Question

In ∆ABC, D and E are points on the sides AB and AC respectively such that

DE∥BC. If AD = 4x-3, AE = 8x-7, BD = 3x-1 and CE = 5x-3, find the value of x.​

Answer 1

Answer:

ANSWER 》The value of x is 1

Step-by-step explanation:

Given: AD = 4x – 3, BD = 3x – 1, AE = 8x – 7 and EC = 5x – 3  Required to find x. 

By using Thales Theorem, [As DE ∥ BC]

  AD/BD = AE/CE 

So, (4x–3)/ (3x–1) = (8x–7)/ (5x–3) 

(4x – 3)(5x – 3) = (3x – 1)(8x – 7) 

4x(5x – 3) -3(5x – 3) = 3x(8x – 7) -1(8x – 7) 

20x² – 12x – 15x + 9 = 24x² – 29x + 7  20x² - 27x + 9 = 24x² - 29x + 7 

⇒ -4x²+ 2x + 2 = 0 

4x² – 2x – 2 = 0 

4x² – 4x + 2x – 2 = 0 

4x(x – 1) + 2(x – 1) = 0 

(4x + 2)(x – 1) = 0 

⇒ x = 1 or x = -2/4

We know that the side of triangle can never be negative. Therefore, we take the positive value.

therefore, x=1

Answer 2

Step-by-step explanation:

We have, DE || BC

Therefore, by basic proportionality theorem,

We have,

AD/DB = AE/EC

= 4x - 3 / 3x -1 = 8x - 7 / 5x - 3

= (4x - 3) (5x - 3) = (8x - 7) (3x - 1)

= 4x(5x - 3) -3(5x - 3) = 8x(3x - 1) -7(3x - 1)

= 20x^2 - 12x -15x + 9 = 24x^2 - 8x - 21x +7

= 20x^2 - 27x + 9 = 24x^2 - 29x +7

= 24x^2 - 29x +7 - 20x^2 + 27x - 9 = 0

= 4x^2 - 2x - 2 = 0

= 2(2x^2 - x - 1) = 0

= 2x^2 - x - 1 = 0

= 2x^2 - 2x + 1x - 1 = 0

= 2x(x - 1) -1(x - 1) = 0

= (x - 1) (2x - 1) = 0

= 2x - 1 = 0 , x - 1 = 0

= x = -1/2 , x = 1

x = -1/2 is not possible

so value of x = 1