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In ΔABC ∠A = 90° and AD ⊥ BC where D lies on BC. If BC = 8 cm, AD = 6 cm, then arΔABC : arΔACD = ?
Answer & Solution
Correct Answer:
Option
C
According to question,
ΔABC ∼ ΔACD
$$\eqalign{ & \therefore \frac{{{\text{area}}\,{\text{of}}\,\Delta ABC}}{{{\text{area}}\,{\text{of}}\,\Delta ACD}} = \frac{{B{C^2}}}{{A{D^2}}} \cr & \frac{{{\text{area}}\,{\text{of}}\,\Delta ABC}}{{{\text{area}}\,{\text{of}}\,\Delta ACD}} = \frac{{{8^2}}}{{{6^2}}} \cr & \frac{{{\text{area}}\,{\text{of}}\,\Delta ABC}}{{{\text{area}}\,{\text{of}}\,\Delta ACD}} = \frac{{64}}{{36}} \cr & \frac{{{\text{area}}\,{\text{of}}\,\Delta ABC}}{{{\text{area}}\,{\text{of}}\,\Delta ACD}} = \frac{{16}}{9} \cr} $$
area ΔABC : area ΔACD = 16 : 9
ΔABC ∼ ΔACD
$$\eqalign{ & \therefore \frac{{{\text{area}}\,{\text{of}}\,\Delta ABC}}{{{\text{area}}\,{\text{of}}\,\Delta ACD}} = \frac{{B{C^2}}}{{A{D^2}}} \cr & \frac{{{\text{area}}\,{\text{of}}\,\Delta ABC}}{{{\text{area}}\,{\text{of}}\,\Delta ACD}} = \frac{{{8^2}}}{{{6^2}}} \cr & \frac{{{\text{area}}\,{\text{of}}\,\Delta ABC}}{{{\text{area}}\,{\text{of}}\,\Delta ACD}} = \frac{{64}}{{36}} \cr & \frac{{{\text{area}}\,{\text{of}}\,\Delta ABC}}{{{\text{area}}\,{\text{of}}\,\Delta ACD}} = \frac{{16}}{9} \cr} $$
area ΔABC : area ΔACD = 16 : 9
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