Chapter 12.7, Problem 21PPS

To determine

To find: The slant height of the largest tent

Answer to Problem 21PPS

The slant height of the largest cone is approximately 48.7cm.

Explanation of Solution

Given:

A can of water repellant says it will seal a canvas that is 50 feet by 60 feet.

Diameter of cone is 30 feet.

Formula used:

The lateral area L of a cone is given by the formula

  $L=\pi rl$

The surface area S of acone is given by the formula

  $S=L+B$

Calculation:

Consider a canvas that is 50 ft. by 30 ft. Also consider the diameter of the cone is 30 ft.

The given figure is a cone.

The objective is to find the slant height of the largest tent that can be made with the given surface.

Let l represent the slant height the cone.

Thus,

  $\begin{array}{l}S=L+B\\ =\pi rl+\pi r^2\\ =\pi \left(rl+r^2\right)\end{array}$

Since radius is half the diameter, so the radius of the cone is $\frac{30}{2}ft$ i.e. $15ft$ .

Put r=15 andl=linthe surface area formula to find the surface area of the cone.

Thus,

  $\begin{array}{l}S=L+B\\ =\pi \left(rl+\pi r^2\right)\\ =\pi \left(15l+{15}^2\right)\\ =15\pi \left(l+15\right)\end{array}$

The area of the canvas is

  $A=50\times 60$

Since, the canvas area is the surface area of the cone, so

  $\begin{array}{l}15\pi \left(l+15\right)=50\times 60\\ \pi \left(l+15\right)=\frac{1}{15}\times 50\times 60\\ =50\times 4\\ =200\\ 3.14\left(l+15\right)\approx 200\left[Since\pi \approx 3.14\right]\\ l+15=\frac{200}{3.14}\\ \approx 63.69\\ l=63.69-15\\ =48.69\\ \approx 48.7\end{array}$

Therefore, the slant height of the largest cone is approximately 48.7cm.

Conclusion:

Therefore, the slant height of the largest cone is approximately 48.7cm.