Thinking about Algebra

Can we find the dimensions of this rectangle?

Ref: R6127

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

$\begin{equation}\label{eq:per} 2x+2y=36, \end{equation}$

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

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

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 $x$ .

Download GeoGebra file

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.

Thinking about Algebra

Can we find the dimensions of this rectangle?

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