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Cuboid Perimeters

Age 14 to 16 Short
Challenge Level

Answer: $35\text{cm}^3$


Label the widths $a, b, c$


This perimeter is equal to $a+b+a+b=2a+2b$




This perimeter is equal to $2a+2c$

Similarly, the third perimeter will be equal to  $2b+2c$

$$2a+2b=12\Rightarrow a+b=6\\
2a+2c=16\Rightarrow a+c=8\\
2b+2c=24\Rightarrow b + c = 12$$ 


Solving by elimination
$\quad\qquad a+b=\ 6\\
\underline{+\ \quad\quad a+c=\ 8\ \ }\\
\quad 2a+b+c=14$


$\quad 2a+b+c=14\\
\underline{-\ \quad\quad b+c=12\ \ }\\
\quad \qquad\quad \ 2a=\ 2$

$\therefore a=1$
So $b=5$, $c=7$, volume $=1\times5\times7=35\text{cm}^3$


Solving by substitution
$a+b=6\Rightarrow b=6-a\\
 a+c=8\Rightarrow c=8-a\\
 \begin{align}b + c = 12\Rightarrow &(6-a)+(8-a)=12\\
\Rightarrow &14-2a=12\\
\Rightarrow &a=1\end{align}$

So $b=5$, $c=7$, volume $=1\times5\times7=35\text{cm}^3$


Adding all of the equations together
$\quad\qquad a+b=\ 6\\
\qquad\quad a+c=\ 8\\
\underline{+\ \quad\quad b+c=12\ \ }\\
\ \ \ 2a+2b+2c=26$

$\therefore a+b+c=13$

$c=13-6=7$
$b=13-8=5$
$a=13-12=1$
So the volume is $1\times5\times7=35\text{cm}^3$.

You can find more short problems, arranged by curriculum topic, in our short problems collection.