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

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For a right-angled triangle with hypotenuse \( h \), the maximum area occurs when \( \theta = 45^\circ \), leading to the formula \( A = \frac{h^2}{4} \).
Updated On: Nov 26, 2025
  • \(\frac{h^2}{2{\sqrt{2}}}\)

  • \( \frac{h^2}{2} \)
  • \( \frac{h^2}{\sqrt{2}} \)
  • \( \frac{h^2}{4} \)
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The Correct Option is D

Solution and Explanation

A right-angled triangle with hypotenuse \( h \) has two legs that can be defined using trigonometric functions. Let the angle be \( \theta \) at vertex \( A \) such that \( \angle OAB = \theta \). Then, the lengths of the legs \( OA \) and \( OB \) are given by \( OA = h \cos\theta \) and \( OB = h \sin\theta \). Step 1: Compute the area The area \( A \) of the triangle is calculated as follows: \[ A = \frac{1}{2} \times OA \times OB \] \[ = \frac{1}{2} \times h \cos\theta \times h \sin\theta \] \[ = \frac{1}{2} h^2 \sin\theta \cos\theta \] Utilizing the trigonometric identity \( \sin\theta \cos\theta = \frac{1}{2} \sin 2\theta \), the area becomes: \[ A = \frac{1}{4} h^2 \sin 2\theta \]
Step 2: Maximize the area
The maximum value of \( \sin 2\theta \) is 1, which occurs when \( 2\theta = 90^\circ \) (i.e., \( \theta = 45^\circ \)). Consequently, the maximum area \( A_{\text{max}} \) is: \[ A_{\text{max}} = \frac{1}{4} h^2 \times 1 = \frac{h^2}{4} \]
Step 3: Verify the correct option
The calculated maximum area is \( \frac{h^2}{4} \), which corresponds to option (D).
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