You can break a regular hexagon into $6$ congruent equilateral triangles. How do you find the area of one equilateral triangle with side $10$, and then use it to get the total base area?

The slant height goes from the apex to the midpoint of a side of the base. What straight-line path on the base connects that midpoint to the center, and what right triangle does this create with the vertical height?

In each equilateral triangle of the hexagon, dropping an altitude creates a $30^\circ$–$60^\circ$–$90^\circ$ triangle. What are the side ratios for this kind of triangle, and how can you use them to find the apothem and then the vertical height?

In Desmos, type the expression for the area of the regular hexagon base: 6*(sqrt(3)/4)*10^2. The value Desmos shows is the base area $B$ in square centimeters.

To find the vertical height, enter sqrt(25^2 - (5*sqrt(3))^2). This uses the Pythagorean theorem with slant height $25$ and apothem $5\sqrt{3}$; the result is the height $h$ of the pyramid.

Now enter (1/3)*[result_from_step_1]*[result_from_step_2], or directly type (1/3)*6*(sqrt(3)/4)*10^2*sqrt(25^2 - (5*sqrt(3))^2). The numerical output is the volume of the pyramid; compare this value with the answer choices to see which expression matches it.

For any pyramid,

where:

• $B$ is the area of the base, and
• $h$ is the vertical height (perpendicular distance from the apex to the base).

In this problem:

• The base is a regular hexagon with side length $10$ cm.
• The given slant height $25$ cm is not the vertical height; it is the distance from the apex to the midpoint of a side of the base.

So we must find:

1. The area $B$ of the regular hexagonal base.
2. The vertical height $h$ using the slant height and the base’s geometry.

A regular hexagon with side length $s$ can be split into $6$ congruent equilateral triangles, each with side $s$.

For one equilateral triangle with side $s=10$:

• The area is

Since there are $6$ such triangles in the hexagon:

So the area of the base is $150\sqrt{3}$ square centimeters.

The apothem of a regular hexagon is the distance from its center to the midpoint of any side. For a regular hexagon made of equilateral triangles:

• Each triangle has side $10$.
• The altitude of that triangle is also the apothem of the hexagon.

In an equilateral triangle of side $10$, the altitude forms a $30^\circ$–$60^\circ$–$90^\circ$ triangle where:

• The hypotenuse is $10$ (the side of the triangle).
• The shorter leg is $5$.
• The longer leg (the altitude) is $5\sqrt{3}$.

Therefore, the apothem of the hexagon is $5\sqrt{3}$ cm.

Now look at the triangle formed by:

• The apex of the pyramid,
• The center of the hexagonal base,
• The midpoint of a side of the base.

This is a right triangle where:

• The slant height $25$ cm is the hypotenuse.
• The apothem $5\sqrt{3}$ cm is one leg (horizontal).
• The vertical height $h$ is the other leg.

Apply the Pythagorean theorem:

So the vertical height is:

Now substitute $B = 150\sqrt{3}$ and $h = 5\sqrt{22}$ into the volume formula:

First combine the numbers and radicals:

So the volume of the pyramid is $250\sqrt{66}$ cubic centimeters, which corresponds to choice B.