- With a ruler I draw the segment [AB] which is 7.5cm long.

- Then with my compasses, I draw an arc with centre A and radius 6cm.

- I draw an arc with centre B and radius 4.5cm. The two arcs intersect at C.

- The triangle ABC seems to be a right angle triangle, let's prove this result.
First we calculate the sum of the two squares : AC² + BC² = 56.25
We compare this result to AB² = 56.25
They are equal so we can apply Pythagora's theorem to conclude that ABC is a right angle triangle.

First we describe the situation with a pattern.
Let L be the length and W be the width.

The rectangle is 6 metres longer than it is wide so L = 6 + W.
The area of the rectangle is 135 m² so L.W = 135.
We substitute 6 + W for L in the previous equation and we have a quadratic equation :
(6 + W).W = 135.
Then we multiply out brackets and collect the terms on one side.
W² + 6W -135 = 0.
The discriminant of this equation is : Δ = 6² -4.(-135) = 576 = 24².
It's positive so we have two real solutions which are : W' = -15 and W'' = 9.
Or W is a length so it can't be negative and then W = 9.
Now we substitute 9 for W in the equality L = 6 + W and we work out the length L = 15.

The dimensions of the rectangle are 9 metres by 15 metres.