Chapter 6.1, Problem 68E

To determine

Find the area of the sector of the circle.

Answer to Problem 68E

  $\frac{\pi }{4}f{t}^2$

Explanation of Solution

Given information:

Circles with radii $1,2\&3ft$ are externally tangent to one another, has radius $1$ that is cut off by the line segments joining the centre of that circle to the centres of the other two circles.

  

Calculation:

Examine that, three circles with radii $1,2\&3ft$ are externally tangent to one another as shown in the figure.

  

It is need to find the area of the sector of thr circle of radius $1$ that is cut off by the line segments joining the centre of that circle to the centres of the other two circles.

Consider the three circles as shown below

  

The three circles have radii of $1,2\&3ft$ .

To calculate the area of the sector in the circle of radius $1ft$ formed by the angle $\theta$ created by the line segments joining the centres of the circles, we must first find the value of $\theta$ .

We notice that points $A,B\&C$ form a triangle with sides

  $\begin{array}{l}AB=3+2\\ AC=3+1\\ BC=2+1\end{array}$

We see that the sides of the triangle are $3,5\&4ft$ and because

  $3^2+4^2=5^2$

Satisfies the Pythagorean Theorem, we know that $\Delta ABC$ is a right triangle.

Hence, $\theta =\frac{\pi }{2}$

In a circle of radius $r$ , the area $A$ of a sector with central angle of $\theta$ radians is

  $A=\frac{1}{2}{r}^2\theta \mathrm{.......}\left(1\right)$

For the circle of radius $1ft$ , we have

  $r=1ft\&\theta =\frac{\pi }{2}$

Substituting $r=1ft\&\theta =\frac{\pi }{2}in\left(1\right),$ it becomes

  $\begin{array}{l}A=\frac{1}{2}\times 1\times \frac{\pi }{2}\\ =\frac{\pi }{4}f{t}^2\end{array}$

Hence, the required area is $\frac{\pi }{4}f{t}^2$