Chapter 11.5, Problem 23PSC

To determine

To calculate: Thetotal area of shaded region.

Answer to Problem 23PSC

The total area of shaded region is $31.33$ .

Explanation of Solution

Given information:

A square is inscribed in an equilateral triangle.

Formula used:

Area of equilateral triangle: $A=\frac{s^2}{4}\sqrt{3}$

where s = side of equilateral triangle.

Area of square: $A=a^2$

where a = side of square

Calculation:

  

  $\Delta ABC$ is an equilateral triangle.

  $\angle C={60}^{\circ }$

  $\Delta EFC$ is a ${30}^{\circ }{60}^{\circ }{90}^{\circ }\Delta$

If EF = x ,

  $\begin{array}{l}\tan {30}^{\circ }=\frac{1}{\sqrt{3}}\\ \frac{FC}{x}=\frac{1}{\sqrt{3}}\end{array}$

FC = $\frac{x}{\sqrt{3}}=\frac{x\sqrt{3}}{3}$

  $\Delta EFC\cong \Delta DGB$ so,

  $\overline{FC}\cong \overline{GB}$ ,

GB = $\frac{12-x}{2}$

  $\begin{array}{l}\frac{x\sqrt{3}}{3}=\frac{12-x}{2}\\ x=\frac{36}{3+2\sqrt{3}}\end{array}$

  $\begin{array}{l}{A}_{ShadedRegion}=A_{Triangle}-A_{Square}\\ =\frac{s^2}{4}\times \sqrt{3}-a^2\\ =\frac{{(12)}^2}{4}\times \sqrt{3}-{\left(\frac{36}{3+2\sqrt{3}}\right)}^2\\ =36\sqrt{3}-\frac{1296}{21+12\sqrt{3}}\\ =1764\sqrt{3}-3024\\ =31.33\end{array}$