA = x
BQ = QA = y (Radius)
CR = BR = z

x + z = 13 ----- (1)
x + y = 5 ----- (2)
y + z = 12 ----- (3)

Adding,
2(x + y + z) = 30
x + y + z = 15
13 + y = 15 (From (1))
y = 2

The radius of the incircle is 2 cm.

Answer 2

$\sf\red{PR=\sqrt{ {PQ}^{2} + {QR}^{2} } }$

$\longrightarrow$$\sf\red{\sqrt{ {12}^{2} + {5}^{2} }}$

$\longrightarrow$$\sf\red{13cm}$

$\sf\green{Area\:(∆ABC)=\frac{1}{2}×PQ×QR}$

$\sf\green{\frac{1}{2}×12×5={30cm}^{2}}$

$\sf\green{s=\frac{12+5+13}{2}=15cm}$

$\therefore$ $\sf{Inradius(r)=\frac{∆}{s}=\frac{30}{15}}$

$\longrightarrow$$\sf{2cm}$