Question

Please help asap!!!!!!!!!!!!!!!!!!!!!!!!

Answer 1

30.96

Explanation:

First you must find the diameter of the circle which is 12

Then you must use the formula A=πr²

So it should be = to A = 3.14 ×144 which is 30.96

Answer 2

`\huge \tt \color{pink}{A}\color{blue}{n}\color{red}{s}\color{green}{w}\color{grey}{e}\color{purple}{r }`

`\large\underline{ \boxed{ \sf{✰\:Notes }}}`

★ Concept ★

• ➣ A square with sides 12m
• ➣then a circle in between
• ➣nd it's diameter is 12m bcz of side of square is equal to radius
• ➣soo here things we to find is the area of circle and then square then by subtracting their will give us the area of shaded part in image

`\rule{70mm}{2.9pt}`

`\large\underline{ \boxed{ \sf{✰\: formule\: used\:✰ }}}`

`{ \boxed{✟\underline{ \boxed{ \sf{\: Area \: of \: circle = \pi {r}^(2) \: }}}✟}}`

`{ \boxed{✟\underline{ \boxed{ \sf{\: Area \: of \: square = side * side \: or \: ( {s}^(2)) \: }}}✟}}`

`\rule{70mm}{2.9pt}`

✛ 1st taking out the the area of circle✛

• ➣ value of pi = 3.14
• ➣ diameter = 12m (have to find radius we know that radius is half of diameter) which is

`\qquad \rm{➛ \: radius(r) = (12)/(2) } \\ \\ \qquad \rm{➛ \: radius(r) = \frac{ \cancel{12}}{ \cancel2} } \\ \\ \qquad \rm{➛ \: radius(r) =6m}`

• ★ Hence radius of circle is 6m

★ let's substitute values now ★

`\\ \qquad \rm{➛Area \: of \: circle = 3.14 * {6}^(2) \:} \\ \\ \qquad \rm{➛Area \: of \: circle = \:3.14 * 36} \\ \\ \qquad \rm{➛Area \: of \: circle = \:113.04 {m}^(2) }`

`\rule{70mm}{2.9pt}`

✛ now finding the area of square✛

• ➣ here two given side are 12m

★ let's substitute values now ★

`\\ \qquad \rm{➛Area \: of \:square = 12×12} \\ \\ \\ \qquad \rm{➛Area \: of \:square = 144 {m}^(2) }`

✞ Now according to given ques we have to find area of shaded part soo let's solve ! ✞

`\\ \rm{➛Area \: of \:shaded \: part = \: area \: of \: circle - area \: of \: squre} \\`

★ Let's substitute value ★

`\\ \rm{➛Area \: of \:shaded \: part =113.04-144} \\ \\ \qquad \rm{➛Area \: of \:shaded \: part =30.96 {m}^(2) }`

`\rule{70mm}{2.9pt}`

★ Hence area of shaded part =

`{ \boxed{✜\underline{ \boxed{ \sf{ \: 30.96 {m}^(2) \green✓\: }}}✜}}`

★ note here we neglect minus (-)

`\rule{70mm}{2.9pt}`

Hope it helps !