ΔABC is in the diagram drawn below - NSC Mathematics - Question 7 - 2024 - Paper 2

The area of triangle ABC can be calculated using the formula for the area of a triangle:

$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$

Here, we can identify the base as BC and the height as AB. Thus, the area is:

$\text{Area} = \frac{1}{2} (BC)(AB) \sin B$

In triangles ADB and ACB, the following is true:

1. ( AD = AC ) (given as part of the problem).
2. ( \angle ADB = \angle ACB = 90° ) (right angles provided by the diagram).
3. Since ( AD = AC ) and both triangles share the common angle B,

This establishes that triangles ADB and ACB are congruent, thus verifying ( AD = AC ).

In triangle BDC, we can use the cosine rule. From the given data, we know:

1. ( CD = k ) and (
2. ( \angle CDB = θ ).

Using the cosine rule:

$BD = \frac{k}{2 \cos θ}$

To find the area of quadrilateral ABCD, we can split it into two triangles, ABC and ACD. The area of triangle ABC is:

$\text{Area}_{ABC} = \frac{1}{2}(BC)(AB)\sin B$

And for triangle ACD, we have the height which can be represented in terms of k and θ as:

$\text{Area}_{ACD} = \frac{1}{2}(k)(AC)\sin θ$

By using trigonometric identities, we can express the areas in terms of k and θ combined.