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, `"Area of ΔABC"/"Area of ΔEBD" = ((BC)/(BD))^2`

`\implies "Area of ΔABC"/9 = (10/5)^2`

`\implies` Area of ΔABC = 4 × 9

`\implies` Area of ΔABC = 36 cm²

Also, `"Area of ΔABC"/"Area of ΔEBD" = (AB)^2/(BE)^2`

`\implies 36/9 = (AB^2)/36`

`\implies AB^2 = (36 xx 36)/9`

`\implies` AB² = 144

`\implies` AB = 12 cm