Let base =b 
Altitude (or perpendicular) =h2−b2 Area, A=21× base × altitude =21×b×h2−b2 ⇒dbd A=21[h2−b2+b⋅2h2−b2−2b] =21[h2−b2h2−2b2] Put dbd A=0,⇒b=2h Maximumarea =21×2h×h2−2h2=4h2
The maximum area of a right angled triangle with hypotenuse h is :
Held on 22 Apr 2013 · Verified 6 Jul 2026.
22h2
2h2
2h2
4h2
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