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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