Q.

In the ∆ABC, DE || BC and AD = 3x − 2, AE = 5x − 4, BD = 7x − 5, CE = 5x − 3, then find the value of x

1 1  
2 7/10  
3 both (a) & (b)  
4 none of these  

Ans.

(3)

Given that, AD = 3x − 2, AE = 5x − 4, BD = 7x − 5, CE = 5x − 3

By Basic Proportionality theorem, we have ADBD=AEEC

⇒3x-27x-5=5x-45x-3⇒(3x-2)(5x-3)=(5x-4)(7x-5)

⇒15x2-19x+6=35x2-53x+20

⇒20x2-34x+14=0⇒10x2-17x+7=0

⇒(x-1)(10x-7)=0⇒x=1,x=710