AP EAMCET · Maths · Application of Derivatives
The maximum area of a right angled triangle with hypotenuse \(h\) is
- A \(h^2 / 2 \sqrt{2}\)
- B \(h^2 / 2\)
- C \(h^2 / \sqrt{2}\)
- D \(h^2 / 4\)
Answer & Solution
Correct Answer
(D) \(h^2 / 4\)
Step-by-step Solution
Detailed explanation
Let a right angled triangle \(\mathrm{OAB}\), with \(\mathrm{h}\) hypoteneous Let \(\angle \mathrm{OAh}=\theta\) now \(\mathrm{OA}=\mathrm{h} \cos \theta\) and \(\mathrm{OB}=\mathrm{h} \sin \theta\)…
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