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

From the given conditions, it is stated that:

$$∠KCF = 120^ ext{o}$$

To find the area of triangle AKF, we can use the formula for the area of a triangle:

$$ext{Area} = \frac{1}{2} \cdot AK \cdot KF \cdot \sin y$$

From our earlier calculation, we have:

$$AK = \frac{\sqrt{3}}{2} x$$

And using the cosine rule for triangle KCF, we find KF:

$$KF^2 = CF^2 + CK^2 - 2 \cdot CF \cdot CK \cdot \cos 120^ ext{o}$$

This gives:

$$KF = \sqrt{x^2 + \left(\frac{x}{2}\right)^2 - 2x \cdot \frac{x}{2} \cdot \left(-\frac{1}{2}\right)}$$

This leads to:

$$KF = \frac{x \sqrt{7}}{2}$$

Plugging it back into the area calculation:

$$ext{Area} = \frac{1}{2} \cdot \left(\frac{\sqrt{3}}{2} x\right) \cdot \left(\frac{x \sqrt{7}}{2}\right) \cdot \sin y = \frac{\sqrt{21}}{8} x^2 \sin y$$