Chapter 12.8, Problem 3GP

To determine

The surface area of new cone is to be estimated when the cone dimensions are reduced by $\frac{1}{2}$ .

Answer to Problem 3GP

The surface area of new cone is $\left(1-\frac{1}{2r}\right)80i{n}^2$ .

Explanation of Solution

Given:

  $A_1=160i{n}^2$

Concept Used:

The surface area of cone is given by the following expression:

  $A=\pi r\left(r+l\right)$

Where,

Radius of cone: $r$

Height of cone: $h$

Length of cone: $l=\sqrt{h^2+r^2}$

Calculation:

The surface area of cone is given as:

  $\begin{array}{l}{A}_1=\pi r\left(r+l\right)\\ 160=\pi r\left(r+l\right)\end{array}$

Now, the cone dimensions are reduced by $\frac{1}{2}$ , the surface area of new cone is calculated as:

  $\begin{array}{l}{A}_2=\pi r\left(r+l\right)\\ A_2=\pi \left(r-\frac{1}{2}\right)\left(\left(r-\frac{1}{2}\right)+\left(l-\frac{1}{2}\right)\right)\\ A_2=\pi \left(r-\frac{1}{2}\right)\frac{1}{2}\left(r+l\right)\end{array}$

Now, the surface area of new cone is:

  $\begin{array}{l}\frac{A_2}{A_1}=\frac{\pi \left(r-\frac{1}{2}\right)\frac{1}{2}\left(r+l\right)}{\pi r\left(r+l\right)}\\ \frac{A_2}{A_1}=\frac{\left(r-\frac{1}{2}\right)\frac{1}{2}}{r}\\ A_2=\left(1-\frac{1}{2r}\right)\frac{1}{2}\times 160\\ A_2=\left(1-\frac{1}{2r}\right)80i{n}^2\end{array}$

Conclusion:

Hence, the surface area of new cone is calculated as $A_2=\left(1-\frac{1}{2r}\right)80i{n}^2$ .