Chapter 11.6, Problem 22PSC

To determine

To calculate:Thearea of shaded region which contains rhombus and circle.

Answer to Problem 22PSC

The area of shaded region is $600-144\pi$ .

Explanation of Solution

Given information:

Diagonals of rhombus are 30 and 40.

Formula used:

Area of a circle: $A=\pi r^2$

r = radius of circle.

Area of rhombus $=\frac{1}{2}(d_1\ast d_2)$

d = diagonal of rhombus.

The below theorem is used:

Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

  

In right angle triangle,

  $a^2+b^2\text{= }{c}^2$

Calculation:

  

Area of rhombus $=\frac{1}{2}(d_1\ast d_2)$

  $\begin{array}{l}=\frac{1}{2}\left(30\ast 40\right)\\ =600\end{array}$

In right angletriangle, we get

  $\begin{array}{l}{\left(Base\right)}^2={\left(15\right)}^2+{\left(20\right)}^2\\ {\left(Base\right)}^2=225+400\\ {\left(Base\right)}^2=625\\ \therefore Base=\sqrt{625}=25\end{array}$

Base of the triangle formed is 25.

  $\begin{array}{l}\frac{1}{2}\ast 15\ast 20=\frac{1}{2}\ast 25\ast h\\ 150=\frac{h}{2}(25)\\ \frac{h}{2}=\frac{150}{25}\\ \frac{h}{2}=6\\ h=12\end{array}$

Area of a circle: $A=\pi r^2$

  $A=\pi {(12)}^2=144\pi$

  $\begin{array}{l}{A}_{shadedarea}=A_{rhombus}-A_{circle}\\ A_{shadedarea}=600-144\pi \end{array}$