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 ^{\frac{2}{3}}+ b ^\frac{2}{3}) ^ \frac{3}{2}

 

Answers (1)

It is given that
A point on the hypotenuse of a triangle is at a distance a and b from the sides of the triangle


Let the angle between  AC and BC is \theta
So, the angle between AD and ED is also \theta
Now,
   CD = b \ cosec\theta
And 
   AD = a \sec\theta
AC = H = AD + CD
      =  a \sec\theta + b \ cosec\theta
\frac{dH}{d\theta} = a \sec\theta\tan\theta - b\cot\theta cosec \theta\\ \frac{dH}{d\theta} = 0\\ a \sec\theta\tan\theta - b\cot\theta cosec \theta =0\\ a \sec\theta\tan\theta = b\cot\theta cosec \theta\\ a\sin^3\theta = b\cos^3\theta\\ \tan^3\theta = \frac{b}{a} \\ \tan\theta = \left ( \frac{b}{a} \right )^\frac{1}{3}
Now,
\frac{d^2H}{d\theta^2} > 0
When  \tan\theta = \left ( \frac{b}{a} \right )^\frac{1}{3}
Hence, \tan\theta = \left ( \frac{b}{a} \right )^\frac{1}{3} is the point of minima
\sec \theta = \frac{a\sqrt{a^\frac{2}{3}+b^\frac{2}{3}}}{a^\frac{1}{3}} and  cosec \theta = \frac{b\sqrt{a^\frac{2}{3}+b^\frac{2}{3}}}{b^\frac{1}{3}}
 

AC = \frac{a\sqrt{a^\frac{2}{3}+b^\frac{2}{3}}}{a^\frac{1}{3}} + \frac{b\sqrt{a^\frac{2}{3}+b^\frac{2}{3}}}{b^\frac{1}{3}} = (a^\frac{2}{3}+b^\frac{2}{3})^\frac{3}{2}
Hence proved

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