The diagram below shows a solar panel, ABCD, which is fixed to a flat piece of concrete slab EFCD - NSC Mathematics - Question 7 - 2019 - Paper 2

The size of KCF is given as 120 degrees.

To calculate the area of triangle AKF, we can use the formula:

$\text{Area} = \frac{1}{2} \times AK \times KF \times \sin(AKF)$

We have already determined that (AK = \frac{\sqrt{3}}{2}x) and (AKF = y). Next, we need to find KF using the cosine rule:

$KF^2 = CF^2 + CK^2 - 2 \cdot CF \cdot CK \cdot \cos(KCF)$

Given that (CF = \frac{x}{2}) and ( \cos(120^{\circ}) = -\frac{1}{2}), we can substitute and solve for KF:

$KF^2 = \left(\frac{x}{2}\right)^2 + CK^2 + (x)(x)$

The area would then be represented as:

$\text{Area} = \frac{1}{2} \cdot \frac{\sqrt{3}}{2}x \cdot KF \cdot \sin(y)$