In the diagram, $A B C D H E F G$ is a truncated square-based pyramid. (Note that $D$ is hidden.) In particular, $A B C D$ is a square with side length $7, E F G H$ is a square with side length 1, and $A E=B F=C G=D H$. Each of the four trapezoidal faces is partially shaded. On each face, only the region that is below both diagonals is unshaded. If the height of the truncated pyramid is 4 , what is the total

![](https://cdn.mathpix.com/cropped/2024_04_17_8918b015730641e4f574g-1.jpg?height=317&width=596&top_left_y=1912&top_left_x=1358)
shaded area?
By symmetry, the area of the shaded region on each of the four trapezoidal faces will be the same. Therefore, the total shaded area will be 4 times the shaded area on trapezoid EFBA. Let $h$ be the height of trapezoid $E F B A$. In other words, $h$ is the distance between the parallel sides $E F$ and $A B$.

Let $P$ be the point of intersection of $E B$ and $F A$.

![](https://cdn.mathpix.com/cropped/2024_04_30_68642ad613c12af1e439g-05.jpg?height=250&width=453&top_left_y=466&top_left_x=890)

Since $E F G H$ is a square with side length 1, then $E F=1$. Similarly, $A B=7$.

Therefore, the area of trapezoid $E F B A$ is $\frac{1}{2}(1+7) h=4 h$.

The shaded area on trapezoid $E F B A$ equals the area of $E F B A$ minus the area of $\triangle A P B$.

Since $E F$ and $A B$ are parallel, then $\angle F E P=\angle P B A$.

This means that $\triangle F P E$ is similar to $\triangle A P B$.

Since $A B: E F=7: 1$, then the ratio of the height of $\triangle A P B$ to the height of $\triangle F P E$ must also be $7: 1$.

Since the sum of the heights of these triangles is the height of the trapezoid, then the height of $\triangle A P B$ is $\frac{7}{8} h$.

Thus, the area of $\triangle A P B$ is $\frac{1}{2} A B \cdot \frac{7}{8} h=\frac{49}{16} h$, because $A B=7$.

Therefore, the shaded area on trapezoid EFBA is $4 h-\frac{49}{16} h=\frac{15}{16} h$, and so the total shaded area on the truncated pyramid is $4 \cdot \frac{15}{16} h=\frac{15}{4} h$.

To complete our solution, we need to calculate $h$.

To do this, we begin by dropping perpendiculars from $E$ and $F$ to points $Q$ and $R$ on $A B$, from $F$ to point $S$ on $B C$, and from $F$ to point $T$ in square $A B C D$.

![](https://cdn.mathpix.com/cropped/2024_04_30_68642ad613c12af1e439g-05.jpg?height=336&width=610&top_left_y=1583&top_left_x=817)

Since $E F$ is parallel to $Q R$ and $\angle E Q R=\angle F R Q=90^{\circ}$, then $E F R Q$ is a rectangle.

Thus, $Q R=1$.

Since $E Q=F R=h$ and $E A=F B$ and each of $\triangle E Q A$ and $\triangle F R B$ is right-angled, then these triangles are congruent, and so $A Q=R B$.

Since $A B=7$ and $Q R=1$, then $A Q=R B=3$. Similarly, $B S=3$.

Now $R B S T$ must be a square of side length 3 , since $E F$ lies directly above $S T$ extended and $F G$ lies directly above $R T$ extended.

Finally, we have $T R=3, F T=4, F R=h$, and $\angle F T R=90^{\circ}$.

By the Pythagorean Theorem, $h^{2}=3^{2}+4^{2}=25$, and so $h=5$, since $h>0$.

This means that the total shaded area is $\frac{15}{4} h=\frac{75}{4}$.

ANSWER: $\frac{75}{4}$