Area of the semi-circle = 9.8125 square units
Solution:
Given ABC is a right triangle.
BC is a semi-circle.
∠BAC = 90°.
AB = 3 units and AC = 4 units.
BC is the hypotenuse of the right triangle ABC.
Using Pythagoras theorem,
In right triangle, square of the hypotenuse is equal to the sum of the squares of the other two sides.
[tex]BC^2=AB^2+AC^2[/tex]
[tex]BC^2=3^2+4^2[/tex]
[tex]BC^2=25[/tex]
Take square root on both sides.
BC = 5 units
The diameter of the semi-circle is 5 units.
Radius = 5 ÷ 2 = 2.5 units
Area of the semi-circle = [tex]\frac{1}{2}\pi r^2[/tex]
[tex]=\frac{1}{2}\times 3.14\times 2.5^2[/tex]
= 9.8125 square units
Area of the semi-circle = 9.8125 square units
