Manhattan Test 2 Quant Question

Can someone help me with this question ?

Hey I think the answer is D, now as both of these triangles share the same hypotenuse…(1)
Let us assume z=3,y=4, then hypotenuse=5
Also
Since its area(∆JKL) > area(∆JLM)
x(KL)>3(4)…(2)
(right angle triangle height * base formula)

So based on these 2 conditions, in ∆JKL, 2 cases can happen
1.my x must be >4 and KL <3, thereby solving the 2 conditions

2.or KL>4 and x< 3
Since both of these cases are possible, where in one case x >y and less than y in other,we don’t have enough information to answer and so we choose D

Correct answer is given as A

JL = \sqrt{y^2 + z^2}. This means that
KL = \sqrt{JL^2 - JK^2} = \sqrt{y^2 + z^2 - x^2}

It’s given that the area of JKL is greater than the area of JLM. This means that

\frac{1}{2}(\sqrt{y^2 + z^2 - x^2} \times x) > \frac{yz}{2}
\sqrt{y^2 + z^2 - x^2} \times x > yz

As z > y,

\sqrt{y^2 + z^2 - x^2} \times x > y^2

(z > y)

\sqrt{z^2 + z^2 - x^2} \times x > y^2

(\sqrt{z^2 + z^2} > \sqrt{z^2 + z^2 - x^2})

\sqrt{y^2 + z^2} \times x > y^2
x > y