Question

$ABCP$ is a quadrant of a circle of radius $14cm$. With $AC$ as diameter, a semicircle is drawn. Find the area of the shaded portion.

• A. $47c{m}^2$
• B. $72c{m}^2$
• C. $98c{m}^2$
• D. None of these
  

Answer 1

The correct option is C $98c{m}^2$

In right angled triangle ABC,

we have
$A{C}^2=A{B}^2+B{C}^2$
$A{C}^2={14}^2+{14}^2$
$AC=\sqrt{2\times {14}^2}=14\sqrt{2}$ $cm$

Now required Area
=Area $APCQA$
=Area $ACQA$ - Area $ACPA$
=Area $ACQA$ - (Area $ABCPA$ - Area of $\Delta$ $ABC$)

$=\frac{1}{2}\times \pi \times {\left(\frac{14\sqrt{2}}{2}\right)}^2-\left[\frac{1}{4}\times \pi (14{)}^2-\frac{1}{2}\times 14\times 14\right]$

$=\frac{1}{2}\times \frac{22}{7}\times 7\sqrt{2}\times 7\sqrt{2}-\frac{1}{4}\times \frac{22}{7}\times 14\times 14+\frac{1}{2}\times 14\times 14$

$=154-154+98$

$=98c{m}^2$