The curved part of this figure is a semicircle.What is the best approximation for the area of this figure?24+18.25π units²12+9.125π units²12+18.25π units²24+9.125π units²

Accepted Solution

Draw the diameter from point (-1, -5) to point (2, 3).

The triangle has base 3 and height 8.

The diameter of the semicircle is the hypotenuse of the triangle.

[tex] d = \sqrt{3^2 + 8^2} [/tex]

[tex] d \approx 8.544 [/tex]

The radius is half the diameter.

[tex] r = \dfrac{d}{2} = \dfrac{8.544}{2} = 4.272 [/tex]

The total area is the area of the triangle plus the area of the semicircle.

[tex] A = \dfrac{bh}{2} + \dfrac{\pi r^2}{2} [/tex]

[tex] A = \dfrac{3 \times 8}{2} + \dfrac{\pi \times 4.272^2}{2} [/tex]

[tex] A = 12 + 9.125 \pi [/tex] [tex] units^2 [/tex]