Question

To find the value of x from the below figure such that \[DE\parallel AB\]

Answer 1

Hint: From the given figure, a line is drawn parallel to one side of the triangle. Hence, this can be solved either by applying Thales theorem or by the property of similar triangles. Line drawn parallel to one side of the triangle, intersects the other two sides in distinct points, then it divides the other two sides in the same ratio. In a similar triangle, corresponding sides are of the same ration. Hence find the corresponding sides of the same ration from the given figure. And then substituting the values on the ratio. Now cross multiplying the ratio and solving further we get the value of x.

Complete step-by-step solution:
If \[DE\parallel AB\], then \[\vartriangle CDE \sim \vartriangle ABC\], that is \[\vartriangle CDE\]is similar to \[\vartriangle ABC\]
Therefore, their corresponding sides are of same ratio
By the property of similar triangles:
Since their corresponding sides are of same ratio
\[\dfrac{{CD}}{{AC}} = \dfrac{{CE}}{{BC}}\]
Substituting the values,
\[ \Rightarrow \dfrac{{x + 3}}{{4x + 22}} = \dfrac{x}{{4x + 4}}\]
Now cross multiply,
\[ \Rightarrow \left( {x + 3} \right)\left( {4x + 4} \right) = x\left( {4x + 22} \right)\]
Further multiplication,
\[ \Rightarrow 4{x^2} + 4x + 12x + 12 = 4{x^2} + 22x\]
Hence,
\[ \Rightarrow 4{x^2} + 16x + 12 = 4{x^2} + 22x\]
Now keeping the higher order on one side and the remaining on the other side,
\[ \Rightarrow 4{x^2} + 16x + 12 - 4{x^2} - 22x = 0\]
Simplifying we get,
\[ \Rightarrow 12 - 6x = 0\]
Solve it for x, rearranging the terms
\[ \Rightarrow 12 = 6x\]
\[ \Rightarrow x = \dfrac{{12}}{6}\]
Hence,
\[ \Rightarrow x = 2\]

The value of x is equal to 2.

Note: Alternative method: Using basic property theorem of Thales theorem
Basic property theorem: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points then the other two sides are divided in the same ratio. That is also known as Thales theorem.
Since \[DE\parallel AB\],
\[\dfrac{{AD}}{{DC}} = \dfrac{{BE}}{{EC}}\]
Applying the values,
\[\dfrac{{3x + 19}}{{x + 3}} = \dfrac{{3x + 4}}{x}\]
Cross multiplying,
\[\left( {3x + 19} \right) \times x = \left( {3x + 4} \right) \times \left( {x + 3} \right)\]
Multiply the terms we get,
\[3{x^2} + 19x = 3{x^2} + 9x + 4x + 12\]
Simplifying we get,
\[3{x^2} + 19x = 3{x^2} + 13x + 12\]
Cancelling the similar terms,
\[19x = 13x + 12\]
\[ \Rightarrow 6x = 12\]
Hence,
\[ \Rightarrow x = 2\]
Hence, the value is \[x = 2\]