Question

In a triangle ABC, DE || BC and AD = (x+3), DB=(3x+19), AE =x and CE = (3x+4), then find x​

Answer 1

In a triangle ABC, DE || BC and AD = (x+3), DB=(3x+19), AE =x and CE = (3x+4), then find x​

Step-by-step explanation:

• In Δ ADE and Δ ABC

∠ ADE = ∠ ABC(corresponding angles in DE || BC)

∠ AED = ∠ ACB(corresponding angles in DE || BC

• By AA similarity criterion, Δ ADE  ≅ Δ ABC

• If two triangles are similar, then the ratio of their corresponding sides are proportional.

• Therefore,

⇒ AD/AB = AE/AC

   AD/(AD+DB) = AE/(AE+EC)

   (x+3) / (x + 3 + 3x + 19) = x / (x + 3x + 4)

   (x + 3) /  4x + 22 = x/ 4x + 4

   (4x + 4)(x + 3) = (x)(4x + 22)

   4x² + 12x + 4x + 12 = 4x² + 22x

   16x +12 = 22x

   22x - 16x =12

   6x = 12

   x = 12/6

   x = 2

Therefore, the value of x is 2.

Answer 2

Given : triangle ABC, DE || BC and AD = (x+3), DB=(3x+19), AE =x and CE = (3x+4

To Find : x

Solution:

triangle ABC, DE || BC

Using Thales theorem ( BPT - Basic proportionality theorem)

AD/DB  = AE/EC

AD = x + 3

DB = 3x + 19

AE = x

CE = 3x + 4

=> (x + 3)/(3x + 19)  = x/(3x + 4)

=> (x + 3)(3x + 4) = x(3x + 19)

=> 3x² + 13x + 12 = 3x² + 19x

=> 6x = 12

=> x = 2

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