# What`s the surface area formula for a rectangular pyramid?

Jan 2, 2016

$\text{SA} = l w + l \sqrt{{h}^{2} + {\left(\frac{w}{2}\right)}^{2}} + w \sqrt{{h}^{2} + {\left(\frac{l}{2}\right)}^{2}}$

#### Explanation:

The surface area will be the sum of the rectangular base and the $4$ triangles, in which there are $2$ pairs of congruent triangles.

Area of the Rectangular Base

The base simply has an area of $l w$, since it's a rectangle.

$\implies l w$

Area of Front and Back Triangles

The area of a triangle is found through the formula $A = \frac{1}{2} \left(\text{base")("height}\right)$.

Here, the base is $l$. To find the height of the triangle, we must find the slant height on that side of the triangle.

The slant height can be found through solving for the hypotenuse of a right triangle on the interior of the pyramid.

The two bases of the triangle will be the height of the pyramid, $h$, and one half the width, $\frac{w}{2}$. Through the Pythagorean theorem, we can see that the slant height is equal to $\sqrt{{h}^{2} + {\left(\frac{w}{2}\right)}^{2}}$.

This is the height of the triangular face. Thus, the area of front triangle is $\frac{1}{2} l \sqrt{{h}^{2} + {\left(\frac{w}{2}\right)}^{2}}$. Since the back triangle is congruent to the front, their combined area is twice the previous expression, or

$\implies l \sqrt{{h}^{2} + {\left(\frac{w}{2}\right)}^{2}}$

Area of the Side Triangles

The side triangles' area can be found in a way very similar to that of the front and back triangles, except for that their slant height is $\sqrt{{h}^{2} + {\left(\frac{l}{2}\right)}^{2}}$. Thus, the area of one of the triangles is $\frac{1}{2} w \sqrt{{h}^{2} + {\left(\frac{l}{2}\right)}^{2}}$ and both the triangles combined is

$\implies w \sqrt{{h}^{2} + {\left(\frac{l}{2}\right)}^{2}}$

Total Surface Area

Simply add all of the areas of the faces.

$\text{SA} = l w + l \sqrt{{h}^{2} + {\left(\frac{w}{2}\right)}^{2}} + w \sqrt{{h}^{2} + {\left(\frac{l}{2}\right)}^{2}}$

This is not a formula you should ever attempt to memorize. Rather, this an exercise of truly understanding the geometry of the triangular prism (as well as a bit of algebra).