More Algebra Than Geometry 2

Let the length of the hypotenuse be $c$ and those of the other two perpendicular sides be $a$ and $b$ . Therefore we have:

$\begin{cases} a + b + c = 32 & \implies a+b = 32-c & ...(1) \\ \dfrac {ab}2 = 20 & \implies ab = 40 & ...(2) \\ a^2 + b^2 = c^2 & \small \color{#3D99F6} \text{Pythagorean theorem} & ...(3) \end{cases}$

$\begin{aligned} (1)^2: \quad (a+b)^2 & = (32-c)^2 \\ {\color{#3D99F6}a^2} + {\color{#D61F06}2ab} + \color{#3D99F6} b^2 & = 32^2 - 2(32)c + c^2 & \small \color{#3D99F6} (3): \quad a^2 + b^2 = c^2 \\ {\color{#3D99F6}c^2} + \color{#D61F06} 80 & = 1024 - 64c + c^2 & \small \color{#D61F06} (2): \quad ab = 40 \\ 64c & = 944 \\ \implies c & = \boxed{14.75} \end{aligned}$

Let a and b be the sides around the right angle, and c be the hypotenuse.

We have this system of equations:

$a^2 + b^2$ = $c^2$

$a + b + c$ =32

ab=40

We need to find out what c equals.

From the first and third equations, we get $a^2 + 2ab + b^2$ = $c^2 + 80$ , which factors down to $(a + b)^2$ = $c^2 + 80$

From the second equation, we get $a + b$ = $32-c$

We can substitute $a + b$ as $32-c$ into $(a + b)^2$ = $c^2 + 80$ , and get $(32-c)^2$ = $c^2 + 80$

We then distribute the left side again, to get $1024-64c + c^2$ = $c^2 + 80$

Our final three steps to solve for c are:

1. Subtract $c^2$ on both sides: $1024-64c$ =80

2. Subtract 1024 on both sides: -64c=-944

3. Divide both sides by -64: $\boxed {c=14.75}$