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

Solution:

Let ΔABC be right-angled at B, AB = x , BC = y and ∠C = θ.

Also, let P be a point on the hypotenuse of the triangle such that P is at a distance of a and b from the sides AB and BC respectively.

We have, AC = √x² + y²

Now,

PC = b cosec θ

AP = a sec θ

Hence,

AC = AP + PC

AC = b cosec θ + a sec θ ...(1)

Therefore,

d (AC) / dθ = - b cosec θ cot θ + a sec θ tan θ

Now,

d (AC)/dθ = 0

⇒ - b cosec θ cot θ + a sec θ tan θ = 0

⇒ a sec θ tan θ = b cosec θ cot θ

⇒ a / cos θ. sin θ / cos θ = b / sin θ. cos θ / sin θ

⇒ a sin³ θ = b cos³ θ

⇒ (a)^1/3 sin θ = (b)^1/3 cos θ

tan θ = (b/a)^1/3

Thus,

sin θ = (b^1/3)/ √(a^2/3 + b^2/3) and cos θ = (a^1/3) / √(a^2/3 + b^2/3).... (2)

It can be clearly shown that d² (AC) / dθ² < 0 when tan θ = (b / a)^1/3

By second derivative test the length of the hypotenuse is the maximum when tan θ = (b / a)^1/3

when tanθ = (b/a)^1/3, we have;

AC = (b / √(a^2/3 + b^2/3)) / (b)^1/3 + (a / √(a^2/3 + b^2/3)) / (a)^1/3 [Using (1) and (2)]

= √(a^2/3 + b^2/3) (b^1/3 + a^1/3)

= (a^2/3 + b^2/3)^3/2

Thus, the maximum length of the hypotenuse is (a^2/3 + b^2/3)^3/2

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NCERT Solutions Class 12 Maths - Chapter 6 Exercise ME Question 12

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

Summary:

Given that a point on the hypotenuse of a triangle is at a distance a and b from the sides of the triangle. Thus, the maximum length of the hypotenuse is (a^2/3 + b^2/3)^3/2