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

Please help asap!!!!!!!!!!!!!!!!!!!!!!!!-example-1

2 Answers

8 votes

Answer:

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

User Jamie Thomas
by
8.3k points
9 votes


\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 !

User Paul Panzer
by
8.7k points

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