Review question

# Can we find the dimensions of this rectangle? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6127

## Solution

Find the dimensions of the rectangle whose perimeter is $\quantity{36}{m}$ and which is such that the square of the length of the diagonal is $\quantity{170}{m^2}$.

Let the sides of the rectangle be of length $x$ and $y$ metres. Then the question tells us that the perimeter of the rectangle is $\quantity{36}{m}$, so $$$\label{eq:per} 2x+2y=36,$$$

and that the square of the length of the diagonal is $\quantity{170}{m^2}$, and so

$$$\label{eq:diag} x^2+y^2=170,$$$

by Pythagoras’ theorem.

Rearranging equation $\eqref{eq:per}$, we find that $y=18-x$, and so equation $\eqref{eq:diag}$ becomes $x^2+(18-x)^2=170,$ which when we multiply out the brackets becomes $x^2+324-36x+x^2=170,$ and so the quadratic we must solve is $2x^2-36x+154=0.$ We can divide this equation through by two and solve $x^2-18x+77=0.$ We can factorise the equation $(x-7)(x-11)=0.$ If $x=7$, we must have $y=11$, and $x=11$ gives us that $y = 7$.

Therefore the dimensions of the rectangle are $\quantity{7}{m}$ by $\quantity{11}{m}$.

This applet might help you visualise how the square on the diagonal varies with $x$.

The red rectangle has perimeter $36$.

When does the area of the square have its minimum value?

Notice how the two solutions for $x$ and $y$ are symmetrical.