Question

Given : $ED=14cm$

Answer 1

R.E.F image

Area of shaded region

$ED=14cm$ (Given)

$ar\left(△EBD\right)=ar\left(△ACDE\right)-ar\left(△ABE\right)-ar\left(△BED\right)$...(1)

$ED=AC=14cm=2AB$ (Since $AB=BC$)

$\Rightarrow AB=BC=7cm=AE=CD$

$\therefore ar\left(ACDE\right)=AC\times CD=14\times 7\Rightarrow 98c{m}^2$

$ar\left(△ABE\right)=\frac{1}{2}\times AB\times BC=\frac{1}{2}\times 7\times 7=\frac{49}{2}c{m}^2$

$ar\left(△BCD\right)=\frac{1}{2}\times BC\times CD=\frac{1}{2}\times 7\times 7=\frac{49}{2}c{m}^2$

$\therefore ae\left(△EBD\right)=98-\frac{49}{2}-\frac{49}{2}=49c{m}^2$

$ar\left(semicircle\right)\frac{\pi r^2}{2}=\frac{\pi }{2}(\frac{7}{2}{)}^2=\frac{\pi }{2}\times (\frac{14}{2}{)}^2$

$=\frac{22}{7}\times \frac{1}{2}\times 7\times 7$

$area\left(semicircle\right)=77c{m}^2)$

$\therefore$ area of shaded region $=\left(1\right)+\left(2\right)$

$\therefore 77+49$

$\therefore 126c{m}^2$