Correct Option: (A) Let the right-angled triangle be $\triangle ABC$, with the right angle at $B$. Let $a$, $b$, and $c$ be the lengths of the sides opposite to vertices $A$, $B$, and $C$ respectively. So, $AB=c$, $BC=a$, and $AC=b$. Let the incircle touch the sides $AB$, $BC$, and $AC$ at points $D$, $E$, and $F$ respectively. Let $r$ be the radius of the incircle. We know that the lengths of tangents from a vertex to the incircle are equal. Thus, $AD = AF$, $BD = BE$, and $CE = CF$. Since $B$ is the right angle, the quadrilateral formed by the incenter $O$, the points of tangency $D$ and $E$ on $AB$ and $BC$, and vertex $B$ ($BDEO$) is a square. This implies that $BD = BE = r$. Given that the point of tangency $F$ on the hypotenuse $AC$ divides it into two segments of lengths $3 \text{ cm}$ and $10 \text{ cm}$. Let $AF = 3 \text{ cm}$ and $CF = 10 \text{ cm}$. Using the tangent property: $AD = AF = 3 \text{ cm}$. $CE = CF = 10 \text{ cm}$. Therefore, the lengths of the sides of the triangle are: $AB = AD + DB = 3 + r$ $BC = BE + EC = r + 10$ $AC = AF + FC = 3 + 10 = 13 \text{ cm}$ Since $\triangle ABC$ is a right-angled triangle at $B$, we can apply the Pythagorean theorem: $AB^2 + BC^2 = AC^2$ $(3+r)^2 + (10+r)^2 = 13^2$ Expand the terms: $(3^2 + 2 \cdot 3 \cdot r + r^2) + (10^2 + 2 \cdot 10 \cdot r + r^2) = 169$ $(9 + 6r + r^2) + (100 + 20r + r^2) = 169$ Combine like terms: $2r^2 + 26r + 109 = 169$ Subtract $169$ from both sides: $2r^2 + 26r + 109 - 169 = 0$ $2r^2 + 26r - 60 = 0$ Divide the entire equation by $2$: $r^2 + 13r - 30 = 0$ Solve the quadratic equation for $r$. We can factorize it: $r^2 + 15r - 2r - 30 = 0$ $r(r+15) - 2(r+15) = 0$ $(r-2)(r+15) = 0$ This gives two possible values for $r$: $r=2$ or $r=-15$. Since the radius $r$ must be a positive length, we take $r=2 \text{ cm}$. Now we can find the lengths of the legs of the right-angled triangle: $AB = 3 + r = 3 + 2 = 5 \text{ cm}$ $BC = 10 + r = 10 + 2 = 12 \text{ cm}$ (As a check, the hypotenuse $AC = 13 \text{ cm}$. We verify $5^2 + 12^2 = 25 + 144 = 169 = 13^2$. The side lengths are correct.) The area of a right-angled triangle is given by the formula $\frac{1}{2} \times \text{base} \times \text{height}$. Area $= \frac{1}{2} \times AB \times BC$ Area $= \frac{1}{2} \times 5 \text{ cm} \times 12 \text{ cm}$ Area $= \frac{1}{2} \times 60 \text{ cm}^2$ Area $= 30 \text{ cm}^2$ The final answer is $\boxed{30 \text{ cm}^2}$.