Question

If the Pyramid of Giza has a height h and a square base of side length a, what would be the area of each of its triangular faces?
(The top vertex of the pyramid is exactly above the center of the base square.)

• A. $\frac{ah}{2}$
• B. $\frac{a}{2}\sqrt{h^2+\frac{a^2}{4}}$
• C. $\frac{a}{4}\sqrt{h^2+a^2}$
• D. $\frac{ah}{4}$

Answer 1

The correct option is B $\frac{a}{2}\sqrt{h^2+\frac{a^2}{4}}$

To make the calculations easier, we make the following markings on the pyramid:


Here:

$A,B,C,D,andEaretheverticesofthepyramid.$

Also,

$AFistheheightof△ACD.$

Now, we need to find the the area of a triangular face such as $△ACD$. So, we need to undestand its position first.

We know that the top vertex A of the pyramid is situated directly above the center of the base square BEDC.

So, taking M as the center of BEDC, we get the height h of the pyramid as the line segment AM.



Now, joining the height of the pyramid AM with the height of $△ACD$, which is AF, we get the right-angled $△AMF.$


Now, using the Pythagorean theorem, we can calculate the length of AF as follows:

$AF=\sqrt{A{M}^2+M{F}^2}=\sqrt{h^2+(\frac{a}{2}{)}^2}=\sqrt{h^2+\frac{a^2}{4}}$

Finally, on calculating the area of $△ACD$, we get,


$Area=\frac{1}{2}\times Base\times Height=\frac{1}{2}\times CD\times AF$

Since CD is a side of the base square, we have,

$CD=a$

$⟹Areaof△ACD=\frac{1}{2}\times CD\times AF=\frac{1}{2}\times a\times \sqrt{h^2+\frac{a^2}{4}}=\frac{a}{2}\sqrt{h^2+\frac{a^2}{4}}$

Hence, the area of each of the triangular faces of the pyramid will be:

$\frac{a}{2}\sqrt{h^2+\frac{a^2}{4}}$