Question

in triangle PQR, XY parallel QR, PX=x-2, XQ=3x, PY=x+2 andYR=9x. find the value of x

Answer 1

Step-by-step explanation:

your answer is X=4 by using basic proportionality theorem

Answer 2

The value of x is  $\frac{2}{13\sqrt{2}}$ .

Step-by-step explanation:

Given,

$PX = x-2 \\XQ = 3x\\PY= x+2 \\YR=9x$

∴ XY parallel QR, we have

$\frac{XQ}{PX} = \frac{PY}{YR}$

$\frac{3x}{x - 2} = \frac{x + 2}{9x}$

$3x \times 9x = (x - 2)(x + 2)$

$27x^{2} = x(x + 2) -2(x +2)$

$27x^{2} = x^{2} + 2x -2x -4$

$27x^{2} - x^{2} = 4$

$26x^{2} = 4$

$x^{2} = \frac{4}{26}$

$x = \sqrt{\frac{4}{26} }$

$x = \frac{2}{13\sqrt{2}}$

So, the value of x is  $\frac{2}{13\sqrt{2}}$ .

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