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{{ printedBook.courseTrack.name }} {{ printedBook.name }} 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.**

Paulina enjoys creating origami objects. After she made the three congruent origami pyramids shown below, she noticed that they could form a cube.

If the volume of a cube is the perfect cube of its side length, what is the volume of each pyramid? Can any three identical pyramids form a cube?

A pyramid is a polyhedron in which one face (the **base**) can be any polygon, and the other faces (the **lateral faces**) are triangles that meet at a common vertex called the **vertex** of the pyramid. Similar to an altitude of a triangle, the **altitude of a pyramid** is a segment from the vertex that is perpendicular to the plane of the base.

The length of the altitude is called the **height** of a pyramid. A pyramid is **regular** if its base is a regular polygon and its lateral faces are congruent, isosceles triangles. In a regular pyramid, the length of the altitude of each lateral face is sometimes referred to as the slant height of a pyramid.

If the vertex of a pyramid is not over the midpoint of its base, it is called an **oblique pyramid**.

When the base area and the height of a pyramid are known, its volume can be calculated.

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.

$V=31 Bh$

Consider 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 $31 .$

Number of Layers | $Volume of PrismSum of Thin Prisms’Volumes $ |
---|---|

$4$ | $≈0.469$ |

$16$ | $≈0.365$ |

$64$ | $≈0.341$ |

$256$ | $≈0.335$ |

$1024$ | $≈0.334$ |

$4096$ | $≈0.333$ |

$∞$ | $31 $ |

Therefore, the volume of a pyramid is one third of the prism with the same base area and height.

$V_{Pyramid}=31 V_{Prism}⇓V=31 Bh $

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.{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.8141079999999999em;vertical-align:0em;\"><\/span><span class=\"mord\"><span class=\"mord text\"><span class=\"mord Roboto-Regular\">m<\/span><\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.8141079999999999em;\"><span style=\"top:-3.063em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">3<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>","answer":{"text":["1.80"]}}

Find the perimeter $p$ and the apothem $a$ of the base. Then substitute their values in the formula $A=21 ap$ to find the area of the base. To find the apothem, use the tangent ratio of half the central angle of regular pentagons.

Start by finding the area of the base of the tent. Then, the formula for the volume of a pyramid will be used.

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=7meters $ 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 $36_{∘}.$ $tan36_{∘}=a0.7 ⇔a=tan36_{∘}0.7 $ Finally, to find the area of the base $B,$ substitute $a=tan36_{∘}0.7 $ and $p=7$ into the area formula for regular polygons.$B=21 ap$

SubstituteII

$a=tan36_{∘}0.7 $, $p=7$

$B=21 (tan36_{∘}0.7 )(7)$

Evaluate right-hand side

MoveRightFacToNum

$ca ⋅b=ca⋅b $

$B=21 (tan36_{∘}4.9 )$

MultFrac

Multiply fractions

$B=2tan36_{∘}4.9 $

UseCalc

Use a calculator

$B=3.372135…$

RoundDec

Round to ${\textstyle 2 \, \ifnumequal{2}{1}{\text{decimal}}{\text{decimals}}}$

$B=3.37$

$V=31 Bh$

SubstituteII

$B=3.37$, $h=1.6$

$V=31 (3.37)(1.6)$

Evaluate right-hand side

Multiply

Multiply

$V=31 ⋅5.392$

MoveRightFacToNumOne

$b1 ⋅a=ba $

$V=35.392 $

CalcQuot

Calculate quotient

$V=1.797333…$

RoundDec

Round to ${\textstyle 2 \, \ifnumequal{2}{1}{\text{decimal}}{\text{decimals}}}$

$V≈1.80$

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.

If the Walter Pyramid is a right pyramid and its slant height is about $258$ feet, help Dominika answer the following questions.

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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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.

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, $2345 =172.5.$

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 $119025$ 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.
The volume of the pyramid is $7617600$ cubic feet.

$V=31 Bh$

SubstituteII

$B=119025$, $h=192$

$V=31 (119025)(192)$

Evaluate right-hand side

Multiply

Multiply

$V=31 ⋅22852800$

MoveRightFacToNumOne

$b1 ⋅a=ba $

$V=322852800 $

CalcQuot

Calculate quotient

$V=7617600$

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.

If the height of the building is $80$ meters and its volume is about $245760$ cubic meters, find the area of the rooftop of the building.

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The volume of a pyramid is one third of the product of its base area and height.

The formula for the volume of a pyramid will be used to find the area of the square rooftop.
$V=31 Bh $
In the formula, $B$ is the area of the base and $h$ is the height of the pyramid. Since $h=80$ meters and $V=245760$ cubic meters, the area of the base $B$ can be found by substituting these values into the formula.
The area of the rooftop is $9216$ square meters.

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.{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.8238736249999999em;vertical-align:-0.009765625em;\"><\/span><span class=\"mord\"><span class=\"mord text\"><span class=\"mord Roboto-Regular\">cm<\/span><\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.8141079999999999em;\"><span style=\"top:-3.063em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">3<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>","answer":{"text":["94.83"]}}

The formula for the area of a regular hexagon with side lengths $a$ is $B=23a_{2}3 .$ Apply the rounding in the last step.

The deck is composed of two solids:

- a regular hexagonal prism with an edge length of $3.5cm$ and a height of $2cm,$ and
- a regular hexagonal pyramid with an edge length of $3cm$ and a height of $4cm.$

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.

$V_{2}=31 Bh$

SubstituteII

$B=2273 $, $h=4$

$V_{2}=31 ⋅2273 ⋅4$

Evaluate right-hand side

MoveRightFacToNum

$ca ⋅b=ca⋅b $

$V_{2}=31 ⋅21083 $

MultFrac

Multiply fractions

$V_{2}=61083 $

SimpQuot

Simplify quotient

$V_{2}=183 $

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 of a regular pyramid can be calculated using the following formula.

$S=21 pℓ+B$

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.

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 $21 sℓ.$ Therefore, the lateral area will be $6$ times $21 sℓ$ because the lateral faces are congruent. $Lateral Area 6(21 sℓ)=21 (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=21 pℓ $ Therefore, the formula for the surface area is obtained.

$Surface AreaS == Lateral Area21 pℓ ++ BaseB $

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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To find the slant height, use the Pythagorean Theorem. Note that only the lateral area of the pyramid is needed.

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, $b=230 =15$ 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. The amount of material needed to cover the roof is $1020$ square feet.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.

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.

Civilizations used these pyramids for different purposes. For example, the pyramids in Mexico were used as places of human sacrifice. Conversely, the Egyptian pyramids were built to be the tombs of pharaohs.

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