Question

Someone answer. Will give brainlest

Answer 1

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

                                $=\frac{1}{2} \times 10\times24$

Area of the triangle = 120 in²

Using Pythagoras theorem,

$\text{Hypotenuse}^2=\text{base}^2+\text{height}^2$

$\text{Hypotenuse}^2=10^2+24^2$

$\text{Hypotenuse}^2=100+576$

$\text{Hypotenuse}^2=676$

Taking square root on both sides, we get

Hypotenuse = 23 inch = diameter

Radius = 23 ÷ 2 = 11.5 in

Area of the semi-circle = $\frac{1}{2}\pi r^2$

                                      $=\frac{1}{2} \pi \times (13.5)^2$

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.