7.1 In the diagram, \( \triangle ABC \) is drawn - NSC Mathematics - Question 7 - 2024 - Paper 2

The area of triangle can be expressed as:
[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} ]
Here, taking ( BC ) as the base and ( AD ) as the height, we have:
[ ext{Area} = \frac{1}{2} (BC)(AD) ]
Substituting the expression we found earlier for ( AD ):
[ \text{Area} = \frac{1}{2} (BC)(AB \sin B) ]
Thus,
[ \text{Area} = \frac{1}{2} (BC)(AB) \sin B ].

In triangles ( ADB ) and ( ACB ):

• We are given that ( AB = AC ) as they are the sides opposite angle ( \alpha ).
• Since ( \angle ABC = \angle ACB = 90^\circ ), by the properties of congruent triangles (SAS or AAS), we can say ( AD \cong AC ).
  Therefore, ( AD = AC ).

Using the cosine rule in triangle ( ABC ):
[ BC^2 = BD^2 + DC^2 - 2(BD)(DC)\cos \theta ]
Given that ( DC = k ) and ( BD = BC ), we can substitute to find ( BD ):
[ k^2 = k^2 + BD^2 - 2(BD)(k)\cos \theta ]
Rearranging yields:
[ 2(BD)(k)\cos \theta = BD^2 ]
Thus, we can isolate ( BD ):
[ BD = \frac{k}{2 \cos \theta} ].

The area of quadrilateral ( ABCD ) can be calculated as:
[ ext{Area} = \text{Area}(ABCD) = \text{Area}(\triangle ABC) + \text{Area}(\triangle ABD) ]
Using the formula for the area of triangles, we find:
[ ext{Area}(ABCD) = \frac{1}{2}(BC)(AD) + \frac{1}{2}(BD)(AD) ]
Substituting the previously calculated areas:
[ ext{Area}(ABCD) = \frac{1}{2}(k)(AD) + \frac{k^2}{2\cos\theta} ]
This can be simplified and expressed appropriately in terms of ( k ) and a trigonometric ratio of ( \theta ).