Question

एक∆ABC,में DE समानान्तर है BC के तथा AD=(x+3) DB=(3x-19), AE=x and CE = (3x+4), तो × का मान ज्ञात करो।​

Answer 1

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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