Quadratic equations from areas and lengths

Areas of rectangles, triangles and borders involve the product of two lengths, so an area condition often gives a quadratic equation. Every length must be positive.

Method

1. Mark the unknown length \(x\) on a sketch and write every other length in terms of \(x\).
2. Use the area or perimeter formula to form an equation, clearing any fraction first.
3. Rearrange to \(ax^2+bx+c=0\), factorise and solve.
4. Reject any root that makes a length zero or negative, then find the length the question asks for.

Example

A right-angled triangle has legs \(x\) cm and \((x+7)\) cm, and its area is \(30\text{ cm}^2\). Find \(x\).

\(\begin{aligned} \dfrac{1}{2}x(x+7) &= 30 \\ x^2+7x &= 60 \\ x^2+7x-60 &= 0 \\ (x+12)(x-5) &= 0 \\ x=-12 \quad &\text{or}\quad x=5 \end{aligned}\)

Reject \(x=-12\) because a length cannot be negative. So \(x=5\), and the legs are \(5\) cm and \(12\) cm.

Check: \(\dfrac{1}{2}\times5\times12=30\). A common error is to forget the \(\dfrac{1}{2}\) and solve \(x(x+7)=30\).