In ΔABC ∠A = 90° and AD ⊥ BC where D lies on BC. If BC = 8 cm, AD = 6 cm, then arΔABC : arΔACD = ?

Answer: Option C

According to question,

ΔABC ∼ ΔACD
$$\begin{array}{rl}& \therefore \frac{\text{area}\text{of}\Delta ABC}{\text{area}\text{of}\Delta ACD}=\frac{B{C}^2}{A{D}^2}\\ & \frac{\text{area}\text{of}\Delta ABC}{\text{area}\text{of}\Delta ACD}=\frac{8^2}{6^2}\\ & \frac{\text{area}\text{of}\Delta ABC}{\text{area}\text{of}\Delta ACD}=\frac{64}{36}\\ & \frac{\text{area}\text{of}\Delta ABC}{\text{area}\text{of}\Delta ACD}=\frac{16}{9}\end{array}$$
area ΔABC : area ΔACD = 16 : 9