Question 158810
draw triangle ABC where AB is the height and BC is the hypotenuse and AC is the base.
A is bottom left, B is top left right above A, C is bottom right.
let z = BC = 20cm = hypotenuse
let x = AC = base
let y = BA = height
let {{{x+y+z = 47}}} (given)
this becomes {{{x+y+20 = 47}}} (z is given as 20)
subtracting 20 from both sides of the equation gets {{{x+y=27}}}
solving for either x or y gets {{{y=27-x}}}
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pythagorean formula states that {{{z = sqrt(x^2 + y^2)}}}
this becomes {{{20 = sqrt(x^2 + y^2)}}}
squaring both sides of the equation, we get
{{{400 = x^2 + y^2}}}  (***** this five star equation used below)
substituting (27-x) for y gets
{{{400 = x^2 + (27-x)^2}}}
this becomes
{{{400 = x^2 + (27*27) -54*x + x^2}}}
this becomes
{{{2*x^2 - 54*x + 329 = 0}}}
i could not find any common multiples so i used the quadratic equation to solve.
the quadratic equation is {{{x=((-b)+sqrt(b^2-4*a*c))/(2*a)}}} and {{{x=((-b)-sqrt(b^2-4*a*c))/(2*a)}}}
in this equation, 
a = 2
b = -54
c = 329
after solving the formula (details left out for sake of brevity)
i got
x = 17.71307489, and x = 9.286925113
one of these becomes y, so the answers appear to be
x = 17.71307489, and y = 9.286925113
plugging these values into the original equation for x + y + z = 47 yielded 47 = 47 proving the value were good for the perimeter equation.
results were 17.71307489 + 9.286925113 + 20 = 47 which became 47 = 47.
plugging these values into the original quadratic equation solving for 400 = 400 = x^2 + y^2(the 5 start equation ***** shown above) yields
17.713^2 + 9.2819^2 = 400 which became 400 = 400 proving that the values for x and y are good.
truncated values are shown but full values were used in the calculations.
answer is
x = 17.7130
y = 9.2869
z = 20
perimeter = 47 checks out ok
z^2 = x^2 + y^2 checks out ok