The area of the shaded region is [tex]$8(\pi \ -\sqrt{3})\ \text{cm}^2[/tex].
Solution:
Given radius = 4 cm
Diameter = 2 × 4 = 8 cm
Let us first find the area of the semi-circle.
Area of the semi-circle = [tex]\frac{1}{2}\times \pi r^2[/tex]
[tex]$=\frac{1}{2}\times \pi\times 4^2[/tex]
[tex]$=\frac{1}{2}\times \pi\times 16[/tex]
Area of the semi-circle = [tex]$8\pi[/tex] cm²
Angle in a semi-circle is always 90º.
∠C = 90°
So, ABC is a right angled triangle.
Using Pythagoras theorem, we can find base of the triangle.
[tex]AC^2+BC^2=AB^2[/tex]
[tex]AC^2+4^2=8^2[/tex]
[tex]AC^2=64-16[/tex]
[tex]AC^2=48[/tex]
[tex]AC=4\sqrt{3}[/tex] cm
Base of the triangle ABC = [tex]4\sqrt{3}[/tex] cm
Height of the triangle = 4 cm
Area of the triangle ABC = [tex]\frac{1}{2}\times b \times h[/tex]
[tex]$=\frac{1}{2}\times 4\sqrt{3} \times 4[/tex]
Area of the triangle ABC = [tex]8\sqrt{3}[/tex] cm²
Area of the shaded region
= Area of the semi-circle – Area of the triangle ABC
= [tex]$8\pi \ \text{cm}^2-8\sqrt{3}\ \text{cm}^2[/tex]
= [tex]$8(\pi \ -\sqrt{3})\ \text{cm}^2[/tex]
Hence the area of the shaded region is [tex]$8(\pi \ -\sqrt{3})\ \text{cm}^2[/tex].
