Question

A right square pyramid has a volume of 324 cubic centimeters and a height of 12 centimeters. What is the length of one side of the base, in centimeters? A) \(18 \sqrt{2}\) B) 18 C) 9 D) \(3 \sqrt{3}\)

Short answer

The length of one side of the base of the right square pyramid is 9 centimeters.

Step-by-step solution

Find the area of the base

Given volume (V) = 324 cubic centimeters, and height (h) = 12 centimeters, use the formula for the volume of a square pyramid to find the area of the base. \[324 = \frac{1}{3}B \cdot 12\] Now, solve the equation for B: \[B = \frac{324 \cdot 3}{12}\]

Calculate B

Calculate B using the equation \[B = \frac{324 \cdot 3}{12} = \frac{972}{12} = 81\] Now that we have the area of the base, we can find the length of one side s of the square base.

Find s

Recall that the area of a square is given by the formula: \[B = s^2\] We already found B = 81. Now, solve the equation for s: \[s^2 = 81\] Take the square root of both sides: \[s = \sqrt{81}\]

Determine the length of one side of the base

Calculate s using the equation \[s = \sqrt{81} = 9\] The length of one side of the base (s) is 9 centimeters. Therefore, the correct answer is (C) 9.

Key concepts

Right Square Pyramid

A right square pyramid is a geometric solid with a square base and four triangular faces that come to a point, called the apex, directly above the center of the base. This symmetry implies that all the triangular faces are congruent isosceles triangles. The 'right' in its name indicates that the apex is directly above the centroid of the base, and the axis from the apex to the center of the base is perpendicular to the base. Understanding the properties of a right square pyramid is essential for solving volume or surface area problems related to this shape.

For students dealing with geometric figures, visualizing a right square pyramid can be beneficial. Imagine the pyramids in Egypt but with a square base, and you'll have a good idea of what a right square pyramid looks like. It's crucial in geometry to identify properties that could simplify calculations – like the isosceles triangles or the symmetrical square base, which lead to straightforward formulas for volume or lateral surface area calculations.

Base Area Calculation

Calculating the base area is fundamental when dealing with pyramids, especially for volume calculations. The base area (\(B\)) of a square pyramid is found by multiplying the length of one side of the square base (\(s\times s\times s\times s\text{ or }s^2\text{ respectively}\). The sides of the base are all equal in length because it's a square, a special case of a rectangle with all four sides of equal length.

To find the side length when you know the area, you would take the square root of the base area. In our problem, we had a base area of 81 square centimeters, and by square-rooting this value, we established that the side length of the square was 9 centimeters. This step is crucial for students to understand since it throws light on reversing the area formula of a square, which is one of the fundamental operations in algebra - finding the square root to undo a square.

Geometry Problem Solving

Problem-solving in geometry often involves a methodical approach, starting with understanding the problem, identifying known and unknown elements, and then deploying relevant formulas. For the problem at hand, the volume of the square pyramid and its height were given, and we had to find the unknown length of a side of the base.

Breaking down the steps systematically, we first applied the volume formula for a square pyramid (\(V=\frac{1}{3}Bh\)), which allowed us to solve for the base area (\(B\)). We then used the base area to find the side length, remembering that for a square, the area is equal to the side length squared (\(s^2\)). Notably, algebra skills are a significant part of geometric problem-solving since they provide the tools necessary to manipulate equations and extract the sought information. Encouraging students to follow these steps can guide them to the solution efficiently, making complex problems more accessible.