Question

adjacent figure is triangular in shape and is semi circular at the bottom .find the total area of this figure

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Answer 1

Answer:

answer for the given problem is given

Answer 2

$\color{red}\huge{\underline{\underline{\bf GIVEN\dag}}}$

• $l = 10m$
• $r = 6m$

$\color{red}\huge{\underline{\underline{\bf SOLUTION \dag}}}$

$\tt About \: triangle:$

$\blue{ \boxed{\bf area _{ \:triangle}= \frac{1}{2} bh}}$

$\tt Here,$

• $\bf b = CB = 12m$
• $\bf h = ?$

$\tt by \: pythagora \: theorem :$

${l}^{2} = {r}^{2} + {h}^{2}$

${10}^{2} = {6}^{2} + {h}^{2}$

${h}^{2} = 100 - 36$

$h = \sqrt{64}$

$\therefore h = 8m$

$\implies area = \frac{1}{2} (12)(8)$

$\therefore area_{ \: triangle} = 48 {m}^{2}$

$\tt About \: semi-circle :$

$\blue{ \boxed{\bf area_{semi-circle} = \frac{\pi {r}^{2} }{2}} }$

$\implies area = \frac{\pi {(6)}^{2} }{2}$

$\implies area = \frac{\pi \times 36}{2}$

$\therefore \bf area_{semi-circle} =56.52 {m}^{2}$

$\tt \: area \: of \: figure :$

$\bf area= area_{semi-circle} + area_{triangle}$

$area = 48 {m}^{2} + 56.52 {m}^{2}$

$\orange{\boxed{\bf area\:of\:figure=104.52m^{2}}}$

HOPE THIS HELPS!!!