Right triangle perimeter

The Pythagorean theorem states that $a^{2} + b^{2} = c^{2}$ . Adding $2ab$ to both sides turns the left hand side equal to the square of $a + b$ . Thus, $a + b = \sqrt{c^{2} + 2ab}$ . Since 756 is $\frac{ab}{2}$ , $2ab$ must then be $756 \times 4 = 3024$ , and since $c = 75$ , $c^{2} = 5625$ . Adding $2ab$ and $c^{2}$ and taking the square root, we get $\sqrt{c^{2} + 2ab} = 93$ . Since $c^{2} = a^{2} + b^{2}$ , we can obtain the equation $\sqrt{a^{2} + 2ab + b^{2}} = 93$ . Since $a^{2} + 2ab + b^{2} = (a + b)^{2}$ , $a + b = 93$ . Now that we know the sum of the lengths of the two legs along with the length of the hypotenuse, we can simply add them up to find that $p = a + b + c = 93 + 75 = 168$ .

Let $a$ and $b$ be the legs of the right triangle. Then

$A=\dfrac{1}{2}ab$

$756=\dfrac{1}{2}ab$

$1512=ab$

$a=\dfrac{1512}{b}$ $(1)$

By pythagorean theorem, we have

$a^2+b^2=75^2$

$a^2+b^2=5625$ $(2)$

Substitute $(1)$ in $(2)$ .

$\left(\dfrac{1512}{b}\right)^2+b^2=5625$

$\dfrac{2286144}{b^2}+b^2=5625$

Multiplying both sides by $b^2$ , we get

$2286144+b^4=5625b^2$

$b^4-5625b^2+2286144=0$

Let $x^2=b^4$ and $x=b^2$ , then

$x^2-5625x+2286144=0$

By using the quadratic formula,

$x=\dfrac{5625}{2}\pm \dfrac{\sqrt{(-5625)^2-4(2286144)}}{2}=\dfrac{5625}{2}\pm \dfrac{4743}{2}$

$x=5184$ and $x=441$

The value of $b$ can be $\sqrt{5184}=72$ or $\sqrt{441}=21$ . The value of $a$ follows, it can be $72$ or $21$ . (depending on the label on the figure or assumption: example: let $a$ be the longer leg and $b$ be the shorter leg or let $a$ be the shorter leg and $b$ be the longer leg ......etc)

The desired perimeter is $72+21+75=\boxed{168}$