Question

a pyramid has a volume of 48 in³ and a height of 9 in. what is its slant height?

Explanation:

1. First, recall the volume formula for a pyramid:

• The volume formula of a pyramid is \(V=\frac{1}{3}Bh\), where \(V\) is the volume, \(B\) is the base - area, and \(h\) is the height.
• We are given that \(V = 48\mathrm{in}^3\) and \(h = 9\mathrm{in}\).
• Substitute the given values into the volume formula:
• \(48=\frac{1}{3}B\times9\).

1. Then, solve for the base - area \(B\):

• Simplify the right - hand side of the equation: \(\frac{1}{3}B\times9 = 3B\).
• So, the equation becomes \(3B = 48\).
• Divide both sides of the equation by 3: \(B=\frac{48}{3}=16\mathrm{in}^2\).

1. Assume the base is a square (since no other information about the base is given). Let the side length of the square base be \(s\).

• For a square base, \(B=s^{2}\). Since \(B = 16\mathrm{in}^2\), then \(s^{2}=16\), and \(s = 4\mathrm{in}\).

1. Now, find the slant height \(l\) using the Pythagorean theorem for the pyramid.

• The height \(h\), half of the side length of the base \(\frac{s}{2}\), and the slant height \(l\) form a right - triangle.
• We know \(h = 9\mathrm{in}\) and \(\frac{s}{2}=\frac{4}{2}=2\mathrm{in}\).
• By the Pythagorean theorem \(l=\sqrt{h^{2}+(\frac{s}{2})^{2}}\).
• Substitute \(h = 9\) and \(\frac{s}{2}=2\) into the formula: \(l=\sqrt{9^{2}+2^{2}}=\sqrt{81 + 4}=\sqrt{85}\approx9.22\mathrm{in}\).

Step1: Use volume formula

\(V=\frac{1}{3}Bh\), substitute \(V = 48\) and \(h = 9\) to get \(48=\frac{1}{3}B\times9\).

Step2: Solve for base - area \(B\)

Simplify \(\frac{1}{3}B\times9\) to \(3B\), then \(B = 16\).

Step3: Find side length of square base

If \(B=s^{2}=16\), then \(s = 4\).

Step4: Use Pythagorean theorem for slant height

\(l=\sqrt{h^{2}+(\frac{s}{2})^{2}}=\sqrt{9^{2}+2^{2}}=\sqrt{85}\approx9.22\).

Answer:

\(\sqrt{85}\text{ in}\approx9.22\text{ in}\)