1.26: Pyramids and Cones

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You may use a calculator throughout this module.

Pyramids

A pyramid is a geometric solid with a polygon base and triangular faces with a common vertex (called the apex of the pyramid). Pyramids are named according to the shape of their bases. The most common pyramids have a square or another regular polygon for a base, making all of the faces identical isosceles triangles. The height, $h$, is the distance from the apex straight down to the center of the base. Two other measures used with pyramids are the edge length $e$, the sides of the triangular faces, and the slant height $l$, the height of the triangular faces.

$Pyramid-variables.png$

Volume of a Pyramid

In general, the volume of a pyramid with base of area $B$ and height $h$ is

\[V=\dfrac{1}{3}Bh$ or $V=Bh\div3 \nonumber \]

If the base is a square with side length $s$, the volume is

\[V=\dfrac{1}{3}s^{2}h$ or $V=s^{2}h\div3 \nonumber \]

Interestingly, the volume of a pyramid is $\dfrac{1}{3}$ the volume of a prism with the same base and height.

Exercises

A pyramid has a square base with sides $16.0$ centimeters long, and a height of $15.0$ centimeters. Find the volume of the pyramid.
$Pyramid-template.gif$
The Great Pyramid at Giza has a height of $137$ meters and a square base with sides $23\overline{0}$ meters long.^[1] Find the volume of the pyramid.

The lateral surface area ($LSA$) of a pyramid is found by adding the area of each triangular face.

Lateral Surface Area of a Pyramid

If the base of a pyramid is a regular polygon with $n$ sides each of length $s$, and the slant height is $l$, then

\[LSA=\dfrac{1}{2}nsl$ or $LSA=nsl\div2 \nonumber \]

If the base is a square, then

\[LSA=2sl \nonumber \]

The total surface area ($TSA$) is of course found by adding the area of the base $B$ to the lateral surface area. If the base is a regular polygon, you will need to use the techniques we studied in Module 23.

Total Surface Area of a Pyramid

\[TSA=LSA+B \nonumber \]

If the base is a square, then

\[TSA=2sl+s^2 \nonumber \]

Exercises

A pyramid has a square base with sides $16.0$ centimeters long, and a slant height of $17.0$ centimeters. Find the lateral surface area and total surface area of the pyramid.
$Pyramid-template-slant-height.gif$
The Great Pyramid at Giza has a slant height of $179$ meters and a square base with sides $23\overline{0}$ meters long. Find the lateral surface area of the pyramid.

Cones

A cone is like a pyramid with a circular base. You may be able to determine the height $h$ of a cone (the altitude from the apex, perpendicular to the base), or the slant height $l$ (which is the length from the apex to the edge of the circular base). Note that the height, radius, and slant height form a right triangle with the slant height as the hypotenuse. We can use the Pythagorean theorem to determine the following equivalences.

$Cone-template-variables.png$

The slant height $l$, height $h$, and radius $r$ of a cone are related as follows:

\[l=\sqrt{r^2+h^2} \nonumber \]

\[h=\sqrt{l^2-r^2} \nonumber \]

\[r=\sqrt{l^2-h^2} \nonumber \]

Just as the volume of a pyramid is $\dfrac{1}{3}$ the volume of a prism with the same base and height, the volume of a cone is $\dfrac{1}{3}$ the volume of a cylinder with the same base and height.

Volume of a Cone

The volume of a cone with a base radius $r$ and height $h$ is

\[V=\dfrac{1}{3}\pi{r^2}h$ or $V=\pi{r^2}h\div3 \nonumber \]

Exercises

The base of a cone has a radius of $5.0$ centimeters, and the vertical height of the cone is $12.0$ centimeters. Find the volume of the cone.
$Cone-radius-5-12.gif$
The base of a cone has a diameter of $6.0$ feet, and the slant height of the cone is $5.0$ feet. Find the volume of the cone.
$Cone-diameter-1.gif$

For the surface area of a cone, we have the following formulas.

Surface Area of a Cone

\[LSA=\pi{rl} \nonumber \]

\[TSA=LSA+\pi{r^2}=\pi{rl}+\pi{r^2} \nonumber \]

$Cone-LSA-net.png$ It’s hard to explain the $LSA$ formula in words, but here goes. The lateral surface of a cone, when flattened out, is a circle with radius $l$ that is missing a wedge. The circumference of this partial circle, because it matched the circumference of the circular base, is $2\pi{r}$. The circumference of the entire circle with radius $l$ would be $2\pi{l}$, so the part we have is just a fraction of the entire circle. To be precise, the fraction is $\dfrac{2\pi{r}}{2\pi{l}}$, which reduces to $\dfrac{r}{l}$. The area of the entire circle with radius $l$ would be $\pi{l^2}$. Because the partial circle is the fraction $\dfrac{r}{l}$ of the entire circle, the area of the partial circle is $\pi{l^2}\cdot\dfrac{r}{l}=\pi{rl}$.

Exercises

The base of a cone has a diameter of $6.0$ feet, and the slant height of the cone is $5.0$ feet. Find the lateral surface area and total surface area of the cone.
$Cone-diameter-1.gif$
The base of a cone has a radius of $5.0$ centimeters, and the vertical height of the cone is $12.0$ centimeters. Find the lateral surface area and total surface area of the cone.
$Cone-radius-5-12.gif$

en.Wikipedia.org/wiki/Great_Pyramid_of_Giza ↵

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