Question

Among all right triangles whose hypotenuse has a length of 12 cm, what is the largest possible perimeter?

Answer 1

Answer:

Largest perimeter of the triangle =  

$P(6\sqrt{2}) = 6\sqrt{2} + \sqrt{144-72} + 12 = 12\sqrt{2} + 12 = 12(\sqrt2 + 1)$

Step-by-step explanation:

We are given the following information in the question:

Right triangles whose hypotenuse has a length of 12 cm.

Let x and y be the other two sides of the triangle.

Then, by Pythagoras theorem:

$x^2 + y^2 = (12)^2 = 144\\y^2 = 144-x^2\\y = \sqrt{144-x^2}$

Perimeter of Triangle = Side 1 + Side 2 + Hypotenuse.

$P(x) = x + \sqrt{144-x^2} + 12$

where P(x) is a function of the perimeter of the triangle.

First, we differentiate P(x) with respect to x, to get,

$\frac{d(P(x))}{dx} = \frac{d(x + \sqrt{144-x^2} + 12)}{dx} = 1-\displaystyle\frac{x}{\sqrt{144-x^2}}$

Equating the first derivative to zero, we get,

$\frac{dP(x))}{dx} = 0\\\\1-\displaystyle\frac{x}{\sqrt{144-x^2}} = 0$

Solving, we get,

$1-\displaystyle\frac{x}{\sqrt{144-x^2}} = 0\\\\x = \sqrt{144-x^2}}\\\\x^2 = 144-x^2\\\\x = \sqrt{72} = 6\sqrt{2}$

Again differentiation P(x), with respect to x, using the quotient rule of differentiation.

$\frac{d^2(P(x))}{dx^2} = \displaystyle\frac{-(144-x^2)^{\frac{3}{2}}-x^2}{(144-x)^{\frac{3}{2}}}$

At x = $6\sqrt{2}$,

$\frac{d^2(V(x))}{dx^2} < 0$

Then, by double derivative test, the maxima occurs at x = $6\sqrt{2}$

Thus, maxima occurs at x = $6\sqrt{2}$ for P(x).

Thus, largest perimeter of the triangle =  

$P(6\sqrt{2}) = 6\sqrt{2} + \sqrt{144-72} + 12 = 12\sqrt{2} + 12 = 12(\sqrt2 + 1)$