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Question
ΔABC is right angled at A and AD⊥BC. If BC = 13cm and AC = 5cm, find the ratio of the areas of ΔABC and ΔADC.
Solution
In ΔABC and ΔADC, we have:
∠𝐵𝐴𝐶= ∠𝐴𝐷𝐶=90°
∠𝐴𝐶𝐵= ∠𝐴𝐶𝐷 (𝑐𝑜𝑚𝑚𝑜𝑛)
By AA similarity, we can conclude that Δ BAC~ Δ ADC.
Hence, the ratio of the areas of these triangles is equal to the ratio of squares of their corresponding sides.
∴ `(ar(ΔBAC))/(ar(ΔADC))=(BC)^2/(AC)^2`
⇒ `(ar (Δ BAC))/(9ar(Δ ADC))=13^2/5^2`
=` 169/25`
Hence, the ratio of areas of both the triangles is 169:25
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