Question

A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle. Show that the minimum length of the hypotenuse is[a^2/3+b^2/3]^3/2

Answer 1

Let ∆ABC is a right - angled triangle at B where AB = x and BC = y.
from Pythagoras theorem,
AC = $\bf{\sqrt{x^2+y^2}}$
Let P is the point on the hypotenuse of triangle ABC in such a way that P is at a distance of a and b from the sides AB and BC respectively.
Let angle BCA = $\theta$

see figure, PC = bcosecθ
AP = asecθ
also, AC = AP + PC
AC = asecθ + bcosecθ
differentiate AC with respect to θ,
d(AC)/dθ = asecθ.tanθ + bcosecθ(-cotθ)
= asecθ.tanθ - bcosecθ.cotθ
now, if d(AC)/dθ = 0
then, asecθ.tanθ - bcosecθ.cotθ = 0
a(sinθ/cos²θ) - b(cosθ/sin²θ) = 0
asin³θ - bcos³θ = 0
asin³θ = bcos³θ
sin³θ/cos³θ = b/a
tan³θ = b/a
tanθ = $\bf{[\frac{b}{a}]^{1/3}}$
so, sinθ = $\bf{\frac{a^{1/3}}{\sqrt{a^{2/3}+b^{2/3}}}}$
and cosθ = $\bf{\frac{b^{1/3}}{\sqrt{a^{2/3}+b^{2/3}}}}$
it is also clear that $\frac{d^2AC}{d\theta^2}$ < 0 when tanθ = (b/a)^{1/3}
Therefore, by second derivative test, the length of the hypotenuse is the maximum when tanθ = (b/a)^{1/3}

now, when tanθ = (b/a)^{1/3} we get ,
AC = bsecθ + acosecθ
= b/cosθ + a/sinθ
= b/$\bf{\frac{b^{1/3}}{\sqrt{a^{2/3}+b^{2/3}}}}$ + a/ $\bf{\frac{a^{1/3}}{\sqrt{a^{2/3}+b^{2/3}}}}$
solve it , we get,
AC = $\bf{[a^{2/3}+b^{2/3}]^{3/2}}$
hence, proved

Answer 2

Answer:

Step-by-step explanation:

Hope you understood.