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| 13 Theory slides |
| 9 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.
Types and Properties of Three-Dimensional Figures:
Geometric Measures of Three-Dimensional Figures:
LaShay and Vincenzo step through the ancient, engraved doors of the Pyramidium House. As they enter, a guidebook enveloped in a mystical glow drifts toward them. It flips open to a page titled The Potion Challenge,
marking the beginning of their enthralling quest.
The book lays out a fascinating challenge: LaShay and Vincenzo are to craft a magical potion, transferring it from a prism-shaped container into two smaller pyramid-shaped vessels.
To escape the enigmatic confines of the Pyramidium House, they must unravel a series of enigmatic challenges, starting with this one. Help them answer the following questions to solve this great challenge.
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.
The applet shows various three-dimensional shapes. Identify if the given 3D shape is 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=31Bh
Find the area of the base using the formula for the area of a triangle. Then, use the formula for the volume of a pyramid to calculate its volume.
B=10.825, h=12
Multiply
Calculate quotient
After successfully solving the riddle of the Secret Chamber's Pyramid, LaShay and Vincenzo enter a room where the walls are constantly moving and reshaping the space. Amidst this changing environment, they notice several geometrically shaped vessels that are morphing in shape and size. One vessel, in the shape of a pentagonal pyramid, catches their attention.
The volume of the vessel, the side length of the base, and its apothem are known. However, the height of the vessel cannot be seen due to the shifting walls. LaShay and Vincenzo need to calculate the height of the vessel to stabilize the room. Help LaShay and Vincenzo solve it!
Calculate the perimeter p of the base of the vessel. Then, find the area of the base using the formula A=21a⋅p. Finally, substitute the area of the base of the vessel and its volume into the formula for the volume of a pyramid and solve for h.
a=5.5, p=40
Multiply
b1⋅a=ba
Calculate quotient
V=1100, B=110
LHS/110=RHS/110
Rearrange equation
LHS⋅3=RHS⋅3
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
Start by finding the perimeter and the base area of the pyramid. Then, substitute this information jointly with the slant height of the pyramid into the formula for the surface area of a pyramid to find the surface area of the inverted pyramid.
Substitute values
Multiply
b1⋅a=ba
Calculate quotient
Add terms
Start by finding the area and the perimeter of the base. Then, plug these values and the surface area into the formula for the surface area of a pyramid. Solve the obtained equation for the slant height ℓ.
c=6, b=3
Calculate power
LHS−9=RHS−9
Rearrange equation
LHS=RHS
Split into factors
a⋅b=a⋅b
Calculate root
a>0
b=6, h=33
Multiply
b1⋅a=ba
Calculate quotient
Substitute values
LHS−93=RHS−93
b1⋅a=ba
Calculate quotient
Rearrange equation
LHS/9=RHS/9
Find the side length and the perimeter of the base. Then substitute these values and the surface area in the formula for the surface area of a pyramid and solve for the slant height. Finally, use the Pythagorean Theorem to find the pyramid's height.
The height and base area of the pyramid are required to calculate its volume. Here, only the surface area and base area are given. In that case, start by determining the side length of the base, the perimeter, the slant height, and finally the height of the pyramid. Then the volume of the pyramid can be determined.
Substitute values
LHS−100=RHS−100
b1⋅a=ba
Calculate quotient
LHS/20=RHS/20
Rearrange equation
AC=19, BC=5
Calculate power
LHS−25=RHS−25
Rearrange equation
LHS=RHS
Split into factors
a⋅b=a⋅b
Calculate root
B=100, h=421
Multiply
b1⋅a=ba
Use a calculator
Round to nearest integer
After completing the last adventure of the Pyramidium House, LaShay and Vincenzo enter the heart of the structure. They recall their initial challenge of needing to pour a potion from a prism container into two smaller pyramid containers.
Using their newfound knowledge and the given dimensions, they can calculate whether the portion in the prism is enough to fill both pyramids.
B=4, h=6
Multiply
b1⋅a=ba
Calculate quotient
A square pyramid has a base side length of 10 meters, a height of 12 meters, and a slant height of 13 meters.
We want to determine the increase in volume of a pyramid when its dimensions are doubled. To do so, we will first calculate the volume of the original pyramid and then calculate the volume of the pyramid with doubled dimensions. Finally, we will calculate the difference between the volumes.
We are given that the side length of a square pyramid is 10 meters. We need to square the side length to calculate the base area. Area of the Base 10^2=100m^2 The base area is 100 square meters. The height of the pyramid is known to be 12 meters. We can substitute these values into the formula for the volume of a pyramid to find its volume.
The volume of the original pyramid is 400 cubic meters.
We are given that all dimensions of the pyramid have been doubled. The original pyramid has a side length of 10 meters and a height of 12 meters. We can represent the new side length of the pyramid with the variable s and its height with the variable h. Let's calculate the values of s and h. s=2* 10 ⇔ s=20m h=2* 12 ⇔ h=24m The side length of the base of the modified pyramid is 20 square meters. Let's square it to find its base area. Area of the Base 20^2=400m^2 We can then use the base area and height of the modified pyramid to find its volume.
The volume of the modified pyramid is 3200 cubic meters.
We have determined that the volume of the initial pyramid is 400 cubic meters, while the volume of the pyramid with doubled dimensions is 3200 cubic meters. We will now calculate the difference between these volumes to determine by how much the new volume increases compared to the original volume. Difference Between Volumes 3200-400=2800m^3 The new volume increases by 2800 cubic meters compared to the original volume.
Let's follow a similar process to determine the increase in surface area of the pyramid when its dimensions are doubled. First, we will calculate the surface area of the original pyramid, and then we will calculate the surface area of the modified pyramid.
The surface area of a pyramid is calculated with the following formula. SA=1/2pl+B Previously, we have found that the base area of the pyramid is 100 square centimeters. However, we still need to calculate the base perimeter. To do so, we will multiply the side length of 10 meters by 4. Perimeter of the Base 4*10=40m The base perimeter is 40 meters. The slant height of the pyramid is 13 meters. We can calculate the surface area of the pyramid by substituting these values into the formula for the surface area.
The surface area of the original pyramid is 360 square meters.
We previously discovered that doubling the dimensions of a pyramid results in a new side length of 20 meters. Additionally, the base area of the modified pyramid is 400 square meters. However, we still need to determine the new perimeter and slant height of the modified pyramid. Let's calculate them. p=4* 20 ⇔ p=80m l=2* 13 ⇔ l=26m We can now use the values we have found to calculate the surface area of the modified pyramid.
The surface area of the modified pyramid is 1440 square meters.
We have calculated that the surface area of the initial pyramid is 360 square meters, and the surface area of the modified pyramid is 1440 square meters. Let's calculate the difference between the new surface area and the initial surface area to find out by how much the new surface area increases compared to the original surface area. Difference Between Surface Areas 1440-360=1080m^2 The new surface area increases by 1080 square meters compared to the original surface area.