Area inside the semi-circle and outside the triangle is (91.125π - 120) in²
Solution:
Base of the triangle = 10 in
Height of the triangle = 24 in
Area of the triangle = [tex]\frac{1}{2} bh[/tex]
[tex]$=\frac{1}{2} \times 10\times24[/tex]
Area of the triangle = 120 in²
Using Pythagoras theorem,
[tex]\text{Hypotenuse}^2=\text{base}^2+\text{height}^2[/tex]
[tex]\text{Hypotenuse}^2=10^2+24^2[/tex]
[tex]\text{Hypotenuse}^2=100+576[/tex]
[tex]\text{Hypotenuse}^2=676[/tex]
Taking square root on both sides, we get
Hypotenuse = 23 inch = diameter
Radius = 23 ÷ 2 = 11.5 in
Area of the semi-circle = [tex]\frac{1}{2}\pi r^2[/tex]
[tex]$=\frac{1}{2} \pi \times (13.5)^2[/tex]
Area of the semi-circle = 91.125π in²
Area of the shaded portion = (91.125π - 120) in²
Area inside the semi-circle and outside the triangle is (91.125π - 120) in².
