AB and CD are two diameter …
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Sia ? 6 years, 6 months ago
Area of a circle with DO as diameter {tex}= \pi r ^ { 2 } {/tex} sq. cm
{tex}= \frac { 22 } { 7 } \times \frac { 7 } { 2 } \times \frac { 7 } { 2 }{/tex}
{tex} = \frac { 77 } { 2 }{/tex} cm2
Area of semi-circle with AB as diameter {tex}= \frac { \pi r ^ { 2 } } { 2 } {/tex}
{tex}= \frac { 22 \times 7 \times 7 } { 7 \times 2 } {/tex}
{tex}= 77 \mathrm { sq } . \mathrm { cm }{/tex}
Area of {tex}\triangle A B C = \frac { 1 } { 2 } \times 14 \times 7 = 49 \mathrm { sq.cm }{/tex}
Area of shaded region = Area of circle + Area of semi-circle - Area of {tex}\triangle{/tex}ABC
{tex}= \frac { 77 } { 2 } + 77 - 49 {/tex}
{tex}= 38.5 + 77 - 49 {/tex}
{tex}=115.5 - 49{/tex}
{tex}= 66.5 \mathrm { sq.cm }{/tex}.
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