In fig ΔABC is a right …
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Sia ? 4 years, 6 months ago
In triangle ABC, by pythagoras theorm,we have,
{tex}B C = \sqrt { A B ^ { 2 } + A C ^ { 2 } }{/tex}
{tex}B C = \sqrt { 9 + 16 }{/tex}
{tex}B C = \sqrt { 25 }{/tex}
= 5 cm
Ar(shaded part) = Ar({tex}\triangle ABC{/tex}) + Ar(semicircle APB) + Ar.( semicircle AQC ) - Ar.(semicircle BAC)
{tex}=\left( \frac { 1 } { 2 } \times 3 \times 4 \right) + \left( \frac { 1 } { 2 } \pi \times 1.5 \times 1.5 \right) + \left( \frac { 1 } { 2 } \pi \times 2 \times 2 \right) - \left( \frac { 1 } { 2 } \pi \times 2.5 \times 2.5 \right){/tex}
{tex}= 6 + \frac { 1 } { 2 } \pi \left( 4 + \frac { 9 } { 4 } - \frac { 25 } { 4 } \right){/tex}
= 6 + 0
= 6 cm2
0Thank You