Question

In the give figure, ABC is a triangle with ∠EDB = ∠ACB. Prove that ΔABC ∼ ΔEBD. If BE = 6 cm, EC = 4 cm, BD = 5 cm and area of ΔBED = 9 cm². Calculate the: 

1. length of AB
2. area of ΔABC

Answer 1

In ΔABC and ΔEBD

∠1 = ∠2   ...(Given)

∠B = ∠B  ...(Common)

`\implies` ΔABC ∼ ΔEBD  ...[By AA axiom of similarity]

Now, \[\frac{\text{Area of } \triangle ABC}{\text{Area of } \triangle EBD} = \left(\frac{BC}{BD}\right)^2\] ⇒ \[\frac{\text{Area of } \triangle ABC}{9} = \left(\frac{10}{5}\right)^2\]

⇒ Area of ΔABC = 4 × 9

⇒ Area of ΔABC = 36 cm²

Also, \[\frac{\text{Area of } \triangle ABC}{\text{Area of } \triangle EBD} = \frac{(AB)^2}{(BE)^2}\]

⇒ `36/9 = (AB^2)/36`

⇒ `AB^2 = (36 xx 36)/9`

⇒ AB² = 144

⇒ AB = 12 cm