Die diagram hieronder toon 'n sonpaneel, ABCD, wat aan 'n plat stuk sementblad, EFCD, vasgeheg is - NSC Mathematics - Question 7 - 2019 - Paper 2

The size of ∠KCF is given as:

$\angle KCF = 120°$

To determine the area of triangle ΔAKF, we use the formula:

$\text{Area} = \frac{1}{2} \cdot AK \cdot KF \cdot \sin \angle AKF$

From previous calculation, we have:

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

For KF, using the cosine rule:

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

Substituting the known values, we can solve for KF. Assuming CF = (\frac{x}{2}), we have:

$KF^2 = \left(\frac{x}{2}\right)^2 + \left(\frac{x}{2}\right)^2 - 2\left(\frac{x}{2}\right) \left(\frac{x}{2}\right) \left(\cos 120°\right)$

Since (\cos 120° = -\frac{1}{2}):

$KF^2 = \frac{x^2}{4} + \frac{x^2}{4} + \frac{x^2}{4} = \frac{3x^2}{4}$

Thus:

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

Now substituting back into the area formula:

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

This simplifies to:

$\text{Area} = \frac{3x^2}{8} \cdot \sin y$