Chapter 12.5, Problem 51SR

To determine

To find the area of the shaded region in the given figure.

Answer to Problem 51SR

26 square feet

Explanation of Solution

Given:

  

Radius of the circle $=3.6$ feet

Formula used:

Area of a circle $=\pi r^2$ , where $r$ is the radius of the circle.

Area of an equilateral triangle $=\frac{\sqrt{3}}{4}{a}^2$ , where $a$ is the side of the triangle.

  $\tan \theta =\frac{perpendicular}{base}$

  $\tan {30}^2=\frac{1}{\sqrt{3}}$

Calculation:

  

As $\Delta DCE$ is and equilateral triangle.

So,

  $\angle DCE={60}^0$

  $\therefore \angle BCA={30}^0$

In $\Delta ABC$ ,

  $AB=$ perpendicular, $BC=$ base, $CA=$ hypotenuse.

  $\Rightarrow \tan \theta =\frac{perpendicular}{base}$

  $\Rightarrow \tan \theta =\frac{AB}{BC}$

  $\Rightarrow \tan {30}^0=\frac{3.6}{BC}$

  $\Rightarrow \frac{1}{\sqrt{3}}=\frac{3.6}{BC}$

On cross- multiplying,

  $\Rightarrow BC=3.6\sqrt{3}$

  $\Rightarrow BC=6.2$ feet

Now in $\Delta DCE$ ,

  $DC=2(BC)$

  $[\because B$ is the mid-point of $DC$

  $]$

  $\Rightarrow DC=a=12.4$ feet.

Area of $\Delta DCE=\frac{\sqrt{3}}{4}{a}^2$

Putting the value of a,

  $\Rightarrow \frac{\sqrt{3}}{4}(12.4)(12.4)$

  $\Rightarrow 66.58$

So, area of the $\Delta DCE=67$ square feet.

Now area of the circle,

  $\Rightarrow \pi r^2$

Putting the value of r ,

  $\Rightarrow 3.14(3.6)(3.6)$

  $\Rightarrow 40.7$

So, area of the circle $=40$ square feet.

Area of the shaded region $=$ Area of $\Delta DCE$

  $-$ Area of the circle

  $\Rightarrow 67-41$

  $\Rightarrow 26$ square feet

Conclusion:

Therefore, the area of the shaded region in the given figure is $26$ square feet.