In the adjoining figure ${\text{AB}}\parallel {\text{CD}}\parallel {\text{EF}}$. The measurements are indicated in centimeters. Find the value of $x$ and $y$ in centimeters.

A) $x = \dfrac{1}{4}$, $y = \dfrac{{20}}{3}$
B) $x = \dfrac{{15}}{4}$, $y = \dfrac{1}{4}$
C) $x = \dfrac{5}{8}$, $y = \dfrac{{20}}{3}$
D) $x = \dfrac{{15}}{4}$, $y = \dfrac{{20}}{3}$

Hint: First find the triangles that are similar in the given figure and follow the congruence rule and then use the property of similar triangles to get the relation between the sides of the triangle. Solve the obtained equation to get the required result.

Complete step-by-step answer:
We have given that ${\text{AB = 6cm, DC = x, DE = y, & EF = 10cm}}$ and AB, CD, and EF are parallel to each other and we have to find the value of $x$ and $y$.
Now, from the figure, we take two triangle ACD and triangle AFE.
These triangles are similar to each other because all the angles are equal as ${\text{CD}}\parallel {\text{EF}}$
So, in triangle ACD and triangle AFE, we can use the property of similarity that the corresponding sides are in the same proportion. That is,
$\dfrac{{{\text{CD}}}}{{{\text{EF}}}} = \dfrac{{{\text{AC}}}}{{{\text{AE}}}}$ ,
Now, put the values, $CD = x$ and $EF = 10$ into the above equation:
$\dfrac{x}{{10}} = \dfrac{{{\text{AC}}}}{{{\text{AE}}}}{\text{ }} \to \left( 1 \right)$
Now, in triangle ABE and in triangle CDE.
These triangles are also similar to each other because all the angles are equal to the corresponding angles of other triangles because ${\text{AB}}\parallel {\text{CD}}$, then using the property of similarity we have,
$\dfrac{{{\text{CD}}}}{{{\text{AB}}}}{\text{ = }}\dfrac{{{\text{DE}}}}{{{\text{BE}}}}{\text{ = }}\dfrac{{{\text{CE}}}}{{{\text{AE}}}}$
Now, we substitute all the given values into the above equation, then we have
$\dfrac{x}{6} = \dfrac{y}{{y + 4}} = \dfrac{{{\text{CE}}}}{{{\text{AE}}}}{\text{ }} \to \left( 2 \right)$
$\dfrac{{{\text{CE}}}}{{{\text{AE}}}} = \dfrac{x}{6}{\text{ }} \to \left( 3 \right)$
We know that:
$CE = AE - AC$
Using the above value, we have
$\dfrac{{{\text{CE}}}}{{{\text{AE}}}}{\text{ = }}\dfrac{{{\text{AE}} - {\text{AC}}}}{{{\text{AE}}}}$
$\dfrac{{{\text{CE}}}}{{{\text{AE}}}}{\text{ = 1}} - \dfrac{{{\text{AC}}}}{{{\text{AE}}}}$
Now, we substitute the values from equation (1) and equation (3). Into the above equation:
$\dfrac{x}{6} = 1 - \dfrac{x}{{10}}$
Now, we solve the equation for the value of $x$.
$\dfrac{x}{6} + \dfrac{x}{{10}} = 1$
$\dfrac{{5x + 3x}}{{30}} = 1$
$\dfrac{{8x}}{{30}} = 1$
$x = \dfrac{{30}}{8} = \dfrac{{15}}{4}$
Now, from equation (2), we have
$\dfrac{x}{6} = \dfrac{y}{{y + 4}}$
Substitute the value of $x$ in the above equation,
$\dfrac{{15}}{{4 \times 6}} = \dfrac{y}{{y + 4}}$
Solve the equation for $y$:
$\dfrac{y}{{y + 4}} = \dfrac{5}{8}$
$8y = 5y + 20$
$3y = 20$
$y = \dfrac{{20}}{3}$
So, the value of $y$ is $\dfrac{{20}}{3}$
Hence, we found out that $x = \dfrac{{15}}{4}$ and $y = \dfrac{{20}}{3}$.

Therefore, the option D is correct.

Note: In this type of question, we just have to remember that if a line cuts two parallel lines it makes the same angles on both sides. If two triangles are similar then their corresponding sides are in the same ratio.