Calculate the area of the designed …

In right triangle ADC,
AC² = AD² + CD².........[By Pythagoras theorem]
= 8² + 8² = 64 + 64 = 128
{tex}\Rightarrow {/tex} AC = {tex}\sqrt {128} {/tex} {tex}\Rightarrow {/tex} AC = {tex}8\sqrt 2 {/tex}
Draw BM {tex} \bot {/tex} AC

Then, AM = MC = {tex}\frac{1}{2}AC{/tex}
= {tex}\frac{1}{2}\left( {8\sqrt 2 } \right) = 4\sqrt 2 {/tex} cm
In right triangle AMB
AB² = AM² + BM² ...........By Pythagoras theorem]
{tex}\Rightarrow {/tex} {tex}{\left( 8 \right)^2} = {\left( {4\sqrt 2 } \right)^2} + B{M^2}{/tex}
{tex}\Rightarrow {/tex} 64 = 32 + BM²
{tex}\Rightarrow {/tex} BM² = 64 - 32
{tex}\Rightarrow {/tex} BM² = 64 - 32
{tex}\Rightarrow {/tex} BM² = 32
{tex}\Rightarrow {/tex} BM ={tex}\sqrt {32} {/tex} = {tex}4\sqrt 2 {/tex} cm
{tex}\therefore {/tex} Area of {tex}\triangle {/tex} ABC={tex}\frac{{AC \times BM}}{2}{/tex}
= {tex}\frac{{8\sqrt 2 \times 4\sqrt 2 }}{2}{/tex} = 32 cm²
{tex}\therefore {/tex} Shaded Area = {tex}\frac{{90}}{{360}}\pi {(8)^2} - 32{/tex}
= {tex}16\pi - 32{/tex}
= {tex}16 \times \frac{{22}}{7} - 32 = \frac{{352}}{7} - 32{/tex}
= {tex}\frac{{352 - 224}}{7} = \frac{{128}}{7}\,c{m^2}{/tex}
{tex}\therefore {/tex} Area of the designed region
= {tex}2 \times \frac{{128}}{7} = \frac{{256}}{7}\,c{m^2}{/tex}