Question

Which value of x would make NO II KJ? 1 6 8 10

Answer 1

We sketch the two similar shapes and they are as shown in the attachment. We want to write corresponding side's length in ratio

$LN : LK = x-3 : (x-3+x+2) = (x-3) : (2x-1)$
$LO : LJ = (x-4) : (x-4+x) = (x-4) : (2x-4)$

Writing the two ratio as fraction gives
$\frac{x-3}{2x-1} = \frac{x-4}{2x-4}$

Cross multipliying gives $(x-3)(2x-4)=(x-4)(2x-1)$

Answer 2

we know that

If NO II KJ

then

triangles LKJ and LNO are similar and the ratio of the corresponding sides are equals

so

$\frac{LO}{LN}= \frac{LJ}{LK}$

we have

$LO=(x-4)\ in$

$LN=(x-3)\ in$

$LJ=LO+OJ=(x-4)+x=2x-4\ in$

$LK=LN+NK=(x-3)+(x-2)=2x-1\ in$

substitute

$\frac{x-4}{x-3}=\frac{2x-4}{2x-1} \\ \\(x-4)*(2x-1)=(2x-4)*(x-3)\\ \\2x^{2} -x-8x+4=2x^{2} -6x-4x+12\\ \\-9x+10x=12-4\\ \\ x=8\ units$

therefore

the answer is the option

$x=8\ units$