Answer:
The answer is the option
[tex](10.5+7.25\pi)\ units^{2}[/tex]
Step-by-step explanation:
we know that
the area of the figure is equal to the area of semicircle plus the area of a triangle
Find the area of semicircle
we know that
The area of semicircle is equal to
[tex]A1=\frac{1}{2}\pi r^{2}[/tex]
Let
[tex]A(-5,-2), B(2,1)[/tex]
The distance AB is equal to the diameter of semicircle
the formula to calculate the distance between two points is equal to
[tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex]
substitute the values
[tex]d=\sqrt{(1+2)^{2}+(2+5)^{2}}[/tex]
[tex]d=\sqrt{(3)^{2}+(7)^{2}}[/tex]
[tex]dAB=\sqrt{58}\ units[/tex] -----> diameter of the semicircle
so the radius is equal to
[tex]r=0.5\sqrt{58}\ units[/tex]
find the area of semicircle
[tex]A1=\frac{1}{2}\pi (0.5\sqrt{58})^{2}[/tex]
[tex]A1=\frac{1}{8}\pi (58)[/tex]
[tex]A1=7.25\pi\ units^{2}[/tex]
Find the area of the triangle
[tex]A2=\frac{1}{2}bh[/tex]
we have
Observing the figure
[tex]b=7\ units[/tex]
[tex]3=3\ units[/tex]
substitute
[tex]A2=\frac{1}{2}*7*3=10.5\ units^{2}[/tex]
Find the area of the figure
[tex]A=A1+A2[/tex]
substitute
[tex]7.25\pi\ units^{2}+10.5\ units^{2}\\ \\(10.5+7.25\pi)\ units^{2}[/tex]
