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# In figure 5, a triangle PQR is drawn to circumscribe a circle of radius 6 cm such the segment QT into which QR is divided by the point of contact T, are of lengths 12 cm and 9 cm respectively. If the area of $\vartriangle PQR=189\ cm^{2}$, then find the lengths of sides PQ and PR."

Given: A circle with centre o of radius 6cm, a triangle PQR circumscribing the circle, T is the point of contact and divides $\displaystyle QR\$such that $QT=12\ cm$ and $TR=9\ cm$. area of the triangle$=189 \ cm^{2}$.

To do: To find out length PQ and PR.

Solution:
Let us join OP, OM, ON, OQ and OR.

$PQ=PN+QN$

$PR=PM+MR$

As PQ and PR tangents to the circle$\ PN=PM$ and$\ QN=QT$ and $MR=TR$

Here area of the given triangle$\ =\ 189\ cm^{2}$,

base of triangle $\vartriangle PQR$,

$QR=QT+TR=12+9=21cm$

As we know area of a triangle
$=\frac{1}{2} \times base\times height$

$189=\frac{1}{2} \times 21\times PT$

$\Rightarrow PT=\frac{( 189\times 2)} {21}$

$=18$ cm

And we know that $PT=PO+OT$

$\Rightarrow PO=PT-OT$

$=18-6$               $( \because \ OT\ is\ radius\ of\ the\ circle\ given\ 6\ cm)$

$=12$ cm

And $ON=OT=OM=$radius of the circle.

In triangle $\vartriangle PON$,

$PN=\sqrt{( OP)^{2} -( ON)^{2} }$

$=\sqrt{\left(( 12)^{2} -( 6)^{2} \ \right)}$

$=\sqrt{( 144-36)}$

$=\sqrt{108}$

$=6\sqrt{3}$

$PQ=PN+QN=PN+QT=6\sqrt 3+12=22.40$ cm

And $PR=PM+MR=PN+TR=6\sqrt 3+9=19.40$ cm

$\therefore$ The lengths of sides PQ and PR are 22.40 cm and 19.40 cm respectively.

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Updated on: 10-Oct-2022

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