Question

First photo question: Given this net of a right circular cone. Find the volume and surface area of the cone.

Answer 1

Problem 1

Ignore the smaller circle for now. We'll be focusing on the portion on the right.

It's a partial circle with central angle theta = 150 degrees and radius r = 24 inches. The 24 inches also happens to be the slant height of the cone when we form the 3D figure.

Let's compute the arc length based on those inputs.

L = arc length

L = (theta/360)*2*pi*r

L = (150/360)*2*pi*24

L = 20pi

This is the distance along the green curve of this portion.

Now imagine we rolled up that partial circle on the right. We'd form the curved slanted roof of the cone. The green curve wraps around to help match up with the circular base. The 20pi calculated earlier is the circumference of this circular base. Use this to find r.

C = 2pi*r

20pi = 2pi*r

r = (20pi)/(2pi)

r = 10

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So far we found that this cone has:

• radius = r = 10
• slant height = s = 24

Now compute the height (h) perpendicular to the base

$h^2+r^2 = s^2\\\\h = \sqrt{s^2-r^2}\\\\h = \sqrt{24^2-10^2}\\\\h = \sqrt{476}$

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We can now compute the volume

$V = (1/3)*\pi*r^2*h\\\\V = (1/3)*\pi*10^2\sqrt{476}\\\\V \approx 2284.7153$

The units of which are cubic inches.

The surface area is of this cone is:

$SA = \pi*r*s + \pi*r^2\\\\SA = \pi*10*24 + \pi*10^2\\\\SA \approx 1068.1415$

The units are in square inches.

For each case, I used the calculator's stored value of pi to get the most accuracy possible. Round the decimal values however you need to, or however your teacher instructs.

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Answers:

• Volume = 2284.7153 cubic inches approximately
• Surface Area = 1068.1415 square inches approximately

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Problem 2

Use the pythagorean theorem to find the slant height

$h^2+r^2 = s^2\\\\s = \sqrt{h^2+r^2}\\\\s = \sqrt{15^2+8^2}\\\\s = \sqrt{289}\\\\s = 17$

This slant height will act as the radius of the partial circle that results when forming the net of the unrolled cone.

The process used in the previous problem was to use the slant height and central angle to find the circumference of the base.

We'll follow that process in reverse to use the circumference and slant height to find the central angle.

But first, we need to find the circumference of the base.

C = 2*pi*r

C = 2*pi*8

C = 16pi

This is the arc length similar to the green arc shown in the previous problem (the figure on the right side of that diagram)

This means we'll plug in L = 16pi and r = 17 (not to be confused with the previous radius of 8 inches) to determine theta

L = (theta/360)*2*pi*r

16pi = (theta/360)*2*pi*17

16pi = theta*(34pi/360)

theta = 16pi*(360/(34pi))

theta = 169.4118 degrees approximately

This is the central angle of the unrolled net. The diagram is basically the same as the previous drawing, where we have a circle on the left to represent the base of the cone. The other portion forms the roof of the cone.

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Let's compute the volume (V) and surface area (SA)

$V = (1/3)*\pi*r^2*h\\\\V = (1/3)*\pi*8^2*15\\\\V \approx 1005.3096$

and,

$SA = \pi*r*s + \pi*r^2\\\\SA = \pi*8*17 + \pi*8^2\\\\SA \approx 628.3185$

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Answers:

• Diagram: Copy the diagram shown in problem 1. However, you'll change the "24 inches" to "17 inches". Also, the central angle is roughly 169.4118 degrees.
• Volume =  1005.3096 cubic inches approximately
• Surface Area = 628.3185 square inches approximately