what is the best approximation for the area of this figure?

Answer:This area can be seen as a semicircle plus a triangle. So, we will find the area of each figure and then add both to get the total area.The area of a triangle is given by the following formula:[tex]A_{triangle}=\frac{1}{2}b.h[/tex]   (1)Where [tex]b[/tex] is the base and  [tex]h[/tex] is the height. [tex]A_{triangle}=\frac{1}{2}(3)(7)[/tex]   [tex]A_{triangle}=10.5 units[/tex]   (2)Now we are going to find the area of the cemicircle, which is the half of the area of a circle:[tex]A_{cemicircle}=\frac{1}{2}\pi r^{2}[/tex]   (3)Where [tex]r[/tex] is the radius, in order to find it we have to calculate the diameter of this semicircle first, and we will do it as follows:We know the points of the ends of the diameter, which are:[tex]P_{1}:(-5,-2)[/tex] and [tex]P_{2}:(2,1)[/tex]We have to use the Pithagorean theorem to calculate the distance between both points (taking into account the x-component and the y-component of each one)[tex]c^{2} =a^{2}+ b^{2}[/tex][tex]c=\sqrt{a^{2}+b^{2}}[/tex][tex]c=\sqrt{(-5-2)^{2}+(-2-1)^{2}}[/tex][tex]c=\sqrt{58}[/tex]>>>>This is the diameter of the semicircleThen, the radius is:[tex]r=\frac{c}{2}=\frac{\sqrt{58}}{2}[/tex]Now we can use the formula written in equation (3):[tex]A_{cemicircle}=\frac{1}{2}\pi (\frac{\sqrt{58}}{2})^{2}[/tex]   [tex]A_{cemicircle}=7.25\pi units[/tex]   (4)Adding (2) and (4):[tex]A_{triangle}+A_{cemicircle}=10.5 units+7.25\pi units[/tex]   [tex]A_{triangle}+A_{cemicircle}=10.5+7.25\pi units^{2}[/tex]>>>>This is the answer