Question

The longest rod that can be placed flat on the bottom of a box is $45\text{cm}$. The box is $9\text{cm}$ longer than it is wide. Find the length and breadth of the box.

Answer 1

Step 1: Finding quadratic equation in $x$

Let the breadth of the box be $x\text{cm}$.

Since it is given that length of box is $9\text{cm}$ longer than the breadth of box.

Therefore, the length of box $=\left(x+9\right)\text{cm}$

Also, given that the longest rod that can be placed flat on the bottom of a box is $45\text{cm}$.

So, the diagonal of box is $45\text{cm}$.

By Pythagoras theorem,

$$\begin{gathered} \Rightarrow x^2+{\left(x+9\right)}^2={45}^2 \\ \Rightarrow x^2+x^2+18x+81=2025 \\ \Rightarrow 2x+18x-1944=0 \end{gathered}$$

Step 2: Finding the length and breadth of the box

By factorization method

$$\begin{gathered} \Rightarrow 2x^2+18x-1944=0 \\ \Rightarrow x^2+9x-972=0 \\ \Rightarrow x^2+36x-27x-972=0 \\ \Rightarrow (x+36)(x-27)=0 \\ \Rightarrow x=-36,27 \end{gathered}$$

The breadth of the box cannot be negative.

So, breadth $\left(x\right)=27\text{cm}$

And length $\left(x+9\right)=36\text{cm}$

Hence, the length and breadth of the box are $36\text{cm}$and $27\text{cm}$ receptively.