Question

Calculate the area of the shaded portion as shown in figure.

• A. $384c{m}^2$
• B. $284c{m}^2$
• C. $304c{m}^2$
• D. None of the above

Answer 1

The correct option is A $384c{m}^2$

In right $\Delta PSQ$, by Pythagoras Theorem
$P{Q}^2=P{S}^2+S{Q}^2=(12{)}^2+(16{)}^2$
$=144+256=400cm$
$\therefore PQ=\sqrt{400}cm=20cm$
Now, $a=20cm,b=48cm$ and $c=52cm$
$\therefore s=\frac{a+b+c}{2}=\frac{20cm+48cm+52cm}{2}$
$=\frac{120cm}{2}=60cm$
Area of $\Delta PQR=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{60(60-20)(60-48)(60-52)}$
$=\sqrt{60\times 40\times 12\times 8}$
$=\sqrt{6\times 10\times 10\times 4\times 6\times 2\times 8}$
$=6\times 10\times 8=480c{m}^2$
and the area of $\Delta PQS=\frac{1}{2}\times PS\times QS$
$=\frac{1}{2}\times 12\times 16=96c{m}^2$
$\therefore$ area of the shaded portion of the triangle
$=480-96=384c{m}^2$