Among all right-angled triangles with hypotenuse 10 cm, the greatest…

The correct answer is Option (2) → 25 cm²

Let one of the two shorter sides be x cm. By Pythagoras’ theorem the other is \(\sqrt{100 - x^{2}}\) cm, and the area is

    \( \displaystyle A = \frac{1}{2}x\sqrt{100 - x^{2}}, \quad 0 \lt x \lt 10 \)

A is largest when \(A^{2}\) is largest, so let

    \( \displaystyle Z = A^{2} = \frac{1}{4}\left(100x^{2} - x^{4}\right) \)

Setting the derivative equal to zero:

    \( \displaystyle \frac{dZ}{dx} = \frac{1}{4}\left(200x - 4x^{3}\right) = x\left(50 - x^{2}\right) = 0 \)

    \( x^{2} = 50 \), so \(x = 5\sqrt{2}\)

Z increases before this point and decreases after it, so it is a maximum. The other side is also \(\sqrt{50} = 5\sqrt{2}\), and

    \( \displaystyle A = \frac{1}{2} \times 5\sqrt{2} \times 5\sqrt{2} = 25 \) cm²

The familiar 6-8-10 triangle has area 24 cm², which is less than 25.