# 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 sin^{3} θ = b cos^{3} 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 d^{2} (AC) / dθ^{2} < 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}

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}

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