Question

ABC is right-angled at A and AD 1 BC. If BC =
13 cm and AC = 5 cm, find the ratio of the areas of
ABC and AADC.​

Answer 1

Answer:

169/25

Step-by-step explanation:

In ∆ABC and ∆ADC, we have: ∠BAC = ∠ADC = 90° ∠ACB = ∠ACD By AA similarity, we can conclude that ∆ BAC~ ∆ ADC. Hence, the ratio of the areas of these triangles is equal to the ratio of squares of their corresponding sides

so

ar(BAC)/ar(ADC)= BC^2/AC^2

= 13^2/5^2

=169/25