Review question

# What is the surface area of this cuboid? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7565

## Solution

The sum of the lengths of the $12$ edges of a cuboid is $\quantity{x}{cm}$. The distance from one corner of the cuboid to the furthest corner is $\quantity{y}{cm}$. What in $\mathrm{cm^2}$ is the total surface area of the cuboid?

(A) $\quad \dfrac{x^2-2y^2}{2} \qquad$ (B) $\quad x^2+y^2 \qquad$ (C) $\quad \dfrac{x^2-4y^2}{4} \qquad$ (D) $\quad \dfrac{xy}{6} \qquad$ (E) $\quad \dfrac{x^2-16y^2}{16}$

Let the sides of the cuboid be $a$, $b$ and $c$.

This means the total edge-length, $x = 4(a+b+c)$.

The length of the longest possible diagonal is, by Pythagoras, $y = \sqrt{a^2 + b^2 + c^2}$.

The surface area we seek is $A=2(ab+bc+ca)$.

We need to find a relationship between $x$, $y$ and $A$. Thinking about the units or dimensions of each quantity, we have $[x]=[y]=[L]$ and $[A]=[L^2]$ which suggests we should be interested in $x^2$ and $y^2$. We have $x^2=16(a+b+c)^2 = 16(a^2+b^2+c^2+2ab+2bc+2ca) = 16(y^2+A).$ Hence we have $A=\frac{x^2-16y^2}{16}$ and the answer is (E).