Explanation:

Let the legs of ABC be $x$ and $y$ and the $z$ be the hypotenuse.
In Right triangle, $x^2 + y^2 = z^2$ .... (1)

We are given:
$x + y + z = 17$ ... (2) and

$x^2 + y^2 + z^2 = 98$ ... (3)

Substituting (1) in (3):
$z^2 + z^2 = 98$
$z^2 = 49$
$z = 7$

So, $x + y + 7 = 17$ i.e. $x + y = 10$

On squaring both sides, $x^2 + y^2 + 2xy = 100$ ... (4)

Substituting (4) in (3):

$100 - 2xy + 7^2 = 98$
$51 = 2xy$
$xy = 25.5$

Area of triangle ABC = $\frac{1}{2}xy = 12.75$

Hence, option A