Question

The maximum area of a right angled triangle with hypotenuse \(h\) is:

• A. \(\dfrac{h^2}{2\sqrt{2}}\)
• B. \(\dfrac{h^2}{2}\)
• C. \(\dfrac{h^2}{\sqrt{2}}\)
• D. \(\dfrac{h^2}{4}\)

Hint:

For a fixed hypotenuse, the area of a right angled triangle is maximum when the two perpendicular sides are equal, meaning the triangle is an isosceles right triangle.

Answer 1

The Correct Option is D

Step 1: Let the perpendicular sides of the right triangle be \(a\) and \(b\).
Since the hypotenuse is \(h\), by Pythagoras theorem, \[ a^2+b^2=h^2 \] The area of the right angled triangle is \[ A=\frac{1}{2}ab \]

Step 2: Use the condition for maximum product.
For fixed value of \[ a^2+b^2, \] the product \(ab\) is maximum when \[ a=b \] So, let \[ a=b \] Then, \[ a^2+a^2=h^2 \] \[ 2a^2=h^2 \] \[ a^2=\frac{h^2}{2} \]

Step 3: Find the maximum area.
The maximum area is \[ A_{\max}=\frac{1}{2}a^2 \] Substituting \[ a^2=\frac{h^2}{2}, \] we get \[ A_{\max}=\frac{1}{2}\cdot \frac{h^2}{2} \] \[ A_{\max}=\frac{h^2}{4} \]

Step 4: Final conclusion.
Hence, the maximum area of the right angled triangle is \[ \boxed{\frac{h^2}{4}} \]