Question

a right pyramid with a square base has a base edge length of 12 meters and slant height of $6\sqrt{2}$ meters. the apothem is \underline{\hspace{1cm}} meters. the hypotenuse of triangle abc is the \underline{\hspace{1cm}}. the height is \underline{\hspace{1cm}} meters. the volume of the pyramid is \underline{\hspace{1cm}} cubic meters.

Explanation:

Part 1: Apothem of the square base

Step1: Recall apothem of square base

For a square base with side length \( s \), the apothem (distance from center to mid - side) is \( \frac{s}{2} \). Given \( s = 12 \) meters.

Step2: Calculate apothem

\( \text{Apothem}=\frac{12}{2}=6 \) meters.

Part 2: Hypotenuse of triangle \( ABC \)

In the right pyramid, the slant height is the hypotenuse of the right triangle formed by the height of the pyramid and the apothem. Triangle \( ABC \) has the slant height as its hypotenuse (since \( AC \) is the height, \( BC \) is the apothem, and \( AB \) is the slant height). So the hypotenuse of \( \triangle ABC \) is the slant height.

Part 3: Height of the pyramid

Step1: Use Pythagorean theorem

We know that for the right triangle formed by the height (\( h \)), apothem (\( a = 6 \)) and slant height (\( l=6\sqrt{2} \)), by Pythagorean theorem \( l^{2}=h^{2}+a^{2} \).

Step2: Substitute values and solve for \( h \)

\( (6\sqrt{2})^{2}=h^{2}+6^{2} \)

\( 72=h^{2} + 36 \)

\( h^{2}=72 - 36=36 \)

\( h = 6 \) meters (we take the positive value as height is a length).

Part 4: Volume of the pyramid

Answer:

s:

• Apothem: \( \boldsymbol{6} \) meters.
• Hypotenuse of \( \triangle ABC \): slant height.
• Height: \( \boldsymbol{6} \) meters.
• Volume: \( \boldsymbol{288} \) cubic meters.