Question

if AD=4x-3, AE=8x-7,DB=3x-1 and CE=5x-3 ,then find the value of X by useing of Thales theorem

Answer 1

here x=1 or x=-1/2 is your answer

Answer 2

Answer:

$Value \:of\:x = 1$

Step-by-step explanation:

In ∆ABC , DE//BC ,

AD = 4x-3, AE=8x-7,

DB=3x-1and CE=5x-3,

$\frac{AE}{EC}=\frac{AD}{DB}$

/* By Thales Theorem */

$\implies \frac{8x-7}{5x-3}=\frac{4x-3}{3x-1}$

$\implies (8x-7)(3x-1)=(4x-3)(5x-3)$

$\implies 8x(3x-1)-7(3x-1)\\=4x(5x-3)-3(5x-3)$

$\implies 24x^{2}-8x-21x+7\\=20x^{2}-12x-15x+9$

$\implies 24x^{2}-29x+7\\=20x^{2}-27x+9$

$\implies 24x^{2}-29x+7-20x^{2}+27x-9=0$

$\implies 4x^{2}-2x-2=0$

$\implies 2x^{2}-x-1=0$

$\implies 2x^{2}-2x+x-1=$

$\implies 2x(x-1)+1(x-1)=0$

$\implies (x-1)(2x+1)=0$

$\implies x-1=0\:or\:2x+1=0$

$\implies x=1\:or\:x=\frac{-1}{2}$

x should not be negative.

Therefore,

$Value \:of\:x = 1$

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