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प्रश्न
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:
- length of AB
- area of ΔABC
उत्तर
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
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