21) The figure on the side is the net of a right pyramid with a base that is an equilateral triangle of side 1 and lateral faces that are three equal right isosceles triangles. The volume of the pyramid is
(A) $\frac{1}{24}$
(B) $\frac{\sqrt{2}}{24}$
(C) $\frac{\sqrt{3}}{24}$
(D) $\frac{1}{12}$
(E) $\frac{\sqrt{5}}{24}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_b5d519ff6c1a8005abb1g-2.jpg?height=299&width=332&top_left_y=466&top_left_x=2415)
21) The answer is (B).

By reconstructing the pyramid, we find that its height is a leg of a right triangle where:

- the hypotenuse is the height relative to the hypotenuse of the isosceles right triangle in the figure;
- the other leg is \( \frac{1}{3} \) of the height of the equilateral triangle in the figure.

The height of the pyramid is therefore equal to

\[
\sqrt{\left(\frac{1}{2}\right)^{2}-\left(\frac{\sqrt{3}}{6}\right)^{2}}=\sqrt{\frac{1}{4}-\frac{1}{12}}=\frac{1}{\sqrt{6}}
\]

and its volume is equal to

\[
V=\frac{1}{3}\left(\frac{1}{2} \cdot 1 \cdot \frac{\sqrt{3}}{2}\right) \frac{1}{\sqrt{6}}=\frac{\sqrt{2}}{24} .
\]