Answer:
Largest perimeter of the triangle =
![P(6\sqrt{2}) = 6\sqrt{2} + \sqrt{144-72} + 12 = 12\sqrt{2} + 12 = 12(\sqrt2 + 1)](https://tex.z-dn.net/?f=P%286%5Csqrt%7B2%7D%29%20%3D%206%5Csqrt%7B2%7D%20%2B%20%5Csqrt%7B144-72%7D%20%2B%2012%20%3D%2012%5Csqrt%7B2%7D%20%2B%2012%20%3D%2012%28%5Csqrt2%20%2B%201%29)
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}](https://tex.z-dn.net/?f=x%5E2%20%2B%20y%5E2%20%3D%20%2812%29%5E2%20%3D%20144%5C%5Cy%5E2%20%3D%20144-x%5E2%5C%5Cy%20%3D%20%5Csqrt%7B144-x%5E2%7D)
Perimeter of Triangle = Side 1 + Side 2 + Hypotenuse.
![P(x) = x + \sqrt{144-x^2} + 12](https://tex.z-dn.net/?f=P%28x%29%20%3D%20x%20%2B%20%5Csqrt%7B144-x%5E2%7D%20%2B%2012)
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}}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%28P%28x%29%29%7D%7Bdx%7D%20%3D%20%5Cfrac%7Bd%28x%20%2B%20%5Csqrt%7B144-x%5E2%7D%20%2B%2012%29%7D%7Bdx%7D%20%3D%201-%5Cdisplaystyle%5Cfrac%7Bx%7D%7B%5Csqrt%7B144-x%5E2%7D%7D%20)
Equating the first derivative to zero, we get,
![\frac{dP(x))}{dx} = 0\\\\1-\displaystyle\frac{x}{\sqrt{144-x^2}} = 0](https://tex.z-dn.net/?f=%5Cfrac%7BdP%28x%29%29%7D%7Bdx%7D%20%3D%200%5C%5C%5C%5C1-%5Cdisplaystyle%5Cfrac%7Bx%7D%7B%5Csqrt%7B144-x%5E2%7D%7D%20%3D%200)
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}](https://tex.z-dn.net/?f=1-%5Cdisplaystyle%5Cfrac%7Bx%7D%7B%5Csqrt%7B144-x%5E2%7D%7D%20%3D%200%5C%5C%5C%5Cx%20%3D%20%5Csqrt%7B144-x%5E2%7D%7D%5C%5C%5C%5Cx%5E2%20%3D%20144-x%5E2%5C%5C%5C%5Cx%20%3D%20%5Csqrt%7B72%7D%20%3D%206%5Csqrt%7B2%7D)
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}}}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5E2%28P%28x%29%29%7D%7Bdx%5E2%7D%20%3D%20%5Cdisplaystyle%5Cfrac%7B-%28144-x%5E2%29%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D-x%5E2%7D%7B%28144-x%29%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D)
At x =
,
![\frac{d^2(V(x))}{dx^2} < 0](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5E2%28V%28x%29%29%7D%7Bdx%5E2%7D%20%3C%200)
Then, by double derivative test, the maxima occurs at x = ![6\sqrt{2}](https://tex.z-dn.net/?f=6%5Csqrt%7B2%7D)
Thus, maxima occurs at x =
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)](https://tex.z-dn.net/?f=P%286%5Csqrt%7B2%7D%29%20%3D%206%5Csqrt%7B2%7D%20%2B%20%5Csqrt%7B144-72%7D%20%2B%2012%20%3D%2012%5Csqrt%7B2%7D%20%2B%2012%20%3D%2012%28%5Csqrt2%20%2B%201%29)