Chapter 15, Problem 24CR

To determine

To Find: The area of shaded region.

Answer to Problem 24CR

  $87.48unit{s}^2$

Explanation of Solution

Given:

  

  $\begin{array}{l}ABCD\text{ is asquare}\\ AB=BC=CD=AD=12\\ \text{Mid-Points of the sides of the square are the centre of arc}\text{.}\\ \text{Let, }E,F,G,H\text{ are the mid-points on }AB,BC,CD,AD\text{ respectively}\text{.}\\ \text{Join }EF,FG,GH,EH\end{array}$

Formula Used:

  $\begin{array}{l}\text{Area of quadrant}=\frac{\pi r^2}{4}\\ \text{Area of square}=sid{e}^2\end{array}$

Calculation:

  $\begin{array}{l}Let,\\ I\&J\text{}\text{ be the point between EF\&FG respectively}\text{.}\\ \text{In }\Delta BIF\\ BF=\frac{BC}{2}=\frac{12}{2}=6\text{Mid-Point of the side of square}\\ \angle IFJ={90}^o\\ \angle IBF=\angle BFI=\angle GJC={45}^o\\ BI=BF\sin {45}^o=6\times \frac{1}{\sqrt{2}}=3\sqrt{2}\\ \text{Length of radius of arc}=3\sqrt{2}\\ \text{Area of shaded region }=\text{ Area of square -Area of four quadrant}\\ ={\left(12\right)}^2-4\times \frac{\pi {\left(3\sqrt{2}\right)}^2}{4}\\ =144-3.14\times 18\\ =144-56.52\\ =87.48unit{s}^2\end{array}$

Conclusion:

Thus, the area of shaded region is $87.48unit{s}^2$