A pyramid has a square base which has an area of $1440 \mathrm{~cm}^{2}$. Each of the pyramid's triangular faces is identical and each has an area of $840 \mathrm{~cm}^{2}$. The height of the pyramid, in $\mathrm{cm}$, is
(A) $30 \sqrt{2}$
(B) 40
(C) $20 \sqrt{6}$
(D) $20 \sqrt{3}$
(E) 30
We label the pyramid with vertices $A, B, C$, and $D$ (the square base) and $T$ the "top" vertex. Let $M$ be the midpoint of side $A B$ on the base, and $O$ the centre of the square base.

Since the pyramid has a square base and each of the four triangular faces is identical, then the "top" vertex of the pyramid lies directly above the centre of the base, by symmetry, and so each of the four triangular faces is isosceles.

Join $T$ to $O, T$ to $M$, and $M$ to $O$.

Then $T O$ is perpendicular to the square base by the symmetry of the pyramid, and so is perpendicular to $O M$.

Therefore, triangle $T O M$ is right-angled at $O$.

Let $H$ be the height of the pyramid, ie. $T O=H$.

Let $s$ be the side length of the base of the pyramid. Then

$M O=\frac{1}{2} s$, since $O$ is the centre of the square and $M$ is the midpoint of $A B$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-226.jpg?height=414&width=479&top_left_y=693&top_left_x=1365)

Let $h$ be the length of $M T$. Since $M$ is the midpoint of $A B$ and $\triangle T A B$ is isosceles, then $T M$ is perpendicular to $A B$.

So by Pythagoras, $H^{2}=h^{2}-\left(\frac{1}{2} s\right)^{2}=h^{2}-\frac{1}{4} s^{2}$.

But the base is square, so its area is $s^{2}=1440$, and the area of each of the triangular faces is $\frac{1}{2} s h=840$, so

$h^{2}=\left(\frac{1680}{s}\right)^{2}=\frac{1680^{2}}{1440}=1960$.

Therefore, $H^{2}=1960-360=1600$, and so $H=40$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-226.jpg?height=447&width=253&top_left_y=1202&top_left_x=1554)

ANSWER: (B)