Chapter 6.1, Problem 4CFU

To determine

To demonstrate: the given situation.

Answer to Problem 4CFU

See explanation

Explanation of Solution

Given information:

If the radius of circle is doubled and the central angle remains same, then the length of the arc is doubled and the area of sector is quadruple.

Calculation:

Let the radius, central angle, length, and area of first circle be $r,\theta ,s,$ and $A$ respectively.

Let the radius, central angle, length, and area of second circle be $r_2,{\theta }_2,s_2,$ and $A_2$ respectively.

According to the question,

  $\begin{array}{l}{r}_2=2r\\ {\theta }_2=\theta \end{array}$

Now evaluate the value of $s_2$ as follows:

  $\begin{array}{c}{s}_2=r_2{\theta }_2\\ s_2=\left(2r\right)\theta \left[\text{Put the value of }{r}_2\text{ and }{\theta }_2\right]\\ s_2=2\left(r\theta \right)\\ s_2=2s\left[\because s=r\theta \right]\end{array}$

Thus, the length of the arc is doubled if the radius of circle is doubled and the central angle remains same.

Now evaluate the value of $A_2$ as follows:

  $\begin{array}{c}{A}_2=\frac{1}{2}{\left(r_2\right)}^2{\theta }_2\\ A_2=\frac{1}{2}{\left(2r\right)}^2\theta \left[\text{Put the value of }{r}_2\text{ and }{\theta }_2\right]\\ A_2=\frac{1}{2}\left(4r^2\right)\theta \\ A_2=4\left(\frac{1}{2}{r}^2\theta \right)\\ A_2=4A\left[\because A=\frac{1}{2}{r}^2\theta \right]\end{array}$

Thus, the area of sector is quadruple if the radius of circle is doubled and the central angle remains same.