✦ Practical Mathematical Problem Solving

Worked Pyramid Volume Examples

Learn how to solve real-world pyramid volume problems step by step. Features famous architectural monuments, camping equipment, and reverse height equations.

Example 1: The Great Pyramid of Giza (Square Base)

Problem: Originally built around 2560 BC, the Great Pyramid of Giza had an original base edge length of $a = 230.4\text{ m}$ and an original perpendicular height of $h = 146.6\text{ m}$. Calculate its original stone volume.

Step 1: Calculate the Square Base Area

$$A_b = a^2 = (230.4)^2 = 53,084.16\text{ m}^2$$

Step 2: Apply the Universal Volume Formula

$$V = \frac{1}{3} A_b h = \frac{1}{3} \times 53,084.16 \times 146.6$$

$$V = \frac{7,782,137.856}{3} \approx \mathbf{2,594,045.95\text{ m}^3}$$

Result: The original volume was approximately 2.59 million cubic meters of limestone and granite.

Example 2: The Louvre Glass Pyramid, Paris

Problem: The modern glass entrance designed by I. M. Pei has a square base width of $a = 35\text{ m}$ and a vertical height of $h = 21.6\text{ m}$. Find its enclosed interior air volume.

Step 1: Compute Base Area

$$A_b = a^2 = 35^2 = 1,225\text{ m}^2$$

Step 2: Compute Volume

$$V = \frac{1}{3} \times 1,225 \times 21.6 = 1,225 \times 7.2 = \mathbf{8,820\text{ m}^3}$$

Result: The glass atrium encloses exactly 8,820 cubic meters of interior space.

Example 3: Reverse Solver - Finding Height from Known Volume

Problem: An architectural hopper has a square base with edge $a = 4\text{ m}$ and must hold a volume of $V = 64\text{ m}^3$. What vertical height $h$ is required?

Step 1: Rearrange the Pyramid Formula for $h$

$$V = \frac{1}{3} A_b h \implies h = \frac{3V}{A_b}$$

Step 2: Substitute Known Values

$$A_b = 4^2 = 16\text{ m}^2$$

$$h = \frac{3 \times 64}{16} = \frac{192}{16} = \mathbf{12\text{ m}}$$

Result: The required perpendicular vertical height is exactly 12 meters.

Example 4: Rectangular Pyramid Attic Pavilion

Problem: A park pavilion has a rectangular pyramid roof with length $l = 12\text{ m}$, width $w = 8\text{ m}$, and perpendicular center height $h = 4.5\text{ m}$. Compute the total enclosed air capacity.

Step 1: Compute Rectangular Base Area

$$A_b = l \times w = 12 \times 8 = 96\text{ m}^2$$

Step 2: Apply Volume Formula

$$V = \frac{1}{3} A_b h = \frac{1}{3} \times 96 \times 4.5 = \mathbf{144\text{ m}^3}$$

Result: The attic encloses exactly 144 cubic meters of air volume.

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