Question

a cube with edge lengths of x units can be deconstructed into 6 pyramids, where each pyramid has the same volume. each pyramid has a square base with side lengths of x units. which equation describes the volume, v, of one of the pyramids in terms of the height, h, in units, of the pyramid?
a. ( v = \frac{1}{6}x^2 = \frac{1}{3}hx ), where ( h = \frac{x}{2} )

b. ( v = \frac{1}{6}x^3 = \frac{1}{3}hx^2 ), where ( h = \frac{x}{2} )

c. ( v = \frac{1}{3}x^3 = \frac{1}{3}hx^2 ), where ( h = x )

d. ( v = \frac{2}{3}x^3 = \frac{1}{3}hx^2 ), where ( h = 2x )

Explanation:

Step1: Find volume of cube

Volume of cube \( V_{cube} = x^3 \) (since edge length is \( x \)).

Step2: Volume of one pyramid

Cube is divided into 6 equal - volume pyramids, so volume of one pyramid \( V=\frac{1}{6}x^3 \).

Step3: Volume formula for pyramid

Volume of a pyramid is \( V = \frac{1}{3}Bh \), where \( B \) is the area of the base. The base is a square with side length \( x \), so \( B=x^2 \). Then \( V=\frac{1}{3}x^2h \).

Step4: Relate the two expressions for pyramid volume

Set \( \frac{1}{6}x^3=\frac{1}{3}x^2h \). Solve for \( h \):
Divide both sides by \( \frac{1}{3}x^2 \) (assuming \( x
eq0 \)), we get \( h = \frac{\frac{1}{6}x^3}{\frac{1}{3}x^2}=\frac{x}{2} \).

Answer:

B. \( V=\frac{1}{6}x^{3}=\frac{1}{3}hx^{2} \), where \( h = \frac{x}{2} \)