23. The edge-length of the solid cube shown is 2 . A single plane cut goes through the points $Y, T, V$ and $W$ which are midpoints of the edges of the cube, as shown.
What is the area of the cross-section?
A $\sqrt{3}$
B $3 \sqrt{3}$
C 6
D $6 \sqrt{2}$
E 8
Solution
B
The plane which goes through the points $Y, T, V$ and $W$ also meets the edges of the cube at the points $U$ and $X$, as shown.
We leave it as an exercise to check that the points $U$ and $X$ are also the midpoints of the edges on which they lie.

The relevant cross-section is therefore the hexagon $T U V W X Y$. We first show that this is a regular hexagon.

Let $P$ and $Q$ be the vertices of the cube, as shown. Applying Pythagoras' Theorem to the right-angled triangle $T P Y$, gives $T Y^{2}=$ $P T^{2}+P Y^{2}=1^{2}+1^{2}=2$. Therefore $T Y=\sqrt{2}$. In a similar way it follows that each edge of the hexagon $T U V W X Y$ has length $\sqrt{2}$.
Applying Pythagoras' Theorem to the right-angled triangle $T P Q$ gives $Q T^{2}=P T^{2}+P Q^{2}=1^{2}+2^{2}=5$. The triangle $T X Q$ has a right angle at $Q$ because the top face of the cube is perpendicular to the edge through $Q$ and $X$. Hence, by Pythagoras' Theorem, $T X^{2}=Q T^{2}+Q X^{2}=5+1^{2}=6$. Hence $T X=\sqrt{6}$.
It follows that in the triangle $T X Y$ we have $T Y=Y X=\sqrt{2}$ and $T X=\sqrt{6}$. We leave it as an exercise to check that it follows that $\angle T Y X=120^{\circ}$. It follows similarly that each angle of $T U V W X Y$ is $120^{\circ}$. Hence this hexagon is regular.
The regular hexagon $T U V W X Y$ may be divided into six congruent equilateral triangles each with side length $\sqrt{2}$.
We leave it as an exercise to check that the area of an equilateral triangle with side length $s$ is $\frac{\sqrt{3}}{4} s^{2}$. It follows that equilateral triangles with side lengths $\sqrt{2}$ have area $\frac{\sqrt{3}}{4}(\sqrt{2})^{2}=\frac{\sqrt{3}}{2}$
Therefore the area of the cross-section is $6 \times \frac{\sqrt{3}}{2}=3 \sqrt{3}$.
Note: The facts you are asked to check in the above solution are given in as Exercises 23.1, 23.2 and 23.3 below.