Question

In figure ABCD is quardant of a circle with radius 28 cm and a semicircle BEC is drawn with BC as diameter. Find the area of the shaded region.

Answer 1

In $△$ BAC, by pythagoras theorem,

$B{C}^2=A{B}^2+A{C}^2$

$B{C}^2=784+784$

$BC=28\sqrt{2}$ $cm$

$\frac{BC}{2}=14\sqrt{2}$ $cm$

Now,

$A=ar\left(BDCEB\right)$

$A=ar\left(BCEB\right)-ar\left(BCDB\right)$

$A=ar\left(BCEB\right)-\left[ar\right(BACDB)-ar(BAC\left)\right]$

$A=\left[\frac{1}{2}\right(\frac{22}{7}\times \left(14\sqrt{2}{)}^2\right)-(\frac{1}{4}\times \frac{22}{7}\times {28}^2-\frac{1}{2}\times 28\times 28)]$

$A=[\frac{1}{2}\times \frac{22}{7}\times 196\times 2-\frac{1}{4}\times \frac{22}{7}\times 28\times 28+\frac{1}{2}\times 28\times 28]$

$A=616-616+392=392$ $c{m}^2$