Given an irregular figure.
The figure consists of a right triangle and a semi-circle
The triangle has a base = b = 12 ft
And a height = h = 6 ft
Area of the triangle =
[tex]\frac{1}{2}bh=\frac{1}{2}\cdot12\cdot6=36ft^2[/tex]
The semi-circle has a radius = r = 6 ft
The area of the semi-circle =
[tex]\frac{1}{2}\cdot\pi\cdot r^2=\frac{1}{2}\cdot3.14\cdot6^2=\frac{1}{2}\cdot3.14\cdot36=56.52ft^2[/tex]
So, the total area =
[tex]36+56.52=92.52[/tex]
Rounding to the nearest tenth
So, the answer will be area = 92.5 ft^2
