Question

The maximum area of a right-angled triangle with hypotenuse \( h \) is: (a) \( \frac{h^2}{2\sqrt{2}} \)

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

Hint:

For a right-angled triangle with hypotenuse \( h \), the maximum area occurs when \( \theta = 45^\circ \), leading to the formula \( A = \frac{h^2}{4} \).

Answer 1

The Correct Option is D

A right-angled triangle with hypotenuse \( h \) has two legs, which can be expressed using trigonometric functions: \[ \text{Let } \angle OAB = \theta. \] \[ \text{Then, } OA = h \cos\theta, \quad OB = h \sin\theta. \] Step 1: Compute the area The area \( A \) of the triangle is given by: \[ A = \frac{1}{2} \times OA \times OB \] \[ = \frac{1}{2} \times h \cos\theta \times h \sin\theta \] \[ = \frac{1}{2} h^2 \sin\theta \cos\theta. \] Using the trigonometric identity: \[ \sin\theta \cos\theta = \frac{1}{2} \sin 2\theta, \] we get: \[ A = \frac{1}{4} h^2 \sin 2\theta. \]
Step 2: Maximize the area Since \( \sin 2\theta \) has a maximum value of 1 when \( 2\theta = 90^\circ \) or \( \theta = 45^\circ \), the maximum area is: \[ A_{\text{max}} = \frac{1}{4} h^2 \times 1 = \frac{h^2}{4}. \]
Step 3: Verify the correct option The correct answer, based on the calculations, is \( \frac{h^2}{4} \), which matches option (D).