Finding the Volume and Surface Area of a Pyramid

Volume and Surface Area of a Pyramid

Volume and Surface Area of a Pyramid

Volume is the space that a figure occupies.

To find the volume of a pyramid take one third of the area of the base – capital B stands for area of the base times the height of the pyramid. The height of a pyramid is the altitude and the altitude is the perpendicular from the vertex perpendicular to the base of the pyramid. Surface area is the area of all the surfaces of a figure. To find surface area of a pyramid take one half times the slant height – l stands for slant height – times the perimeter of the base – p is for perimeter of the base plus the area of the base. Again, there’s that capital B that stands for the area of the base.

Let’s put this into practice by looking at an example. Find the volume and surface area of the square pyramid. Since it’s called a square pyramid, that means that our base of our pyramid is a square which will help us when we have to find the area of the base – that’ll help us find it knowing that it’s a square. Let’s start with volume. Volume is one third the area of the base times the height of the pyramid. And now we need to substitute the area of our base and our height. Volume is one third times – now since they told us that our base is a square to find the area of a square we just take the side and square it and the side of our base is ten inches so ten inches squared will give us the area of our base times the height of the pyramid and the height is the altitude or the perpendicular segment from the vertex to the base. So that’s twelve inches. I need to follow my order of operations so I’m going to have to square ten inches squared first. Volume is one third times ten inches squared. Ten inches squared is 100. Inches squared is inches squared times twelve inches. Now all I have left is multiplication so I can multiply in any order. I’m going to take a third of twelve inches first. A third of twelve inches is four inches times a hundred inches squared.

Finally, we multiply four inches times a hundred inches squared inches times inches squared is inches cubed. So the volume of a pyramid is four hundred inches cubed. Now to find the surface area. The surface area of a pyramid is half the slant height times the perimeter of the base plus the area of the base. And now we’ll need to substitute our different values for those variables. The surface area is one half times slant height of the lateral faces which are triangles so it is the height of your triangle – one of your lateral faces and in this case that’s thirteen inches – so one half times thirteen inches. P again stands for perimeter of the base and to find the perimeter of any shape you can just add all the sides but for a square since the sides are all the same we can do ten inches times four which is forty inches plus the area of the base. So we’re going to do the same thing we did on volume. The base is a square. To find the area of a square that’s side squared and our side is ten inches and ten inches is squared. Then we simplify. I’m going to take half of forty inches since that’s easier than taking half of thirteen inches. Half of forty inches is twenty inches times thirteen inches plus ten inches squared is 100 inches squared. And I’m going to multiply twenty inches times thirteen inches. Ten times thirteen would be one thirty but we double that since we’re multiplying by twenty. That’s two hundred sixty plus one hundred inches squared and we combine like terms. Two hundred sixty inches squared plus one hundred inches squared is three hundred sixty inches squared and that’s our surface area of our pyramid.

Practice Questions

Question #1:

 
Calculate the surface area of the following square pyramid.

95 cm2

100 cm2

105 cm2

110 cm2

Answer:

The formula \(SA=B+\frac{1}{2}pl\) can be used to calculate the surface area of a square pyramid.
\(B\) represents the area of the base.
\(p\) represents the perimeter of the base.
\(l\) represents the slant height.
When the appropriate values are plugged into the formula, \(SA=B+\frac{1}{2}pl\) becomes \(SA=25+\frac{1}{2}(20)(8)\) which simplifies to 105 cm2. The surface area of the square pyramid is 105 cm2.

Question #2:

 
Calculate the surface area of the following equilateral triangular pyramid.

357 in2

350 in2

257 in2

332 in2

Answer:

The formula for calculating the surface area of a triangular pyramid is very similar to the formula for a square pyramid. The only difference occurs when determining the area of the base shape. The formula \(SA=B+\frac{1}{2}pl\) will still apply, but B will be calculated by multiplying \(\frac{1}{2}bh\). This can be plugged into the formula for B, which results in the following: \(SA=(\frac{1}{2}bh)+\frac{1}{2}pl\). From here, we can simply plug in all of the appropriate values so that \(SA=(\frac{1}{2}bh)+\frac{1}{2}pl\) becomes \(SA=(\frac{1}{2})(14)(12)+\frac{1}{2}(42)(13)\) which simplifies to \(SA=357\text{ in}^2\). The surface area of the equilateral triangular pyramid is \(357\text{ in}^2\).

Question #3:

 
Calculate the volume of the pyramid.

2,670 in3

3,670 in3

4,670 in3

5,670 in3

Answer:

The formula \(V=\frac{1}{3}Bh\) can be used to calculate the volume of a pyramid.
B represents the area of the base shape.
h represents the height of the pyramid (not the slant height). When the appropriate values are plugged into the formula, \(V=\frac{1}{3}Bh\) becomes \(V=\frac{1}{3}(2{,}130)(27)\) which simplifies to \(V=5{,}670\). The volume of the pyramid is 5,670 in3.

Question #4:

 
Leo has a triangular glass pyramid that he is hoping to fill with some beach sand. On his vacation to Florida he collects 6.5 cm3 of sand. The triangle at the base of the glass pyramid has an area of 5 cm2, and the height of the pyramid is 6 cm. If he collects 6.5 cm3 of sand, will the pyramid be large enough to hold all of the sand?
Triangular glass pyramid filled with sand and seashells

Yes, the pyramid is large enough

No, the pyramid is too small.

Answer:

The volume of the triangular based pyramid can be calculated using the formula \(V=\frac{1}{3}Bh\) where B is the area of the base triangle, and h is the height of the pyramid. When B and h are plugged in, \(V=\frac{1}{3}Bh\) becomes \(V=\frac{1}{3}(5)(6)\) which simplifies to \(V=10\). The volume of the pyramid is 10 cm3, which is enough room for the 6.5 cm3 of sand.

Question #5:

 
Some of Leonardo da Vinci’s first sketches of parachute designs were in the shape of square based pyramids. For a parachute design with rectangular base side lengths of six feet and eight feet, and a height of seven feet, how much air would be captured in the pyramid once it is deployed fully?

112 ft3

109 ft3

115 ft3

118 ft3

Answer:

The formula \(V=\frac{1}{3}Bh\) can be used to calculate the volume of a pyramid. B represents the area of the base shape, and h represents the height of the pyramid. When the appropriate values are plugged into the formula \(V=\frac{1}{3}Bh\) becomes \(V=\frac{1}{3}(48)(7)\), which simplifies to \(V=112\). The volume of the fully deployed parachute would be 112 ft3.

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by Mometrix Test Preparation | Last Updated: June 2, 2021