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11.3: Surface Area of Pyramids and Cones

Difficulty Level: At Grade Created by: CK-12

Learning Objectives

  • Find the surface area of pyramids and cones.

Review Queue

  1. A rectangular prism has sides of 5 cm, 6 cm, and 7 cm. What is the surface area?
  2. A cylinder has a diameter of 10 in and a height of 25 in. What is the surface area?
  3. A cylinder has a circumference of \begin{align*}72 \pi \ ft\end{align*}72π ft. and a height of 24 ft. What is the surface area?
  4. Draw the net of a square pyramid.

Know What? A typical waffle cone is 6 inches tall and has a diameter of 2 inches. What is the surface area of the waffle cone? (You may assume that the cone is straight across at the top)

Parts of a Pyramid

Pyramid: A solid with one base and the lateral faces meet at a common vertex.

The edges between the lateral faces are lateral edges.

The edges between the base and the lateral faces are base edges.

Regular Pyramid: A pyramid where the base is a regular polygon.

All regular pyramids also have a slant height which is the height of a lateral face. A non-regular pyramid does not have a slant height.

Example 1: Find the slant height of the square pyramid.

Solution: The slant height is the hypotenuse of the right triangle formed by the height and half the base length. Use the Pythagorean Theorem.

\begin{align*}8^2+24^2 &= l^2\\ 640 &= l^2\\ l = \sqrt{640} & = 8\sqrt{10}\end{align*}82+242640l=640=l2=l2=810

Surface Area of a Regular Pyramid

Using the slant height, which is labeled \begin{align*}l\end{align*}l, the area of each triangular face is \begin{align*}A=\frac{1}{2} bl\end{align*}A=12bl.

Example 2: Find the surface area of the pyramid from Example 1.

Solution: The four triangular faces are \begin{align*}4 \left(\frac{1}{2} bl \right)=2(16)\left(8\sqrt{10}\right)=256 \sqrt{10}\end{align*}. To find the total surface area, we also need the area of the base, which is \begin{align*}16^2 = 256\end{align*}. The total surface area is \begin{align*}256 \sqrt{10}+256 \approx 1065.54 \ units^2\end{align*}.

From this example, we see that the formula for a square pyramid is:

\begin{align*}SA &= (area \ of \ the \ base)+4(area \ of \ triangular \ faces)\\ SA &= B+n\left(\frac{1}{2} bl\right)\end{align*}

\begin{align*}B\end{align*} is the area of the base and \begin{align*}n\end{align*} is the number of triangles.

Surface Area of a Regular Pyramid: If \begin{align*}B\end{align*} is the area of the base, then \begin{align*}SA=B+\frac{1}{2} nbl\end{align*}.

The net shows the surface area of a pyramid. If you ever forget the formula, use the net.

Example 3: Find the area of the regular triangular pyramid.

Solution: “Regular” tells us the base is an equilateral triangle. Let’s draw it and find its area.

\begin{align*}B=\frac{1}{2} \cdot 8 \cdot 4\sqrt{3}=16\sqrt{3}\end{align*}

The surface area is:

\begin{align*}SA=16\sqrt{3}+\frac{1}{2} \cdot 3 \cdot 8 \cdot 18=16\sqrt{3}+216 \approx 243.71\end{align*}

Example 4: If the lateral surface area of a square pyramid is \begin{align*}72 \ ft^2\end{align*} and the base edge is equal to the slant height. What is the length of the base edge?

Solution: In the formula for surface area, the lateral surface area is \begin{align*}\frac{1}{2} nbl\end{align*}. We know that \begin{align*}n = 4\end{align*} and \begin{align*}b = l\end{align*}. Let’s solve for \begin{align*}b\end{align*}.

\begin{align*}\frac{1}{2} nbl &= 72 \ ft^2\\ \frac{1}{2} (4) b^2 &= 72\\ 2b^2 &= 72\\ b^2 &= 36\\ b &= 6\end{align*}

Surface Area of a Cone

Cone: A solid with a circular base and sides taper up towards a vertex.

A cone has a slant height, just like a pyramid.

A cone is generated from rotating a right triangle, around one leg, in a circle.

Surface Area of a Right Cone: \begin{align*}SA=\pi r^2+\pi rl\end{align*}.

Area of the base: \begin{align*}\pi r^2\end{align*}

Area of the sides: \begin{align*}\pi rl\end{align*}

Example 5: What is the surface area of the cone?

Solution: First, we need to find the slant height. Use the Pythagorean Theorem.

\begin{align*}l^2 &= 9^2+21^2\\ &= 81+441\\ l &= \sqrt{522} \approx 22.85\end{align*}

The surface area would be \begin{align*}SA=\pi 9^2+\pi (9)(22.85) \approx 900.54 \ units^2\end{align*}.

Example 6: The surface area of a cone is \begin{align*}36 \pi\end{align*} and the radius is 4 units. What is the slant height?

Solution: Plug in what you know into the formula for the surface area of a cone and solve for \begin{align*}l\end{align*}.

\begin{align*}36 \pi &= \pi 4^2+\pi 4l\\ 36 &= 16+4l \qquad When \ each \ term \ has \ a \ \pi, \ they \ cancel \ out.\\ 20 &= 4l\\ 5 &= l\end{align*}

Know What? Revisited The standard cone has a surface area of \begin{align*}\pi+ \sqrt{35}\pi \approx 21.73 \ in^2\end{align*}.

Review Questions

  • Questions 1-10 use the definitions of pyramids and cones.
  • Questions 11-19 are similar to Example 1.
  • Questions 20-26 are similar to Examples 2, 3, and 5.
  • Questions 27-31 are similar to Examples 4 and 6.
  • Questions 32-25 are similar to Example 5.

Fill in the blanks about the diagram to the left.

  1. \begin{align*}x\end{align*} is the ___________.
  2. The slant height is ________.
  3. \begin{align*}y\end{align*} is the ___________.
  4. The height is ________.
  5. The base is _______.
  6. The base edge is ________.

Use the cone to fill in the blanks.

  1. \begin{align*}v\end{align*} is the ___________.
  2. The height of the cone is ______.
  3. \begin{align*}x\end{align*} is a __________ and it is the ___________ of the cone.
  4. \begin{align*}w\end{align*} is the _____________ ____________.

For questions 11-13, sketch each of the following solids and answer the question. Your drawings should be to scale, but not one-to-one. Leave your answer in simplest radical form.

  1. Draw a right cone with a radius of 5 cm and a height of 15 cm. What is the slant height?
  2. Draw a square pyramid with an edge length of 9 in and a 12 in height. Find the slant height.
  3. Draw an equilateral triangle pyramid with an edge length of 6 cm and a height of 6 cm. What is the height of the base?

Find the slant height, \begin{align*}l\end{align*}, of one lateral face in each pyramid or of the cone. Round your answer to the nearest hundredth.

Find the area of a lateral face of the regular pyramid. Round your answers to the nearest hundredth.

Find the surface area of the regular pyramids and right cones. Round your answers to 2 decimal places.

  1. A regular tetrahedron has four equilateral triangles as its faces.
    1. Find the height of one of the faces if the edge length is 6 units.
    2. Find the area of one face.
    3. Find the total surface area of the regular tetrahedron.
  2. If the lateral surface area of a cone is \begin{align*}30 \pi \ cm^2\end{align*} and the radius is 5 cm, what is the slant height?
  3. If the surface area of a cone is \begin{align*}105 \pi \ cm^2\end{align*} and the slant height is 8 cm, what is the radius?
  4. If the surface area of a square pyramid is \begin{align*}40 \ ft^2\end{align*} and the base edge is 4 ft, what is the slant height?
  5. If the lateral area of a square pyramid is \begin{align*}800 \ in^2\end{align*} and the slant height is 16 in, what is the length of the base edge?
  6. If the lateral area of a regular triangle pyramid is \begin{align*}252 \ in^2\end{align*} and the base edge is 8 in, what is the slant height?

The traffic cone is cut off at the top and the base is a square with 24 in sides. Round answers to the nearest hundredth.

  1. Find the area of the entire square. Then, subtract the area of the base of the cone.
  2. Find the lateral area of the cone portion (include the 4 inch cut off top of the cone).
  3. Subtract the cut-off top of the cone, to only have the lateral area of the cone portion of the traffic cone.
  4. Combine your answers from #27 and #30 to find the entire surface area of the traffic cone.

Review Queue Answers

  1. \begin{align*}2(5 \cdot 6) + 2(5 \cdot 7) + 2(6 \cdot 7) = 214 \ cm^2\end{align*}
  2. \begin{align*}2(15 \cdot 18) + 2(15 \cdot 21) + 2(18 \cdot 21) = 1926 \ cm^2\end{align*}
  3. \begin{align*}2 \cdot 25 \pi + 250 \pi = 300 \pi \ in^2\end{align*}
  4. \begin{align*}36^2 (2 \pi) + 72 \pi (24) = 4320 \pi \ ft^2\end{align*}

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CK.MAT.ENG.SE.1.Geometry-Basic.11.3