Question

PLEASE PLEASE ANSWER!

Answer 1

The area of the shaded region is $8(\pi \ -\sqrt{3})\ \text{cm}^2$.

Solution:

Given radius = 4 cm

Diameter = 2 × 4 = 8 cm

Let us first find the area of the semi-circle.

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

                                      $=\frac{1}{2}\times \pi\times 4^2$

                                      $=\frac{1}{2}\times \pi\times 16$

Area of the semi-circle = $8\pi$ cm²

Angle in a semi-circle is always 90º.

∠C = 90°

So, ABC is a right angled triangle.

Using Pythagoras theorem, we can find base of the triangle.

$AC^2+BC^2=AB^2$

$AC^2+4^2=8^2$

$AC^2=64-16$

$AC^2=48$

$AC=4\sqrt{3}$ cm

Base of the triangle ABC = $4\sqrt{3}$ cm

Height of the triangle = 4 cm

Area of the triangle ABC = $\frac{1}{2}\times b \times h$

                                          $=\frac{1}{2}\times 4\sqrt{3} \times 4$

Area of the triangle ABC =  $8\sqrt{3}$ cm²

Area of the shaded region

                   = Area of the semi-circle – Area of the triangle ABC

                   = $8\pi \ \text{cm}^2-8\sqrt{3}\ \text{cm}^2$

                   = $8(\pi \ -\sqrt{3})\ \text{cm}^2$

Hence the area of the shaded region is $8(\pi \ -\sqrt{3})\ \text{cm}^2$.