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Calculate the area of the shaded portion.
"
Given :
PS $= 12$ cm, QS $= 16$ cm, QR $= 48$ cm, PR $= 52$ cm and ∠PSQ $= 90°$.
To find :
We have to find the area of the shaded portion.
Solution :
Area of triangle PQR $=$ Area of triangle PQS$+$Area of the shaded portion.
In the triangle PQS,
$ PQ^{2} =PS^{2} +QS^{2}$
$=( 12)^{2} +( 16)^{2} \ cm^{2}$
$=144+256\ cm^{2}$
$=400\ cm^{2}$
$PQ=\sqrt{400} \ cm$
$PQ=20\ cm $
We know that,
Area of a triangle of sides of lengths a, b and c is $\sqrt{s( s-a)( s-b)( s-c)}$
where $s=\frac{a+b+c}{2}$
Therefore,
Area of the triangle PQR $=\sqrt{60( 60-52)( 60-48)( 60-20)}$
$where\ s=\frac{52+48+20}{2} =\frac{120}{2} =60$
Area of PQR
$ =\sqrt{60( 8)( 12)( 40)}$
$=\sqrt{3\times 5\times 4\times 4\times 2\times 4\times 3\times 5\times 2\times 4}$
$=\sqrt{( 3\times 5\times 4\times 4\times 2)^{2}}$
$=\sqrt{( 480)^{2}}\ =480\ cm^{2} $
Area of the right angled PQS $=\frac{1}{2}\times16\times12 cm^2$
$=8\times12 cm^2$
$=96 cm^2$
Area of the shaded portion$=$Area of triangle PQR$-$Area of triangle PQS
$=(480-96) cm^2$
$=384 cm^2$
The area of the shaded portion is $384 cm^2$.
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