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puteri [66]
4 years ago
7

Can someone help me with this?

Mathematics
1 answer:
Andrei [34K]4 years ago
3 0
Check the picture below.

notice, the figure is really just a triangle on top of a semi-circle.

now, the triangle has a base and height of 3 each, and the semi-circle has a diameter of 3√(2), so its radius is half that.

\bf \begin{array}{llll}
\textit{area of a circle}\\\\
A=\pi r^2
\end{array}\qquad 
\begin{array}{llll}
\textit{area of a semi-circle}\\\\
A=\cfrac{\pi r^2}{2}
\end{array}\qquad 
\begin{array}{llll}
\textit{area of a triangle}\\\\
A=\cfrac{1}{2}bh
\end{array}\\\\
-------------------------------\\\\

\bf d=3\sqrt{2}\qquad r=\cfrac{3\sqrt{2}}{2}\impliedby \textit{radius of the semi-circle}\\\\
-------------------------------\\\\
\stackrel{\textit{area of triangle}}{\cfrac{3\cdot 3}{2}}+\stackrel{\textit{area of semi-circle}}{\cfrac{\pi \left( \frac{3\sqrt{2}}{2} \right)^2}{2}}\implies \cfrac{9}{2}+\cfrac{\frac{\pi \cdot 3^2\cdot 2}{2}}{2}\implies \cfrac{9}{2}+\cfrac{9\pi }{2}
\\\\\\
\cfrac{9+9\pi }{2}

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