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In the figure, $ A B C $ is a right triangle right-angled at $ B $ such that $ B C=6 \mathrm{~cm} $ and $ A B=8 \mathrm{~cm} $ Find the radius of its incircle."
Given:
In the figure, \( A B C \) is a right triangle right-angled at \( B \) such that \( B C=6 \mathrm{~cm} \) and \( A B=8 \mathrm{~cm} \).
To do:
We have to find the radius of its incircle.
Solution:
In right-angled triangle $ABC$,
$\angle B = 90^o, BC = 6\ cm, AB = 8\ cm$
Let $r$ be the radius of incircle whose centre is $O$ and touches the sides $AB, BC$ and $CA$ at $P, Q$ and $R$ respectively.
$AP$ and $AR$ are the tangents to the circle.
This implies,
$AP = AR$
Similarly,
$CR = CQ$ and $BQ = BP$
$OP$ and $OQ$ are radii of the circle.
$OP\ perp\ AB$ and $OQ\ \perp\ BC$ and $\angle B = 90^o$
$BPOQ$ is a square.
$BP = BQ = r$
$AR = AP = AB - BD = 8 - r$
$CR = CQ = BC - BQ = 6 - r$
$AC^2 = AB^2 + BC^2$ (By Pythagoras theorem)
$= 8^2 + 6^2$
$= 64 + 36$
$= 100$
$= (10)^2$
$AC = 10\ cm$
$AR + CR = 10\ cm$
$8 - r + 6 - r = 10$
$14 - 2r = 10$
$2r = 14 - 10$
$2r = 4$
$r = 2\ cm$
The radius of the incircle is 2 cm.