$( A)\ 11$
$( B)\ 10$
$( C)\ 14$
$( D)\ 15$"
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In fig. the sides AB, BC and CA of a triangle ABC, touches a circle at P, Q and R respectively. If $PA=4$ cm, $BP=3$ cm and $AC=11$ cm, then the length of BC$( in\ cm)$ is: 

$( A)\ 11$
$( B)\ 10$
$( C)\ 14$
$( D)\ 15$"

Given: A triangle ABC touching a circle on P, Q and R, 
$AP=4\ cm$, $BP=3\ cm$ and $AC=11\ cm$.
 
To do: To find out the length of BC.

Solution: here $\vartriangle ABC$ touches a circle on P, Q and R.

$\therefore $ AP and AR, BP and BQ and CQ and CR are tangents to the circle from the points A, B and C.

And we know that tangents drawn from an external point to the circle are equal in length.

$\therefore \ AP=AR\ \dotsc \dotsc \dotsc ( 1)$

$BP=BQ\ \ \dotsc \dotsc \dotsc \dotsc ( 2)$

$CQ=CR\ \ \ \dotsc \dotsc \dotsc \dotsc .( 3)$

And here given $PA=4\ cm,\ BP=3\ cm$ and $AC=11\ cm$

From $( 1)$

$AP=AR=4\ cm$

And $AC=11\ cm\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ( given)$

$\Rightarrow CR=AC-AR=11-4=7\ cm$

$\Rightarrow CQ=7\ cm\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ( \because \ CR=CQ)$

$And\ BP=BQ\ from\ ( 2)$

$\therefore \ BQ=3\ cm$

$\therefore \ BC=BQ+CQ=3+7=10\ cm$

$\therefore$ Option $( B)$ is correct.

 

Updated on: 10-Oct-2022

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