Chapter 6.6, Problem 36E

To determine

Find the area of the shaded figure.

Answer to Problem 36E

  $2.45sq.units$

Explanation of Solution

Given information:

Given,

  

Calculation:

We have to find the area of the shaded figure given below,

  

First we redraw the figure by marking points $A,B,C,DonitsvertexandsideAB.$

  

From the above figure, we see that the area of the given figure is difference of the areas of the triangles $ABC$ and $ABD$ .

So, $A=A_{ABC}-A_{ABD}$

Since we have the two sides $a=4,b=4and\theta ={60}^{\circ }$ , we use to use the formula $A=\frac{1}{2}ab\sin \theta$ , to find the area of the triangle $ABD$ .

So,

  $\begin{array}{l}{A}_{ABC}=\frac{1}{2}\times 4\times 4\times \sin {60}^{\circ }\\ =8\sin {60}^{\circ }\end{array}$

On using a calculator, we have $\sin {60}^{\circ }=\frac{\sqrt{3}}{2}$ .

So,

  $\begin{array}{l}{A}_{ABC}=8\times \frac{\sqrt{3}}{2}\\ =4\sqrt{3}\\ =4\times 1.732\\ =6.928sq.units\end{array}$

Now, to calculate the area of the triangle $ABD$ , calculate the side $AB$ by using the Law of Cosines in the triangle $ABC$ ,which is given as

  $c^2=a^2+b^2-2ab\cos C$

We have $a=4,b=4,C={60}^{\circ }andletAB=c$ .

So,

  $\begin{array}{l}A{B}^2=4^2+4^2-2\times 4\times 4\times \cos {60}^{\circ }\\ =16+16-32\cos {60}^{\circ }\end{array}$

we have $\cos {60}^{\circ }=\frac{1}{2}$

So,

  $\begin{array}{l}A{B}^2=32-32\times \frac{1}{2}\\ =16\end{array}$

On taking square root both sides, we get,

  $\begin{array}{l}AB=\sqrt{16}\\ =4\end{array}$

Now, we have all the sides of the triangle $ABD$ as $a=3,b=3andc=4$

Use Heron’s formula to find its area.

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

where $s$ is the semi perimeter of the triangle, which is,

  $s=\frac{a+b+c}{2}$

On substituting $a=3,b=3$ and $c=4$

We get,

  $\begin{array}{l}s=\frac{3+3+4}{2}\\ =\frac{10}{2}\\ =5\end{array}$

Also,

  $\begin{array}{l}{A}_{ABD}=\sqrt{5\left(5-3\right)\left(5-3\right)\left(5-4\right)}\\ =\sqrt{5\times 2\times 2\times 1}\\ =2\sqrt{5}\\ =2\times 2.24\\ 4.48sq.units\end{array}$

Now, substituting $A_{ABD}=4.48and{A}_{ABC}=6.93inA=A_{ABC}-A_{ABD}$ , we get,

  $\begin{array}{l}A=6.93-4.48\\ A=2.45sq.units\end{array}$

Hence, the area of the given figure is $2.45sq.units$ .