Question

In the figure, the radius of the partial circle with center A is 4 feet.

Answer 1

1)24.50 ;))))))))))
2)45.68 ;)))))))))

Answer 2

Step 1

Find the perimeter of the figure

we know that

the perimeter of a circle is equal to the circumference

$C=2\pi r$

where

C is the circumference of the circle

r is the radius of the circle

we have

$r=4\ ft$

substitute

$C=2(3.14)4=25.12\ ft$

In the figure

the triangle ABC is a right triangle

Applying the Pythagoras Theorem find the length side CB

$CB^{2}=AB^{2} +AC^{2}$

we have

$AC=AB=4\ ft$

substitute the values

$CB^{2}=4^{2} +4^{2}$

$CB^{2}=32$

$CB=\sqrt{32}\ ft$

$CB=5.66\ ft$

The perimeter of the figure is equal to $\frac{3}{4}$ of the circumference plus the length side CB

so

$P=\frac{3}{4}25.12+5.66=24.50\ ft$

therefore

the answer Part a) is

the perimeter of the figure is $24.50\ ft$

Step 2

Find the area of the figure

we know that

the area of a circle is equal to

$A=\pi r^{2}$

where

r is the radius of the circle

we have

$r=4\ ft$

substitute

$A=3.14(4^{2})=50.24\ ft^{2}$

the area of the triangle ABC is equal to

$A=\frac{1}{2} bh$

in this problem

$b=BC=\sqrt{32}\ ft$

$h=BC/2=\sqrt{32}/2\ ft$

substitute the values

$A=\frac{1}{2}(\sqrt{32})(\sqrt{32}/2)=8\ ft^{2}$

the area of the figure is equal to $\frac{3}{4}$ of the area of the circle plus the area of the triangle ABC

$A=\frac{3}{4}*50.24+8=45.68\ ft^{2}$

therefore

the answer Part b) is

the area of the figure is $45.68\ ft^{2}$