Question

question 3
a pyramid has a height of 5 inches and a volume of 60 cubic inches. select all figures that could be the base for this pyramid.
a. square with side length 6 inches
b. a 3 - inch - by - 4 - inch rectangle
c. a 4 - inch - by - 9 - inch rectangle
d. a circle with radius 4 inches
e. a right triangle with one side 5 inches and the hypotenuse 13 inches
f. a hexagon with an area of 36 square inches

Explanation:

Step1: Calculate the base area of the 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. Given \(V = 60\) cubic inches and \(h=5\) inches. We can solve for \(B\) by rearranging the formula: \(B=\frac{3V}{h}\).
Substitute \(V = 60\) and \(h = 5\) into the formula: \(B=\frac{3\times60}{5}=36\) square inches.

Step2: Calculate the area of each figure

• Option a:

The area of a square with side length \(s\) is \(A = s^{2}\). For \(s = 6\) inches, \(A=6^{2}=36\) square inches.

• Option b:

The area of a rectangle with length \(l\) and width \(w\) is \(A=l\times w\). For \(l = 3\) inches and \(w = 4\) inches, \(A=3\times4 = 12\) square inches.

• Option c:

For a rectangle with \(l = 4\) inches and \(w = 9\) inches, \(A=4\times9=36\) square inches.

• Option d:

The area of a circle is \(A=\pi r^{2}\). For \(r = 4\) inches, \(A=\pi\times4^{2}=16\pi\approx16\times3.14 = 50.24\) square inches.

• Option e:

For a right - triangle with hypotenuse \(c = 13\) inches and one side \(a = 5\) inches, using the Pythagorean theorem \(c^{2}=a^{2}+b^{2}\), we can find the other side \(b=\sqrt{13^{2}-5^{2}}=\sqrt{169 - 25}=\sqrt{144}=12\) inches. The area of a right - triangle is \(A=\frac{1}{2}ab\). Substitute \(a = 5\) and \(b = 12\), \(A=\frac{1}{2}\times5\times12=30\) square inches.

• Option f:

The area is given as \(A = 36\) square inches.

Answer:

A. Square with side length 6 inches, C. A 4 - inch - by - 9 - inch rectangle, F. A hexagon with an area of 36 square inches