Question

If the sum of the lengths of the hypothesis and another side of a right angle is given, show that the area of the triangle is maximum when the angle between these sides is $\frac{\pi }{3}$

Answer 1

Let the hypothesis of the right triangle be $x$ and the height be $y$

Hence its base is $\sqrt{x^2-y^2}$ by hypothesis theorem

$Area=\frac{1}{2}\times \sqrt{x^2-y^2}\times y$

$x+y=p\left(say\right)$

Substituting this in the area we get

$Area=\frac{1}{2}\times \sqrt{(p-y{)}^2-y^2\times y}$

$=\frac{1}{2}y\sqrt{p^2+y^2-2py-y^2}$

$=\frac{1}{2}y\sqrt{p^2-2py}$

Squaring on both sides we get

$(Area{)}^2=\frac{1}{4}{y}^2(p^2-2py)$

$=\frac{1}{4}{p}^2{y}^2-\frac{1}{2}p{y}^3$

for maximum of minimum area

$\frac{dy}{dx}=0$

Hence area of is maximum when

$x=\frac{2P}{3}$ and $y=\frac{P}{3}$

$\cos \theta =\frac{y}{x}=\frac{\frac{p}{3}}{2\frac{p}{3}}$

$\cos \theta =\frac{1}{2}$

$\theta =\frac{\pi }{3}$ or ${60}^0$