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.

Answer 1

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