Solution — 2016 Question 5 Quadratic Equations & Inequalities

(a)

By Pythagoras’ Theorem, height of $△ A B C$ :

\begin{matrix} h^{2} & = {(2 x)}^{2} - x^{2} \\ = 3 x^{2} \\ h & = \sqrt{3} x \end{matrix}

\begin{matrix} Area of △ A B C & = \frac{1}{2} (2 x) (\sqrt{3} x) \\ = x^{2} \sqrt{3} \end{matrix}

By Pythagoras’ Theorem, height of $△ D E F$ :

\begin{matrix} h^{2} & = y^{2} - {(\frac{y}{2})}^{2} \\ = \frac{3}{4} y^{2} \\ h & = \frac{\sqrt{3}}{2} y \end{matrix}

\begin{matrix} Area of △ D E F & = \frac{1}{2} y (\frac{\sqrt{3}}{2} y) \\ = \frac{y^{2} \sqrt{3}}{4} \end{matrix}

\begin{matrix} x^{2} \sqrt{3} - \frac{y^{2} \sqrt{3}}{4} & = 2 \sqrt{3} \\ x^{2} - \frac{y^{2}}{4} & = 2 \\ 4 x^{2} - y^{2} & = 8 ∎ & (1) \end{matrix}

(b)

Considering the perimeter,

\begin{matrix} 2 (2 x) + 2 y + 2 x - y & = 10 \\ 6 x + y & = 10 \\ y & = 10 - 6 x & (2) \end{matrix}

Substituting (2) into (1)

\begin{matrix} 4 x^{2} - y^{2} & = 8 \\ 4 x^{2} - {(10 - 6 x)}^{2} & = 8 \\ 4 x^{2} - 100 + 120 x - 36 x^{2} & = 8 \\ - 32 x^{2} - 100 + 120 x & = 8 \\ 32 x^{2} + 108 - 120 x & = 0 \\ 32 x^{2} - 120 x + 108 & = 0 \\ 8 x^{2} - 30 x + 27 & = 0 \\ (2 x - 3) (4 x - 9) & = 0 \end{matrix}

$x = \frac{3}{2} or x = \frac{9}{4}$

Since $y = 10 - 6 x > 0,$ $x = \frac{9}{4}$ is rejected.

When $x = \frac{3}{2} ∎,$

\begin{matrix} y & = 10 - 6 (\frac{3}{2}) \\ = 1 ∎ \end{matrix}