Question

Two different right cones are being considered by a design team to hold one liter (1000

cubic centimeters) of oil. Design A has a height that is double the diameter of the base.

Design B has a height that is triple the diameter of the base. Which design has a smaller

surface area? What percent less is that surface area?

Answer 1

The surface area of Design A is smaller than the surface area of Design B.

The area of Design A is 94.65% of the Design B.

Explanation:

The area of a right cone is given by the sum of the circle area of the base and the lateral area:

`A = \pi r^(2) + \pi rL` (1)

Where:

r: is the radius

L: is the slant height

The slant height is related to the height and to the radius by Pitagoras:

`L^(2) = H^(2) + r^(2)`

`L = \sqrt{H^(2) + r^(2)}` (2)

By entering equation (2) into (1) we have:

`A = \pi r^(2) + \pi r(\sqrt{H^(2) + r^(2)})`

Now, let's find the area of the two cases.

Design A: height that is double the diameter of the base, H= 2D = 4r

`A_(1) = \pi r^(2) + \pi r(\sqrt{(4r)^(2) + r^(2)}) = \pi r^(2)(1+ √(17))`

The volume of the cone is:

`V = (1)/(3)\pi r^(2)H`

We can find "r":

`V = (1)/(3)\pi r^(2)(4r) = (4)/(3)\pi r^(3)`

`r = \sqrt[3]{(3V)/(4\pi)} = \sqrt[3]{(3*1000)/(4\pi)} = 6.20 cm`

The area is:

`A_(1) = \pi (6.20)^(2)(1+ √(17)) = 618.7 cm^(2)`

Design B: height that is triple the diameter of the base, H = 3D = 6r

The radius is:

`r = \sqrt[3]{(3V)/(6\pi)} = \sqrt[3]{(3*1000)/(6\pi)} = 5.42 cm`

The area is:

`A_(2) = \pi r^(2) + \pi r(\sqrt{(6r)^(2) + r^(2)}) = \pi r^(2)(1 + √(37)) = \pi (5.42)^(2)(1 + √(37)) = 653.7 cm^(2)`

Hence, the surface area of Design A is smaller than the surface area of Design B.

The percent of the surface area of Design A is less than Design B by:

`\% A = (618.7 cm^(2))/(653.7 cm^(2))* 100 = 94.65 \%`

Therefore, the area of Design A is 94.65% of the Design B.

I hope it helps you!