As the measure of different angles formed by the ladder with the ground and the wall is not known, a unique comparison cannot be formed between x and y

Note: - As shown in the figure on the right in a triangle having angles 30 ,
$\(60 \& 90\)$, the corresponding opposite sides are $\(\mathrm{a}, \mathrm{a} \sqrt{3} \& 2 \mathrm{a}\)$ respectively.

Similarly the right triangle having angles $\(90,45 \& 45\)$, the corresponding opposite sides are $\(\mathrm{a} \sqrt{ } 2\)$, a \& a respectively.

We can use the above two right triangle rules to find the sides in the given figure.
Let us consider different possibilities

As shown in the figure above, let $A C$ be the original position of the ladder and triangle $A B C$ be $\(30^{\circ}-60^{\circ}-90^{\circ}\)$ triangle.

After the ladder slides downwards, let DE be the new position of the ladder such that triangle DBE is a $\(45^{\circ}-90^{\circ}-45^{\circ}\)$ triangle.

Now let $\(\mathrm{BC}=1\)$ then the length of $\(\mathrm{AB} \& \mathrm{AC}\)$ would be $\(\sqrt{3} \& 2\)$ respectively. Next as the length of the ladder remains the same we have $\(\mathrm{AC}=\mathrm{DE}=2\)$, so we get $\(\mathrm{DB}=\mathrm{BE}=\sqrt{2}\)$.


Thus, we get $\(x=A B-D B=\sqrt{3}-\sqrt{2} \& y=B E-B C=\sqrt{2}-1\)$. As $\(\sqrt{3}-\sqrt{2}<\sqrt{2}-1 \Rightarrow x<y\)$

Let $\(\mathrm{DB}=1\)$, then we get $\(\mathrm{BE}=\sqrt{3} \& \mathrm{DE}=2\)$

Also as $\(\mathrm{AC}=2\)$, we get $\(\mathrm{AB}=\mathrm{BC}=\sqrt{2}\)$

So, we have $\(\mathrm{x}=\mathrm{x}=\mathrm{AB}-\mathrm{DB}=\sqrt{2}-1 \& \mathrm{y}=\mathrm{BE}-\mathrm{BC}=\sqrt{3}-\sqrt{2}\)$


As $\(\sqrt{2}-1>\sqrt{3}-\sqrt{2} \Rightarrow x>y\)$

Hence a unique comparison cannot be formed between column A \& column B quantities, so the answer is (D).