Question

In the fig. LM is parallel to AB. If AB =x-3, AC =2x, BM = x-2 and BC =2x+3 ,then find the value of x. Plz answer this question​

Answer 1

Answer:

In △ABC, we have

 LM∣∣AB

∴  LCAL=MCBM              [By Thaley's Theorem] 

         

⇒    AC−ALAL=BC−BMBM

⇒   2x−(x−3)x−3=(2x+3)−(x−2)x−2

⇒     x+3x−3=x+5x−2

⇒ (x−3)(x+5)=(x−2)(x+3)

⇒   x2+2x−15=x2+x−6

⇒ x=9

Dear I think this helps you