Step 1
Find the perimeter of the figure
we know that
the perimeter of a circle is equal to the circumference
[tex]C=2\pi r[/tex]
where
C is the circumference of the circle
r is the radius of the circle
we have
[tex]r=4\ ft[/tex]
substitute
[tex]C=2(3.14)4=25.12\ ft[/tex]
In the figure
the triangle ABC is a right triangle
Applying the Pythagoras Theorem find the length side CB
[tex]CB^{2}=AB^{2} +AC^{2}[/tex]
we have
[tex]AC=AB=4\ ft[/tex]
substitute the values
[tex]CB^{2}=4^{2} +4^{2}[/tex]
[tex]CB^{2}=32[/tex]
[tex]CB=\sqrt{32}\ ft[/tex]
[tex]CB=5.66\ ft[/tex]
The perimeter of the figure is equal to [tex]\frac{3}{4}[/tex] of the circumference plus the length side CB
so
[tex]P=\frac{3}{4}25.12+5.66=24.50\ ft[/tex]
therefore
the answer Part a) is
the perimeter of the figure is [tex]24.50\ ft[/tex]
Step 2
Find the area of the figure
we know that
the area of a circle is equal to
[tex]A=\pi r^{2}[/tex]
where
r is the radius of the circle
we have
[tex]r=4\ ft[/tex]
substitute
[tex]A=3.14(4^{2})=50.24\ ft^{2}[/tex]
the area of the triangle ABC is equal to
[tex]A=\frac{1}{2} bh[/tex]
in this problem
[tex]b=BC=\sqrt{32}\ ft[/tex]
[tex]h=BC/2=\sqrt{32}/2\ ft[/tex]
substitute the values
[tex]A=\frac{1}{2}(\sqrt{32})(\sqrt{32}/2)=8\ ft^{2}[/tex]
the area of the figure is equal to [tex]\frac{3}{4}[/tex] of the area of the circle plus the area of the triangle ABC
[tex]A=\frac{3}{4}*50.24+8=45.68\ ft^{2}[/tex]
therefore
the answer Part b) is
the area of the figure is [tex]45.68\ ft^{2}[/tex]
