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