In the diagram below, ΔPQR is drawn with PQ = 20 − 4x, RQ = x and ∠Q = 60° - NSC Mathematics - Question 7 - 2016 - Paper 2

To find the maximum area, we differentiate the area function:

$A = 5\sqrt{3}x - \sqrt{3}x^2$

Taking the derivative: $\frac{dA}{dx} = 5\sqrt{3} - 2\sqrt{3}x$

Setting the derivative to zero for maximization:

$5\sqrt{3} - 2\sqrt{3}x = 0$

Solving for $x$: $2\sqrt{3}x = 5\sqrt{3}$ $x = \frac{5}{2}$

Thus, the value of $x$ for maximum area is $\frac{5}{2}$.

We already found that $x = \frac{5}{2}$. Now, using the length expression for PQ and RQ:

$PQ = 20 - 4\left(\frac{5}{2}\right) = 20 - 10 = 10$ $RQ = \frac{5}{2}$

Using the cosine rule to calculate PR:

$PR^2 = PQ^2 + RQ^2 - 2 \cdot PQ \cdot RQ \cdot \cos(60°)$ $= 10^2 + \left(\frac{5}{2}\right)^2 - 2 \cdot 10 \cdot \frac{5}{2} \cdot \frac{1}{2}$ $= 100 + \frac{25}{4} - 25$ $= 100 - 25 + \frac{25}{4}$ $= 75 + \frac{25}{4} = \frac{300}{4} + \frac{25}{4} = \frac{325}{4}$

Therefore, $PR = \sqrt{\frac{325}{4}} = \frac{\sqrt{325}}{2}$.

From triangle ABC, we have:

$\sin(\beta) = \frac{h}{AB}$ Thus, $AB = \frac{h}{\sin(\beta)}$

Using the cosine rule in triangle ABD: $AD = \frac{AB \cdot \sin(2\beta)}{\sin(90 - \beta)}$ Substituting: $AD = \frac{h}{\sin(\beta)} \cdot 2\sin(\beta)\cos(\beta)$ $= \frac{h \cdot 2\cos(\beta)}{\sin(\beta)}$

Expanding gives: $AD = \frac{2h \cdot \cos(\beta)}{\sin(\beta)} = 2h \cot(\beta)$

Thus, the distance AD is expressed as $2h \cot(\beta)$.