ABC is a triangle. AB = 5 cm, AC = 41 cm and BC = 8 cm. AD is perpendicular to BC. What is the area (in cm2) of triangle ABD?

Answer: Option B

Solution

Answer: Option B

Solution:

$$\begin{array}{rl}& \text{Let }\angle ACB=\alpha \\ & \angle DAC=90^{\circ }-\alpha \\ & \cos \alpha =\frac{{\left(\sqrt{41}\right)}^2+{\left(8\right)}^2-{\left(5\right)}^2}{2\sqrt{41}\times 8}=\frac{DC}{\sqrt{41}}\\ & \frac{41+64-25}{16\sqrt{41}}=\frac{DC}{\sqrt{41}}\\ & \frac{80}{16}=DC\\ & DC=5\text{ cm}\\ & BD=8-5=3\text{ cm}\end{array}$$

$$\text{Area of }\Delta ABD=\frac{1}{2}\times 4\times 3=6\text{ c}{\text{m}}^2$$