Question

Can someone help me with this?

Answer 1

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.

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

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