OA Explanation

The length of AP is maximum when the given triangle in an Isosceles Right angled traingle.
(Refer to the figure below)

Ratio of sides in an Isosceles Right angled traingle is \(x : x : x\sqrt{2}\)
So, \(20 = x\sqrt{2}\)
i.e. \(x = 10\sqrt{2}\)

Now, let us equate the area;
\(\frac{1}{2}(x)(x) = \frac{1}{2}(20)(AP)\)
\((10\sqrt{2})(10\sqrt{2}) = 20(AP)\)
\(AP = 10\)

Col. A: 10
Col. B: 10

Hence, option C