Question

This figure is made up of a quadrilateral and a semicircle.

Answer 1

Answer:

The correct option is B. The area of the figure is 40.4 units².

Step-by-step explanation:

The line AB divides the figure in two parts one is a rectangle and another is semicircle.

The distance formula is

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

The length of AB is

$AB=\sqrt{(2+3)^2+(4-2)^2}=\sqrt{25+4}=\sqrt{29}$

The length of AD is

$AD=\sqrt{(-1+3)^2+(-3-2)^2}=\sqrt{4+25}=\sqrt{29}$

Since AB=AD, therefore ABCD is a square. The area of the of square is

$A_1=a\times a=\sqrt{29}\times \sqrt{29} =29$

The area of square is 29 units².

The area of a semicircle is

$A_2=\frac{\pi}{2}r^2$

Since AB is the diameter of the semicircle, therefore the radius of the semicircle is

$r=\frac{d}{2}=\frac{\sqrt{29}}{2}$

The area of the semicircle is

$A_2=\frac{3.14}{2}(\frac{\sqrt{29}}{2})^2=40.3825$

The area of the figure is

$A=A_1+A_2=29+11.2825=40.3825\approx 40.4$

Therefore the area of the figure is 40.4 units². Option B is correct.