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| 11 Theory slides |
| 8 Exercises - Grade E - A |
| Each lesson is meant to take 1-2 classroom sessions |
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.
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=31Bh
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.
Tadeo is getting ready to go camping. He has a pyramid-shaped tent with a regular pentagonal base.
Find the perimeter p and the apothem a of the base. Then substitute their values in the formula A=21ap 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 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.
a=tan36∘0.7, p=7
ca⋅b=ca⋅b
Multiply fractions
Use a calculator
Round to 2 decimal place(s)
B=3.37, h=1.6
Multiply
b1⋅a=ba
Calculate quotient
Round to 2 decimal place(s)
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.
The volume of a pyramid is one third of the product of its base area and height.
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.
The formula for the area of a regular hexagon with side lengths a is B=23a23. Apply the rounding in the last step.
The deck is composed of two solids:
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.
B=2273, h=4
ca⋅b=ca⋅b
Multiply fractions
Simplify quotient
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=21pℓ+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.
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 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.
A pyramid has a height of 9 feet and a square base with a side length of 5 feet.
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
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
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.
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.
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
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.
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.
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.
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.
What is the volume of the regular pentagonal pyramid? Round the answer to the nearest cubic inch.
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.
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.
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).
Now we can find the volume of the pyramid.
The volume of the solid is about 342 cubic inches.