Chapter 11.3, Problem 33PPS

To determine

The area of the shaded region.

Answer to Problem 33PPS

The area of the shaded region is 53.5 sq. ft. (approx.).

Explanation of Solution

Given: A triangle of side $5\sqrt{2}$ meters enclosed within a circle as shown below-

  

Concept Used: The formula for the area of circle with radius r is given by-

Area of circle $=\pi r^2$

The formula for the area of a triangle with base b and height h is given by-

Area of triangle $=\frac{1}{2}bh$

Also, any triangle enclosed within a semicircle is a right angled triangle.

The area of any shaded region can be found by subtracting the unshaded region from the total region.

Calculations: From the figure it is obvious that the given triangle is isosceles and lies within the semi-circle. Hence, it is a right angled isosceles triangle. Thus, the hypotenuse of the triangle which is also the diameter of the circle can be determined using Pythagoras theorem as follows-

Hypotenuse $=\sqrt{{\left(5\sqrt{2}\right)}^2+{\left(5\sqrt{2}\right)}^2}$ meters

  $=\sqrt{2\times {\left(5\sqrt{2}\right)}^2}$ meters

  $=5\sqrt{2}\times \sqrt{2}$ meters

  $=10$ meters

Thus, radius of the circle is given by-

Radius = $\frac{10}{2}=5$ meters

Total region in this case is a circle whose diameter is equal $5$ meters.

Thus area of the total region is given by the area of the circle which is-

Total area = Area of circle

  $=\pi {\left(5\right)}^2$ $m^2$

  $=3.14\times 25$ $m^2$

  $=78.5$ $m^2$

The unshaded region is an isosceles right angled triangle whose base and height both are equal to $5\sqrt{2}$ meters. Therefore, area of the unshaded region is given by-

Unshaded area = Area of isosceles triangle

  $=\frac{1}{2}\times {\left(5\sqrt{2}\right)}^2$ $m^2$

  $=\frac{1}{2}\times 5\times 2$ $m^2$

  $=25$ $m^2$

Therefore, total area of the shaded region is-

Shaded area = Area of circle $-$ Area of triangle

  $=78.5-25$ $m^2$

  $=53.5$ $m^2$

Thus, area of the shaded region is 53.5 $m^2$ (approx.).