In the figure,\[RS\parallel DB\parallel PQ\]. If \[CP=PD=11\text{ }cm\] and \[DR=RA=3\text{ }cm\]. Then the values of X and Y are respectively.

a) 12,10
b) 14,6
c) 10,7
d) 16,8

Hint: We will try to show that \[\Delta ASR\] is similar to \[\Delta ABD\], By using the properties of a similar triangle we will find the relation between X and Y and finally we will select the best answer possible in the given option.

Complete step-by-step answer:
It is given in the question that in figure \[RS\parallel DB\parallel PQ\]. Also, \[CP=PD=11\text{ }cm\] and \[DR=RA=3\text{ }cm\] then we have to find out the value of X and Y resp.

Now in \[\Delta ASR\] and \[\Delta ABD\], we have
$\angle A=\angle A$, Common angle in both triangle
We know that \[\angle ASR\] and \[\angle ABQ\] because both are corresponding angle b/w parallel lines RS and BD respectively, hence,\[\angle ASR\text{ }=\angle ABQ\] (both are corresponding angle)
Therefore, we can say that \[\Delta ASR\text{ }\cong \Delta ABQ\] by AA similarity.
Here AA similarity means Angle-Angle similarity. It means we can say that in \[\Delta ASR\] and \[\Delta ABD\], we have $\dfrac{AR}{AD}=\dfrac{AS}{AB}=\dfrac{RS}{DB}$
We know that \[AR=3cm\] and\[AD=6\text{ }cm\]. So on putting these values, we have, \[\dfrac{RS}{DB}=\dfrac{3}{6}\]→$\dfrac{Y}{X}=\dfrac{1}{2}$
On cross multiplication, we get \[X=2Y\].Now, if we compare this obtained relation \[X=2Y\], we have only one similar option i.e. option \[D\text{ }X=16cm\] and \[Y=8\text{ }cm\] all the other option do not show \[X=2Y\] thus neglected.
Therefore, the correct option is option d).

Note: Students may assume \[\Delta ABD\] and \[\Delta ARS\] are right triangles and they may use Pythagoras theorem to find the value of X and Y respectively. But this is wrong because it is not specified anywhere in the questions and in the figure that they are right angle triangles.