Math, asked by aparna6866, 11 months ago

In triangle abc, DE is parallel to BC such that AD = x+3, DB =2x-3, AE = x-2 and AC = 3x+3 ; find the value of x.​

Answers

Answered by nagvanshi20srishti
2

Answer:

The value of x is 2.

Step-by-step explanation:

By Basic Proportionality Theorem or Thales Theorem :

If a line is parallel to a side of a triangle which intersects the other side into two distinct points, then the line divides those sides in proportion.

Therefore, AD/DB =AE/EC

x+2 / 3x+16 = x / 3x+5

x(3x+16) = (x+2) / (3x+5)

3x^2 + 16 = 3x^2 + 5x +6x + 10

16x = 11x + 10

16x - 11x = 10

5x = 10

x = 10/5

x = 2

, 2.

Similar questions