Question

A point on the hypotenuse of a right angled
triangle is at distance a and b from the sides making right
angle, (a,b constant). then hypotenuse has
minimum length ${\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)}^{\frac{2}{3}}.$

• A. True
• B. False

Answer 1

The correct option is A True

Let $\Delta ABC$ be right angled at & let $AB=x$ and $BC=y.$ Let P be a point in the hypotenuse of the triangle such that P is at a distance of a and b from the sides AB and BC respectively.

Let $\angle C=\theta$

We have $AC=\sqrt{x^2+y^2}$ Ref. image

Now

$PC=bcosec\theta$

And, $AP=a\sec \theta$

$AC=AP+PC$

$AC=bcosec\theta +a\sec \theta \to$ (1)

$\therefore \frac{d\left(AC\right)}{d\theta }=-bcosec\theta .\cot \theta +a\sec \theta .cotan\theta$

$\therefore \frac{d\left(AC\right)}{d\theta }=0$

$a\sec \theta \tan \theta =bcosec\theta .\cot \theta$

$\frac{a}{\cos \theta }.\frac{\sin \theta }{\cos \theta }=\frac{b}{\sin \theta }.\frac{\cos \theta }{\sin \theta }$

$a{\sin }^2\theta =b{\cos }^2\theta$

$(a{)}^{1/3}\sin \theta =(b{)}^{1/3}\cos \theta$

$\tan \theta ={\left(\frac{b}{a}\right)}^{\frac{1}{3}}$

$\therefore \sin \theta =\frac{(b{)}^{1/3}}{\sqrt{a^{2/3}+b^{2/3}}}$ and $\cos \theta =\frac{{\cos }^{1/3}}{\sqrt{a^{2/3}+b^{2/3}}}$ ___(2)

It can be clearly shown that $\frac{d^2\left(AC\right)}{d{\theta }^2}<0$ when

$\tan \theta ={\left(\frac{b}{a}\right)}^{\frac{1}{3}}$

Therefore by second derirative test, the length of the hypotenuse is the maximum when

$\tan \theta ={\left(\frac{b}{a}\right)}^{1/3}$

Now when $\tan \theta ={\left(\frac{b}{a}\right)}^{1/3}$ we have :

$AC=\frac{b\sqrt{a^{2/3}+b^{2/3}}}{b^{1/3}}+\frac{a\sqrt{a^{2/3}+b^{2/3}}}{a^{1/3}}$

$=\sqrt{a^{2/3}+b^{2/3}}(b^{2/3}+a^{2/3})$

$=(a^{2/3}+b^{2/3}{)}^{3/2}$

Hence the maximum length of the hypotenuses is $(a^{2/3}+b^{2/3}{)}^{3/2}$.