Answer:
Part a) [tex]114.3\ yd^{2}[/tex]
Part b) [tex]49.5\ yd^{2}[/tex]
Step-by-step explanation:
Part a) we know that
The area of the shaded figure is equal to the area of semicircle plus the area of a triangle
Find the area of semicircle
The area of a semicircle is equal to
[tex]A=\frac{1}{2}\pi r^{2}[/tex]
we have
[tex]r=5\ yd[/tex]
substitute
[tex]A=\frac{1}{2}\pi (5)^{2}=12.5 \pi\ yd^{2}[/tex]
Find the area of triangle
The area of the triangle is equal to
[tex]A=\frac{1}{2}(b)(h)[/tex]
we have
[tex]b=15\ yd[/tex]
[tex]h=5(2)=10\ yd[/tex] -----> the height of triangle is equal to the diameter of semicircle
substitute
[tex]A=\frac{1}{2}(15)(10)=75\ yd^{2}[/tex]
Find the area of the shaded figure
[tex]12.5 \pi\ yd^{2}+75\ yd^{2}=(12.5 \pi+75)\ yd^{2}[/tex]
assume
[tex]\pi=3.14[/tex]
[tex](12.5(3.14)+75)=114.3\ yd^{2}[/tex]
Part b) we know that
The area of the shaded figure is equal to the area of the triangle minus the area of the circle
Find the area of the circle
The area of the circle is equal to
[tex]A=\pi r^{2}[/tex]
we have
[tex]r=5\ yd[/tex]
substitute
[tex]A=\pi (5)^{2}=25 \pi\ yd^{2}[/tex]
Find the area of triangle
The area of the triangle is equal to
[tex]A=\frac{1}{2}(b)(h)[/tex]
we have
[tex]b=16\ yd[/tex]
[tex]h=16\ yd[/tex]
substitute
[tex]A=\frac{1}{2}(16)(16)=128\ yd^{2}[/tex]
Find the area of the shaded figure
[tex]128\ yd^{2}-25 \pi\ yd^{2}=(128-25 \pi)\ yd^{2}[/tex]
assume
[tex]\pi=3.14[/tex]
[tex](128-25(3.14))=49.5\ yd^{2}[/tex]
