1. Analyze the Triangle and Circle:

- The image shows a triangle ABC inscribed in a circle.
- AC is given as the diameter of the circle, and its length is 10 units.
- There is a right angle symbol at vertex B , indicating that $\(\angle A B C=90^{\circ}\)$. This is consistent with the theorem that an angle inscribed in a semicircle is a right angle.

2. Identify the Base and Height of the Triangle:

- Since $\angle A B C=90^{\circ}$, triangle ABC is a right-angled triangle.
- In a right-angled triangle, the two legs can serve as the base and height for calculating the area. Let BC be the height ( $h$ ) and AB be the base ( $b$ ).

3. Express the Area of the Triangle:

- The area of a triangle is given by the formula: Area $\(=\frac{1}{2} \times$ base $\times\)$ height.
- Area of $\(\triangle A B C=\frac{1}{2} \times A B \times B C\)$.

4. Consider the Maximum Area of a Right-Angled Triangle with a Fixed Hypotenuse:

- The hypotenuse of $\(\triangle A B C\)$ is AC , which is the diameter of the circle, with a length of 10 units.
- For a right-angled triangle with a fixed hypotenuse, its area is maximized when the two legs ( AB and BC ) are equal.

- If $\mathrm{AB}=\mathrm{BC}$, then by the Pythagorean theorem: $\(A B^2+B C^2=A C^2 A B^2+A B^2=10^2\)$ $\(2 \times A B^2=100 A B^2=50 A B=\sqrt{50}=5 \sqrt{2}\)$
- In this maximum area case, $\(A B=B C=5 \sqrt{2}\)$.
- The maximum area would be: Area $\(=\frac{1}{2} \times(5 \sqrt{2}) \times(5 \sqrt{2})=\frac{1}{2} \times(25 \times 2)=\frac{1}{2} \times 50=\)$ 25 square units.

5. Compare Quantity A and Quantity B:
- Quantity A: Area of the triangle in sq. units.
- Quantity B: 25

Since the base and height (legs AB and BC ) can vary as long as point B remains on the semicircle, the area of the triangle will vary. The maximum possible area of the triangle is 25 square units, which occurs when triangle $A B C$ is an isosceles right-angled triangle (i.e., when $B$ is at the midpoint of the arc).

For any other position of B on the semicircle (excluding A or C , which would make it a degenerate triangle with area 0 ), the legs AB and BC would not be equal, and the area would be less than 25 square units.

- For example, if $\(\mathrm{AB}=6\)$, then $\(B C=\sqrt{10^2-6^2}=\sqrt{100-36}=\sqrt{64}=8\)$. Area $\(=\frac{1}{2} \times$ $6 \times 8=24\)$. In this case, Quantity A (24) is less than Quantity B (25).
- If $\(\mathrm{AB}=8\)$, then $\(B C=\sqrt{10^2-8^2}=\sqrt{100-64}=\sqrt{36}=6\)$. Area $\(=\frac{1}{2} \times 8 \times 6=24\)$. Quantity $A$ (24) is less than Quantity B (25).

Since the area of the triangle can be equal to or less than 25 , we cannot definitively say whether Quantity A is always greater, always less, or always equal to Quantity B. It can be equal to 25 (at its maximum) or less than 25.

Therefore, the relationship cannot be determined from the information given.
The final answer is The relationship cannot be determined from the information given.