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| | 11 Theory slides |
| | 8 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Paulina enjoys creating origami objects. After she made the three congruent origami pyramids shown below, she noticed that they could form a cube.
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.
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
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
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.
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.
a= 0.7/tan 36^(∘), p= 7
a/c* b = a* b/c
Multiply fractions
Use a calculator
Round to 2 decimal place(s)
The area of the base is about 3.37 square meters.
Now that the base area and height are known, substitute these values into the formula for the volume of a pyramid.
B= 3.37, h= 1.6
Multiply
1/b* a = a/b
Calculate quotient
Round to 2 decimal place(s)
The volume of the tent is about 1.80 cubic meters.
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.
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.
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.
B= 119 025, h= 192
Multiply
1/b* a = a/b
Calculate quotient
The volume of the pyramid is 7 617 600 cubic feet.
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 245 760 cubic meters, find the area of the rooftop of the building.
V= 245 760, h= 80
LHS * 3=RHS* 3
.LHS /80.=.RHS /80.
Rearrange equation
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.
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.
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)
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.
B= 27sqrt(3)/2, h= 4
a/c* b = a* b/c
Multiply fractions
Simplify quotient
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.
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.
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
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.
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.
p= 120, l= 17
Multiply
1/b* a = a/b
Calculate quotient
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.
A frustum is the part of a pyramid that lies between the base and a plane parallel to the base.
Write a formula for the volume of the given frustum.
Let's extend the sides of the frustum such that all of them intersect at one point forming a whole pyramid.
To determine the volume of the frustum, we must subtract the volume of the small pyramid that is sitting on top of the frustum from the volume of the whole pyramid. To do that, we must first find the height of the smaller pyramid.
To find the height of the small pyramid, we will analyze the cross-section of the whole pyramid and a plane that goes through the height and is parallel to one of the sides of the base. We will label the height of the upper triangle as c.
Since FE is parallel to AB, it must be that △ FCE and △ ACB are similar triangles. With this information, we can write a proportionality. CG/FE=CD/AB ⇒ c/y=c+ h/x Let's solve for c.
Now, we will find the volume of the two pyramids.
Now that we found the height of the smaller pyramid, we can calculate its volume V_SP. The base is a square with an area of y^2. Let's use the formula for the volume of a pyramid.
Next, we will find the volume of the whole pyramid, V_(BP). Let's first find its height.
Now we have everything we need to calculate V_(BP). Note that the area of its base is x^2.
Finally, we can find the volume of the frustum.
The volume of the frustum is (x^2+xy+y^2)h3.