Question

The curved part of this figures is a semicircle.

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


a) 28+16.25π units²

b) 14+8.125π units²

c) 28+8.125π units²

d) 14+16.25π units²

Answer 1

Option b: $(14+8.125 \pi) \text { units }^{2}$ is the area of the figure.

Explanation:

The area of the figure is equal to the area of the triangle plus area of the semicircle.

Area of the triangle:

The formula for area of the triangle is given by

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

where $b=(3-(-4))=7 \text { units }$

and $h=(2-(-2))=4 \text { units }$

Substituting the values, we have,

$A=\frac{1}{2}(7\times4)=14units^2$

Thus, the area of the triangle is $14 u n i t s^{2}$

Area of the semicircle:

The formula for area of the semicircle is given by

$A=\frac{1}{2} \pi r^{2}$

We shall determine the radius using the coordinates $(-4,-2)$ and $(3,2)$

Substituting these coordinates in the equation $d=\sqrt{(y_2-y_1)^{2}+(x_2-x_1)^{2}}$ , we get,

$d=\sqrt{(2+2)^{2}+(3+4)^{2}}$

$d=\sqrt{(4)^{2}+(7)^{2}}$

$d=\sqrt{65} \text { units }$

where d is the diameter.

The radius r is given by $r=\left\frac{\sqrt{65} }{2}\right\text { units }$

Substituting the value of r in the formula $A=\frac{1}{2} \pi r^{2}$, we get,

$A=\frac{1}{2} \pi\left\left(\frac{\sqrt{65}}{2}\right) ^{2}$

$A=\frac{65}{8} \pi \text { units }^{2}$

$A=8.125 \pi \text { units }^{2}$

Thus, the area of the semicircle is $8.125 \pi units ^{2}$

Area of the figure = Area of the triangle + Area of semicircle

Area of the figure = $14 units ^{2}+8.125 \pi units ^{2} =( 14+8.125 \pi )units ^{2}$

Thus, the area of the figure is $(14+8.125 \pi) \text { units }^{2}$

Hence, Option b is the correct answer.