Maximum Area of a Right-Angled Triangle

When a right-angled triangle is inscribed in a circle, its hypotenuse must be a diameter of the circle. Let the circle have a radius $r$. Then the diameter, which is the hypotenuse of the right-angled triangle, has a length of $2r$.

Let the two legs of the right-angled triangle be $a$ and $b$. The hypotenuse is $c$. By the Pythagorean theorem, we have: \[ a^2 + b^2 = c^2 \] Since the hypotenuse is the diameter, $c = 2r$. So, \[ a^2 + b^2 = (2r)^2 = 4r^2 \]

Triangle Area Calculation

The area of a right-angled triangle is given by half the product of its legs: \[ \text{Area } A = \frac{1}{2}ab \] We want to maximize this area $A$, subject to the constraint $a^2 + b^2 = 4r^2$.

We can use the AM-GM (Arithmetic Mean - Geometric Mean) inequality. For non-negative numbers $a^2$ and $b^2$, the AM-GM inequality states: \[ \frac{a^2 + b^2}{2} \ge \sqrt{a^2 b^2} \] Substituting $a^2 + b^2 = 4r^2$: \[ \frac{4r^2}{2} \ge \sqrt{(ab)^2} \] \[ 2r^2 \ge |ab| \] Since $a$ and $b$ are lengths, they are positive, so $ab$ is positive: \[ 2r^2 \ge ab \]

Now substitute this inequality back into the area formula: \[ A = \frac{1}{2}ab \] \[ A \le \frac{1}{2}(2r^2) \] \[ A \le r^2 \]

The maximum value of $ab$ is $2r^2$, which occurs when $a^2 = b^2$. Since $a$ and $b$ are positive, this means $a = b$.

Condition for Maximum Area

The maximum area occurs when the triangle is an isosceles right-angled triangle. In this case, $a = b$, and from $a^2 + b^2 = 4r^2$, we get $a^2 + a^2 = 4r^2$, so $2a^2 = 4r^2$, which means $a^2 = 2r^2$. Then $a = \sqrt{2r^2} = r\sqrt{2}$. Since $a=b$, $b = r\sqrt{2}$.

The legs are $r\sqrt{2}$ and $r\sqrt{2}$. The area is $\frac{1}{2} \times r\sqrt{2} \times r\sqrt{2} = \frac{1}{2} \times r^2 \times 2 = r^2$.

Alternatively, using calculus, we could express $b = \sqrt{4r^2 - a^2}$ and substitute into the area formula $A(a) = \frac{1}{2}a\sqrt{4r^2 - a^2}$. Then find the maximum by taking the derivative with respect to $a$ and setting it to zero. This method also confirms the maximum area is $r^2$.

Therefore, the maximum area of a right-angled triangle inscribed in a circle of radius $r$ is $r^2$.