Chapter 11, Problem 27RP

a.

To determine

To find: Thearea of the shaded segment which involves quadrant of a circle.

Answer to Problem 27RP

Area of the shaded region is $\mathrm{10.28.}$

Explanation of Solution

Given Information:

Radius of the quadrant $\left(r\right)=6$

Formula used:

Area of a quadrant $\left(A\right)=\frac{\pi r^2}{4}$

  $"r"$ is the radius of a circle

Area of a triangle $\left(A_1\right)=\frac{1}{2}\times b\times h$

  $"b"$ is the base of the triangle

  $"h"$ is the height of the triangle

Calculation:

Radius of a quadrant $\left(r\right)=6$

Base of a triangle $\left(b\right)=6$

Height of a triangle $\left(h\right)=6$

We know that, Area of a quadrant $\left(A\right)=\frac{\pi r^2}{4}$

  $\begin{array}{l}\therefore A=\frac{1}{4}\times \frac{22}{7}\times 6^2\\ \therefore A=28.28\end{array}$

As, Area of a triangle $\left(A_1\right)=\frac{1}{2}\times b\times h$

  $\begin{array}{l}\therefore A_1=\frac{1}{2}\times 6\times 6\\ A_1=18\end{array}$

Area of the shaded segment $=A-A_1$

Area of the shaded segment $=28.28-18$

Area of the shaded segment $=10.28$

Hence, area of the shaded region is $\mathrm{10.28.}$

b.

To determine

To calculate: The area of the shaded segment with radius 6 and angle ${60}^{\circ }.$

Answer to Problem 27RP

Area of the shaded segment is $\mathrm{3.27.}$

Explanation of Solution

Given Information:

Radius of a circle $\left(r\right)=6$

Angle subtended by an arc $\left(\theta \right)={60}^{\circ }$

Formula used:

Area of a sector $\left(A\right)=\frac{\theta }{{360}^{\circ }}\times \pi r^2$

  $"r"$ is the radius of the circle

  $"\theta "$ is the angle subtended by an arc

Area of an equilateral triangle $\left(A_1\right)=\frac{\sqrt{3}}{4}{a}^2$

  $"a"$ is the side of an equilateral triangle

Calculation:

Radius of a circle $\left(r\right)=6$

Angle subtended by an arc $\left(\theta \right)={60}^{\circ }$

We know that, Area of a sector $\left(A\right)=\frac{\theta }{{360}^{\circ }}\times \pi r^2$

  $\begin{array}{l}\therefore A=\frac{{60}^{\circ }}{{360}^{\circ }}\times \frac{22}{7}\times 6^2\\ \therefore A=\frac{132}{7}\\ \therefore A=18.85\end{array}$

Angle made by an arc is ${60}^{\circ }$

So, the triangle made by the two radii and a chord is an equilateral triangle.

As, Area of an equilateral triangle $\left(A_1\right)=\frac{\sqrt{3}}{4}{a}^2$

  $\begin{array}{l}\therefore A_1=\frac{\sqrt{3}}{4}\times 6^2\\ \therefore A_1=15.58\end{array}$

Area of the shaded segment $=A-A_1$

Area of the shaded segment $=18.85-15.58$

Area of the shaded segment $=3.27$

Hence, area of the shaded segment is $\mathrm{3.27.}$