4. Properties of Pyramids
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Chapter 3
4.
Properties of Pyramids
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Why
Pyramids, with their unique shape, have been a subject of intrigue both in mathematics and in ancient civilizations. In geometry, pyramids are defined by their base, which can be any polygon, and a vertex point. The volume of a pyramid is derived from its base area and height. The surface area, on the other hand, is a combination of its base area and the areas of its triangular sides. Pyramids have been used in various real-life scenarios. For instance, the Louvre in Paris boasts a glass pyramid structure, and ancient civilizations like the Egyptians and Aztecs built pyramid-like structures for different purposes. Understanding the properties of pyramids aids in exploring these historical and architectural marvels.
Lesson Settings & Tools
| | 11 Theory slides |
| | 8 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Properties of Pyramids
Slide of 11
Pyramid-shaped structures can be found in different countries such as Egypt and Mexico. In this lesson, the formulas for the volume and surface area of a pyramid will be used to explore some of these real-life locations.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
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Explore
Relating the Volumes of Pyramids and Cubes
Paulina enjoys creating origami objects. After she made the three congruent origami pyramids shown below, she noticed that they could form a cube.
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Discussion
Definition of a Pyramid
A pyramid is a polyhedron that has a base, which can be any polygon, and faces that are triangular and meet at a vertex called the apex. The triangular faces are called lateral faces. The altitude of a pyramid is the perpendicular segment that connects the apex to the base, similar to the altitude of a triangle.
The length of the altitude is the height of the pyramid. If a pyramid has a regular polygon as its base and congruent, isosceles triangles as its lateral faces, it is called a regular pyramid. The altitude of each lateral face in a regular pyramid is also known as the slant height of the pyramid.
If the apex of the pyramid is over the center of its base, it is called a right pyramid. Otherwise, it is called an oblique pyramid.
When the base area and the height of a pyramid are known, its volume can be calculated.
Rule
Volume of a Pyramid
The volume of a pyramid is one third of the product of its base area and height.
The base area B is the area of the polygon opposite the vertex of the pyramid, and the height h is measured perpendicular to the base.
V = 1/3Bh
Proof
Informal JustificationConsider a pyramid and a prism that have the same base area and height.
A pyramid can be modeled as a stack of prisms. The sum of the volumes of the small prisms will be greater than the pyramid's volume. However, as the number of prisms increases and they get thinner, the sum will approximate the volume of the pyramid.
Furthermore, the ratio of the sum of the volumes of each small prism to the volume of the prism will be approximated to 13.
| Number of Layers | Sum of Thin Prisms' Volumes/Volume of Prism |
|---|---|
| 4 | ≈ 0.469 |
| 16 | ≈ 0.365 |
| 64 | ≈ 0.341 |
| 256 | ≈ 0.335 |
| 1024 | ≈ 0.334 |
| 4096 | ≈ 0.333 |
| ∞ | 1/3 |
Therefore, the volume of a pyramid is one third of the prism with the same base area and height.
V_(Pyramid)= 1/3 V_(Prism) [0.6em] ⇓ [0.6em] V = 1/3 Bh
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Example
Finding the Volume of a Pentagonal Pyramid
Tadeo is getting ready to go camping. He has a pyramid-shaped tent with a regular pentagonal base.
The tent has a height of 1.6 meters and its base has side lengths of 1.4 meters. Find the volume of the tent. Round the answer to the two decimal places.
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Hint
Find the perimeter p and the apothem a of the base. Then substitute their values in the formula A= 12ap to find the area of the base. To find the apothem, use the tangent ratio of half the central angle of regular pentagons.
Solution
Start by finding the area of the base of the tent. Then, the formula for the volume of a pyramid will be used.
Area of the Base
The base of the tent is a regular pentagon with side lengths 1.4 meters. Therefore, its perimeter p is 5 times 1.4. p = 5* 1.4= 7 meters Now, draw the apothem of the pentagonal base. The measure of each central angle of a regular pentagon is 72^(∘) and is bisected by the apothem.
The apothem is perpendicular to any side of the polygon and bisects it. As a result, a right triangle with a leg of 0.7 meters is formed.
The value of a can be written in terms of the tangent ratio of 36^(∘). tan 36^(∘) = 0.7/a ⇔ a = 0.7/tan 36^(∘) Finally, to find the area of the base B, substitute a= 0.7tan 36^(∘) and p= 7 into the area formula for regular polygons.
B = 1/2ap
SubstituteII
a= 0.7/tan 36^(∘), p= 7
B = 1/2 ( 0.7/tan 36^(∘))( 7)
▼
Evaluate right-hand side
MoveRightFacToNum
a/c* b = a* b/c
B = 1/2 (4.9/tan 36^(∘))
MultFrac
Multiply fractions
B = 4.9/2 tan 36^(∘)
UseCalc
Use a calculator
B = 3.372135 ...
RoundDec
Round to 2 decimal place(s)
B = 3.37
The area of the base is about 3.37 square meters.
Volume of the Tent
Now that the base area and height are known, substitute these values into the formula for the volume of a pyramid.
V=1/3 B h
SubstituteII
B= 3.37, h= 1.6
V=1/3( 3.37)( 1.6)
▼
Evaluate right-hand side
Multiply
Multiply
V=1/3* 5.392
MoveRightFacToNumOne
1/b* a = a/b
V=5.392/3
CalcQuot
Calculate quotient
V= 1.797333 ...
RoundDec
Round to 2 decimal place(s)
V≈ 1.80
The volume of the tent is about 1.80 cubic meters.
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Example
Finding Height and Volume of the Walter Pyramid
Dominika goes to watch a basketball match at the Walter Pyramid in Long Beach, California. She is amazed by the appearance of the arena. She finds out that the arena was built on a square base with side lengths of 345 feet.
External credits: Summum
a What is the height of the pyramid? Round the answer to the nearest foot.
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b What is the volume of the pyramid?
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Hint
a To find the height of the pyramid, use the Pythagorean Theorem.
b The volume of a pyramid is one third of the product of its base area and height.
Solution
a To find the height of the pyramid, the Pythagorean Theorem will be used. Since the pyramid is a right pyramid, the vertex of the pyramid is over the center of its base. Subsequently, the distance between the center and a side is half its side length, 3452=172.5.
As can be seen, the slant height of the pyramid is the hypotenuse of the right triangle ABC. Now, the height of the pyramid, or AB, can be found using the Pythagorean Theorem.
The height of the pyramid is approximately 192 feet.
b Since the base of the pyramid is a square, its area is the square of its side length.
The base is 119 025 square feet. Now that the base area and height are known, the formula for the volume of a pyramid can be used to find the volume.
V=1/3 B h
SubstituteII
B= 119 025, h= 192
V=1/3( 119 025)( 192)
▼
Evaluate right-hand side
Multiply
Multiply
V=1/3* 22 852 800
MoveRightFacToNumOne
1/b* a = a/b
V=22 852 800/3
CalcQuot
Calculate quotient
V= 7 617 600
The volume of the pyramid is 7 617 600 cubic feet.
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Example
Finding the Base Area of the Slovak Radio Building
Architects might enjoy turning things upside down. Maya is interested in architecture and follows some online magazines about it. After reading an article about the Slovak Radio Building in Bratislava, Slovakia, she wonders about the area of the square rooftop.
External credits: Thomas Ledl
If the height of the building is 80 meters and its volume is about 245 760 cubic meters, find the area of the rooftop of the building.
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Hint
The volume of a pyramid is one third of the product of its base area and height.
Solution
The formula for the volume of a pyramid will be used to find the area of the square rooftop.
V = 1/3B h
In the formula, B is the area of the base and h is the height of the pyramid. Since h=80 meters and V=245 760 cubic meters, the area of the base B can be found by substituting these values into the formula.
V=1/3Bh
SubstituteII
V= 245 760, h= 80
245 760=1/3B* 80
▼
Solve for B
MultEqn
LHS * 3=RHS* 3
737 280 = B * 80
DivEqn
.LHS /80.=.RHS /80.
9216 = B
RearrangeEqn
Rearrange equation
B =9216
The area of the rooftop is 9216 square meters.
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Example
Finding the Volume of a Composite Solid That Includes a Pyramid
Designers and inventors also benefit from pyramids. An object attracts Mark's attention on a school trip to a maritime museum. The guide explains that it is called a deck prism, which was invented to illuminate the cabins below deck before electric lighting. Mark buys a replica of the deck prism, which is composed of a prism and pyramid, each with a regular hexagonal base.
Find the volume of the deck. Round the answer to the two decimal places.
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Hint
The formula for the area of a regular hexagon with side lengths a is B= 3a^2sqrt(3)2. Apply the rounding in the last step.
Solution
The deck is composed of two solids:
- a regular hexagonal prism with an edge length of 3.5 cm and a height of 2 cm, and
- a regular hexagonal pyramid with an edge length of 3 cm and a height of 4 cm.
Therefore, the volume of the deck is the sum of the volumes of the above solids. The volume of each solid will be found one at a time.
Prism's Volume
The base of the prism is a regular hexagon with a side length of 3.5 cm. Recall the formula for the area of a regular hexagon with side lengths a. B =3a^2sqrt(3)/2 Substitute 3.5 for a into the formula and evaluate its value.
Now the volume can be found by multiplying the base area by the height. V_1 = 36.75sqrt(3)/2 * 2 ⇒ V_1 = 36.75sqrt(3)
Pyramid's Volume
The base of the pyramid is a regular hexagon with side lengths 3 cm. Recall the formula for the area of a regular hexagon with side lengths a. B =3a^2sqrt(3)/2 Substitute 3 for a into the formula and evaluate its value.
Now the volume can be found. Recall that the volume of a pyramid is one third of the product of its base area and height.
V_2 =1/3Bh
SubstituteII
B= 27sqrt(3)/2, h= 4
V_2 = 1/3 * 27sqrt(3)/2 * 4
▼
Evaluate right-hand side
MoveRightFacToNum
a/c* b = a* b/c
V_2 = 1/3 * 108sqrt(3)/2
MultFrac
Multiply fractions
V_2 = 108sqrt(3)/6
SimpQuot
Simplify quotient
V_2 = 18 sqrt(3)
Deck Prism's Volume
The sum of the volumes found will give the volume of the deck prism.
The deck prism has a volume of about 94.83 cubic centimeters.
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Discussion
Surface Area of a Pyramid
The surface area of a pyramid is just as important as its volume.
Consider a regular pyramid with an edge length s and a slant height l.
The surface area SA of a regular pyramid can be calculated using the following formula.
SA= 1/2pl + B
In this formula, p is the perimeter of the base, B is the base area, and l is the slant height. In the case that the pyramid is not regular, the area of each lateral face has to be calculated one by one and then added to the area of the base.
Proof
A regular pyramid's surface area can be seen as two separate parts: the lateral area and the base. Surface Area = Lateral Area+ Base
Since for a regular pyramid, its base can be any n-sided regular polygon, the lateral area is the sum of the area of n congruent triangles. For example, consider a regular hexagonal pyramid with an edge length s and a slant height l. Take a look at its net.
As can be seen, the area of each lateral face is 12sl. Therefore, the total lateral area will be 6 times 12sl, because there are 6 congruent lateral faces. Lateral Area [0.8em] 6( 1/2sl ) = 1/2 ( 6s)l Notice that 6s is the perimeter of the base, which can be denoted by p. Then, the lateral area can be expressed as follows. Lateral Area= 1/2 pl Therefore, the formula for the surface area is obtained.
ccccc Surface Area & = & Lateral Area & + & Base [0.8em] SA & =& 1/2pl & + & B
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Example
Finding the Lateral Area of a Roof
Maya's father decides to cover the roof of their house with waterproof insulation material. Maya's father asks Maya to calculate how many square feet of insulation material is needed.
The roof is a square pyramid with a height of 8 feet and base side length of 30 feet. Help Maya calculate the area.
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Hint
To find the slant height, use the Pythagorean Theorem. Note that only the lateral area of the pyramid is needed.
Solution
Maya needs to calculate the lateral area of the square pyramid. To do so, she first needs to calculate the slant height l of the pyramid, which can be found by using the Pythagorean Theorem.
The height h of the pyramid is the distance between the vertex and the center of the base, and b is half the base side length. Therefore, b= 302=15 feet.
Since a negative value does not make sense in this context, only the principal root is considered. Therefore, the slant height is 17 feet. Next, the perimeter of the base will be found. Since the base is a square, its perimeter p is 4 times the base side length. p = 4 * 30 =120 Finally, the lateral area of the pyramid can be found by substituting p=120 and l = 17 into the formula.
LA = 1/2pl
SubstituteII
p= 120, l= 17
LA = 1/2( 120)( 17)
▼
Evaluate right-hand side
Multiply
Multiply
LA = 1/2 * 2040
MoveRightFacToNumOne
1/b* a = a/b
LA = 2040/2
CalcQuot
Calculate quotient
LA = 1020
The amount of material needed to cover the roof is 1020 square feet.
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Pop Quiz
Practice Finding the Surface Area and Volume of Pyramids
The applet shows some right pyramids with different regular polygonal bases. Use the given information to answer the question. If necessary, round the answer to two decimal places.
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Closure
Pyramid-Shaped Structures in the Real World
Pyramid-shaped structures can be seen in many countries around the world. The Aztecs, Mayans, and ancient Egyptians were some of the earliest civilizations to build pyramid-shaped structures. The Aztecs and Mayans built their pyramids with tiered steps and a flat top, whereas the pyramids built by the Egyptians fit the mathematical definition of a pyramid.
External credits: Eric Baetscher
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Properties of Pyramids
Exercise 2.1
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Let's calculate the volume of the original pyramid.
The volume of a pyramid is the base area multiplied by the height and divided by 3. V = 1/3( 5^2)( 9) ⇒ V=75 ft^3
Doubling the Height
Next, we will double the height, meaning h becomes 18 feet. Let's calculate the new volume. Note that the side length of the base remains the same.
As we can see, the volume of the second pyramid is 2 times greater than the first pyramid.
From the previous section we know the volume of the original pyramid to be 75 square feet. This time we want to double the side length of the square base. It used to be 5 inches, which means it becomes 10 inches when doubled. Therefore, we can identify the base area as follows. B=10^2
From here, let's find the volume.
As we can see, the volume of the new pyramid is 4 times greater than the original.
In this part, let's consider a pyramid with a square base with side length of s and height h.
We can write an expression for the volume of the pyramid as follows. V=1/3s^2h
Doubling the Height
Next, let's find the volume of a second pyramid with the same base but where the height is 2h. We will label this volume as V_(2h).
As we can see, the volume of the pyramid is doubled, regardless of the height and the side length. Now let's find the volume of another pyramid with a height h and a side length of the base 2s. We will label this V_(2s).
If we double the side length of the base, the volume quadruples. Therefore, the answers we found to the other exercises are true for any square pyramid.
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Solution
A cube has a side length of 6 inches and a pyramid-shaped hole where the vertices of the base of the pyramid intercept the midpoints of the cube's sides.
What is the volume of the composite solid if the hole goes all the way to the bottom of the cube?
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Let's first calculate the volume of the cube without the hole. Recall that the side length of the cube is 6 inches. We will label this volume V_C. V_C=( 6)^3 ⇒ V_C=216 in^3
Analyzing the Top Side View
Next, we will analyze the top side view of the given composite solid. Recall that a cube has square faces. Since the vertices of the pyramid's base intercept the midpoints of the cube's face, the pyramid must have a square base. Below we see the top face view of the cube and of the pyramid-shaped hole.
Now we can calculate the base area.
Base Area
Notice that the sides of the pyramid's base are also the hypotenuses of 45-45-90 triangles. In this type of triangle, the hypotenuse is sqrt(2) times greater than its legs. Since the legs are half of the cube's side length, the hypotenuse must be 3sqrt(2) inches.
Now we can calculate the area of the pyramid's base by squaring its side.
The base area is 18 square inches.
Volume of the Pyramid
To calculate the volume of a pyramid, we multiply the base area with the height and divide the product by three. Let's call this volume V_P. V_P=1/3Bh We have calculated the base area and the height is the same as the cube's side length. If we substitute this into the formula, we can calculate the volume.
The volume of the hole is 36 cubic inches.
Volume of Composite Solid
Now that we know the volume of the cube and of the pyramid-shaped hole, we can determine the volume of the composite solid, which we will label V_(CS).
The volume of the composite solid is 180 cubic inches.
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What is the volume of the regular pentagonal pyramid? Round the answer to the nearest cubic inch.
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To find the volume of a pyramid, we multiply the base area with the height and divide by 3. V=1/3Bh To determine the volume of our pyramid, we must determine both the base area and the height.
Base Area
From the diagram, we see that the base area is a regular pentagon with a side length of 10 inches. To calculate the area of this figure, we will divide the pentagon into five congruent isosceles triangles.
To find the base area, we should first find the area of one of the triangles. The angle sum of a pentagon is 540^(∘). Since the pentagon is regular, each vertex has an angle measure of 540^(∘)5=108^(∘). Furthermore, the base angles of each triangle is half of 108^(∘).
To determine the area of △ AFB, we first need to find its height a. We can determine this by using the tangent ratio on one of the right triangles.
The height of the triangle is 5tan 54^(∘). We will keep the height in exact form. Now we can find the area of the triangle.
Finally, we will multiply the area of one triangle by 5 to get the area of the pentagon. 5A= 5(25tan 54^(∘)) ⇓ 5A=125tan 54^(∘) The area of the pyramid's base is 125tan 54^(∘) square inches.
Height
Examining the pyramid, we see that the pyramid's height is one of the legs of a right triangle. The second leg is one of the legs in the isosceles triangles we identified earlier.
We can find the length of c by using the Pythagorean Theorem.
Let's add this to the diagram.
To find the height, we must use the tangent ratio.
The height is tan 35^(∘)sqrt(25+25(tan 54^(∘))^2).
Volume of the Pyramid
Now we can find the volume of the pyramid.
The volume of the solid is about 342 cubic inches.
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Properties of Pyramids
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