Question

In the given figure, DE || BC such that AD = x cm, DB = (3x + 4) cm, AE = (x + 3) cm and EC = (3x + 19) cm. Find the value of x.

Answer 1

In ΔADE and ΔABC
∠𝐴𝐷𝐸 = ∠𝐴𝐵𝐶 (𝐶𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑎𝑛𝑔𝑙𝑒𝑠 𝑖𝑛 𝐷𝐸 ∥ 𝐵𝐶)
∠𝐴𝐸𝐷 = ∠𝐴𝐶𝐵 (𝐶𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑎𝑛𝑔𝑙𝑒𝑠 𝑖𝑛 𝐷𝐸 ∥ 𝐵𝐶
By AA similarity criterion, ΔADE ~ ΔABC
If two triangles are similar, then the ratio of their corresponding sides are proportional  

 

∴ `(AD)/(AB)=(AE)/(AC)` 

⇒`( AD)/(AD+DB)=(AE)/(AE+EC)` 

⇒` x/(x+3x+4)=(x+3)/(x+3+3x+19)` 

⇒ `x/(4x+4)=(x+3)/(x+3+3x+19)` 

⇒ `x/(2x+2)=(x+3)/(2x+11)` 

⇒ `2x^2+11x=2x^2+2x+6x+6` 

⇒ 3x=6 

⇒ x=2 

Hence, the value of x is 2.