Question 129·Hard·Geometry and Trigonometry

The $13$-unit edges go from the apex to the vertices of the square base. In a right square pyramid, the apex is directly above the center of the base. How can these facts form a right triangle involving the height?

You need the distance from the center of the square base to a vertex. Use the fact that the base is a square of side $10$; what is the length of its diagonal, and then half of that diagonal?

Set up a right triangle where the hypotenuse is $13$, one leg is the distance from the center of the base to a vertex, and the other leg is the unknown height. Use the Pythagorean theorem to solve for the height, then plug it into the volume formula.

In Desmos, first enter the expression for the height using the Pythagorean theorem:

• Type: sqrt(13^2 - (5*sqrt(2))^2)

Desmos will display the numeric value of the pyramid's vertical height.

Now use the volume formula with base area $100$:

• In a new line, type: (1/3)*100*sqrt(13^2 - (5*sqrt(2))^2)

The resulting value is the volume of the pyramid; if you wish, you can compare this decimal to the simplified radical form in the answer choices.

For any pyramid,

where $B$ is the area of the base and $h$ is the vertical height (altitude) from the apex to the base.

The base is a square with side $10$, so its area is

So the volume will be

once we find the height $h$.

Because it is a right square pyramid, the apex is directly above the center of the square base. The vertical height $h$ drops straight down to the center of the base.

To use the given slant edge (length $13$ from apex to a vertex), consider the right triangle formed by:

• the apex,
• the center of the base,
• and a vertex of the base.

In this triangle:

• one leg is the vertical height $h$,
• the other leg is the distance from the center of the square to a vertex,
• the hypotenuse is the slant edge of length $13$.

So we need the distance from the center of the square to a vertex.

The segment from one vertex of the square to the opposite vertex is a diagonal of the square. For a square with side $s$, the diagonal has length $s\sqrt{2}$.

Here, $s=10$, so the full diagonal is

The center of the square is the midpoint of this diagonal, so the distance from the center to any vertex is half the diagonal:

This $5\sqrt{2}$ is the horizontal leg in our right triangle with hypotenuse $13$ and vertical leg $h$.

In the right triangle with legs $h$ and $5\sqrt{2}$ and hypotenuse $13$, apply the Pythagorean theorem:

Compute each term:

So

Now we know the vertical height of the pyramid.

Substitute $B=100$ and $h=\sqrt{119}$ into the volume formula $V=\frac{1}{3}Bh$:

So the volume of the pyramid is

which corresponds to choice D.