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JEE Mains · Maths · STD 12 - 6. Application of derivatives
The maximum area of a right angled triangle with hypotenuse \(h\) is
- A \(\frac{{{h^2}}}{{2\sqrt 2 }}\)
- B \(\frac{{{h^2}}}{{2}}\)
- C \(\frac{{{h^2}}}{{\sqrt 2 }}\)
- D \(\frac{{{h^2}}}{{4}}\)
Answer & Solution
Correct Answer
(D) \(\frac{{{h^2}}}{{4}}\)
Step-by-step Solution
Detailed explanation
Let base \(=b\) Altitude (or perpendicur) \( = \sqrt {{h^2} - {b^2}} \) Area, \(\mathrm{A}=\frac{1}{2} \times\) base \(\times\) altitude \(=\frac{1}{2} \times b \times \sqrt{h^{2}-b^{2}}\)…
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