Chapter 6.6, Problem 37E

To determine

To find: the shaded area enclosed between the circles.

Answer to Problem 37E

The shaded area enclosed between the circles is $1.49{\text{cm}}^{\text{2}}$

Explanation of Solution

Given information:

The given radii of the three circles are,

  $\begin{array}{l}{a}_1=4\\ a_2=5\\ a_3=6\end{array}$

Concept used:

Law of Cosines:

If ABC is a triangle with sides $a$ , $b$ and $c$ , then

  $\begin{array}{l}\cos A=\frac{b^2+c^2-a^2}{2bc}\\ \cos B=\frac{a^2+c^2-b^2}{2ac}\\ \cos C=\frac{a^2+b^2-c^2}{2ab}\end{array}$

There are three circles and their radii

  $\begin{array}{l}{r}_1=4\text{cm}\\ r_2=5\text{cm}\\ r_3=6\text{cm}\end{array}$

To calculate the shaded area enclosed between the circles a triangle enclosed between the circles.

Now, the sides of triangle $ABC$ are

  $\begin{array}{c}a=r_1+r_2\\ a=4+5\\ a=9\text{cm}\end{array}$

And,

  $\begin{array}{l}b=r_2+r_3\\ b=5+6\\ b=11\text{cm}\end{array}$

And,

  $\begin{array}{c}c=r_3+r_1\\ c=6+4\\ c=10\end{array}$

The area of a triangle is,

  $A_1=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$

Where,

  $\begin{array}{c}s=\frac{a+b+c}{2}\\ s=\frac{9+11+10}{2}\\ s=\frac{30}{2}\\ s=15\end{array}$

The area of triangle is calculated as,

  $\begin{array}{c}{A}_1=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}\\ =\sqrt{15\left(15-9\right)\left(15-11\right)\left(15-10\right)}\\ A_1=42.42\end{array}$

The angle $A$ of triangle is calculated as,

  $\begin{array}{l}\cos A=\frac{b^2+c^2-a^2}{2bc}\\ \cos A=\frac{{11}^2+{10}^2-9^2}{2\left(11\right)\left(10\right)}\\ A={50.47}^{\circ }\end{array}$

The angle $B$ of triangle is calculated as,

  $\begin{array}{l}\cos B=\frac{a^2+c^2-b^2}{2ac}\\ \cos B=\frac{9^2+{(10)}^2-{11}^2}{2\left(9\right)\left(10\right)}\\ B=70.52°\end{array}$

The angle $C$ of triangle is calculated as,

  $\begin{array}{l}A+B+C=180°\\ 50.47°+70.52°+C=180°\\ C=59.01°\end{array}$

Area of sector $S=\pi r^2\frac{\theta }{{360}^{\circ }}$

  $\begin{array}{c}{S}_A=\pi \times {\left(4\right)}^2\times \frac{50.47}{{360}^{\circ }}\\ =7.04\end{array}$

  $\begin{array}{c}{S}_B=\pi \times {\left(5\right)}^2\times \frac{70.52}{{360}^{\circ }}\\ =15.37\end{array}$

  $\begin{array}{c}{S}_C=\pi \times {\left(6\right)}^2\times \frac{59.01}{{360}^{\circ }}\\ =18.52\end{array}$

Therefore, the shaded area is calculated as,

  $\begin{array}{c}\text{the}\text{shaded}\text{area }=A_1-\left(S_A+S_B+S_C\right)\\ =42.42-\left(7.04+15.37+18.52\right)\\ =42.42-40.93\\ =1.49{\text{cm}}^2\end{array}$

Therefore, the shaded area enclosed between the circles is $1.49{\text{cm}}^{\text{2}}$ .