Question

What values of x will make $ DE\parallel AB $ in the given figure?

Answer 1

Hint: : To solve this question, i.e., to find the value of x. We will use the converse of basic proportionality theorem, i.e., if a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side., here the line DE is dividing two sides of the triangle. So, the ratio we get is \[\dfrac{{CD}}{{DA}} = \dfrac{{CE}}{{EB}},\] we will put the value of sides in this ration, and then on making a quadratic equation, we will get our required answer, and with that we will also get $ DE\parallel AB $ .

Complete step-by-step answer:
We have been given a figure where we need to find the values of x that will make $ DE\parallel AB $ .
We know that, by the converse of basic proportionality theorem, that if \[\dfrac{{CD}}{{DA}} = \dfrac{{CE}}{{EB}},\] then, $ DE\parallel AB $ .
So, now let us take the condition, \[\dfrac{{CD}}{{DA}} = \dfrac{{CE}}{{EB}},\]
\[ \Rightarrow \dfrac{{x + 3}}{{8x + 9}} = \dfrac{x}{{3x + 4}}\]
\[
\;\left( {3x + 4} \right)\left( {x + 3} \right) = x\left( {8x + 9} \right) \\
\;3{x^2} + 13x + 12 = 8{x^2} + 9x \\
\;5{x^2} - 4x - 12 = 0 \\
\;5{x^2} - 10x + 6x - 12 = 0 \\
\;5x\left( {x - 2} \right) + 6\left( {x - 2} \right) = 0 \\
\;\left( {5x + 6} \right)\left( {x - 2} \right) = 0 \\
\]
\[ \Rightarrow \;x = \dfrac{{ - 6}}{5},\;2\]
So, neglecting the negative value, Therefore, \[x{\text{ }} = {\text{ }}2.\]

If a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio, this is called the basic proportionality theorem.
And If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side, this is called the converse of basic proportionality theorem.

Note: We can check if the value of x is correct or not.
Taking L.H.S. of \[\dfrac{{x + 3}}{{8x + 9}} = \dfrac{x}{{3x + 4}}.\]
\[\dfrac{{x + 3}}{{8x + 9}}\]\[ = \] $ \dfrac{{2 + 3}}{{8(2) + 9}} = \dfrac{5}{{16 + 9}} = \dfrac{5}{{25}} = \dfrac{1}{5} $
Now, taking R.H.S. of \[\dfrac{{x + 3}}{{8x + 9}} = \dfrac{x}{{3x + 4}}.\]
\[\dfrac{x}{{3x + 4}} = \dfrac{2}{{3(2) + 4}} = \dfrac{2}{{6 + 4}} = \dfrac{2}{{10}} = \dfrac{1}{5}\]
Thus, LHS \[ = \] RHS
Hence, the value of x is correct.