Question

11)
12)
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(diagrams of various pyramids with labeled dimensions: 11) pyramid with 10 mi, 6 mi, 5.2 mi; 12) pyramid with 11 yd, 6 yd, 6 yd; 13) pyramid with 11 ft, 10 ft, 10 ft; 14) pyramid with 6 cm, 6 cm, 8 cm; 15) pyramid with 7 cm, 5 cm, 5 cm; 16) pyramid with 10 in, 6 in, 4.1 in; 17) pyramid with 10 mi, 8 mi, 6 mi, 10 mi; 18) pyramid with 8 in, 6 in, 6 in; 19) pyramid with 12 ft, 4.1 ft, 6 ft; 20) pyramid with 11 ft, 6 ft, 5.2 ft)

Explanation:

Let's solve for the volume of a pyramid (assuming we take problem 12 as an example, but the process is similar for others). The formula for the volume of a pyramid is \( V=\frac{1}{3}Bh \), where \( B \) is the area of the base and \( h \) is the height.

Step 1: Find the base area (\( B \))

For problem 12, the base is a square with side length \( 6 \) yd. The area of a square is \( B = s^2 \), so \( B=6\times6 = 36 \) square yards.

Step 2: Identify the height (\( h \))

The height of the pyramid is given as \( 11 \) yd.

Step 3: Calculate the volume (\( V \))

Using the formula \( V=\frac{1}{3}Bh \), substitute \( B = 36 \) and \( h = 11 \):
\( V=\frac{1}{3}\times36\times11 \)
First, \( \frac{1}{3}\times36 = 12 \), then \( 12\times11 = 132 \).

(Note: If we take a different problem, say problem 13: base is a square with side \( 10 \) ft, height \( 11 \) ft. \( B = 10\times10 = 100 \), \( V=\frac{1}{3}\times100\times11=\frac{1100}{3}\approx366.67 \) cubic feet. The process is consistent for each pyramid problem: find base area (depending on base shape: square, rectangle, etc.), then apply \( V=\frac{1}{3}Bh \).)

Answer:

For problem 12, the volume is \( 132 \) cubic yards (or adjust based on the specific problem chosen, following the same steps for base area and height substitution).