Question

Prove that the area of right-angled triangle of given hypotenuse is maximum when the triangle is isosceles.

Answer 1

Let h be the hypotenuse of the right-angled triangle and x be its altitude.

So, base = `sqrt(h^2 - x^2)`   ...[Using pythagoras theroem]

Now, area of triangle (A) = `1/2 xx sqrt(h^2 - x^2) xx x`

∴ `(dA)/dx = 1/2 [x{1/2 (h^2 - x^2)^(-1//2) xx (-2x)} + sqrt(h^2 - x^2)]`

= `1/2 [(-x^2)/sqrt(h^2 - x^2) + sqrt(h^2 - x^2)]`

= `1/2 [(-x^2 + h^2 - x^2)/sqrt(h^2 - x^2)]`

= `1/2 [(h^2 - 2x^2)/sqrt(h^2 - x^2)]`

And `(d^2A)/(dx^2) = 1/2 [(sqrt(h^2 - x^2). (-4x) - (h^2 - 2x^2) {1/2 (h^2 - x^2)^(-1//2) (-2x)})/((sqrt(h^2 - x^2))^2)]`

= `1/2 [(-4x (sqrt(h^2 - x^2)) + ((h^2 - 2x^2)x)/sqrt(h^2 - x^2))/((h^2 - x^2))]`

= `1/2 [(-4x (h^2 - x^2) + (h^2 - 2x^2)x)/((h^2 - x^2).sqrt(h^2 - x^2))]`

= `1/2 [(-4xh^2 + 4x^3 + h^2x - 2x^3)/((h^2 - x^2)^(3//2))]`

= `1/2 [(2x^3 - 3xh^2)/(h^2 - x^2)^(3//2)]`

For maximum and minimum value,

Put `(dA)/dx = 0`

⇒ `1/2 [(h^2 - 2x^2)/(sqrt(h^2 - x^2))] = 0`

⇒ h² – 2x² = 0

⇒ `x = h/sqrt(2)`

∴ `((d^2A)/(dx^2))_(x = h/sqrt(2)) = 1/2 [(2 xx h^3/(2sqrt(2)) - 3 xx h/sqrt(2) xx h^2)/((h^2 - h^2/2)^(3//2))]`

= `1/2 [(h^3 - 3h^3)/((h^2/2)^(3//2) sqrt(2))]`

= `(-2h^3)/(2 xx h^3/(2sqrt(2)) xx sqrt(2))`

= –2 < 0

Thus, A is maximum at `x = h/sqrt(2)`

Now, Base = `sqrt(h^2 - x^2)`

= `sqrt(h^2 - h^2/2)`

= `sqrt(h^2/2)`

= `h/sqrt(2)`

And Altitude = x = `h/sqrt(2)`

Base = `sqrt(h^2 - x^2)`

= `sqrt(h^2 - (h/sqrt(2))^2`

= `h/sqrt(2)`

Since, Base = Altitude = `h/sqrt(2)`

Hence, the triangle is isosceles.

Hence Proved.