Question

In trianle ABC pq is parallel to bc
and ap=3x-19,pb=x-5,aq=x-3,qc=3cm
find x value
anybody solve it fastly

Answer 1

shh to h na ........??

Answer 2

Answer: The value of x is 8 or 9.

Step-by-step explanation:

Given : In triangle ABC,

PQ ║ BC,

Such that, P∈ AB and Q ∈ AC,

Also, AP = 3x-19, PB = x-5, AQ = x-3, QC = 3 cm

To find : The value of x.

Since, PQ ║ BC

By the alternative interior angle theorem,

$\angle APQ\cong \angle ABC$

$\angle AQP\cong \angle ACB$

Also,

$\angle PAC\cong \angle BAC$     ( Reflexive )

By AAA similarity postulate,

$\triangle APQ\sim \triangle ABC$

Since, the corresponding sides of the similar triangle are in same proportion,

$\implies \frac{AP}{AB}=\frac{AQ}{AC}$

$\implies \frac{AP}{AP+PB}=\frac{AQ}{AQ+QC}$

$\implies \frac{3x-19}{3x-19+x-5}=\frac{x-3}{x-3+3}$

$\implies \frac{3x-19}{2x-24}=\frac{x-3}{x}$

$\implies 3x^2-19x=4x^2-24x-12x+72$

$\implies x^2-17x+72=0$

By solving this,

We get, x = 8 or 9.

Hence, The value of x is 8 or 9.