Question 158810: a right triangle has hypotenuse with a measure of 20 cm and a perimeter of 47cm. find the measure of the remaining sides. Answer by gonzo(654) (Show Source):
You can put this solution on YOUR website! 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 (given)
this becomes (z is given as 20)
subtracting 20 from both sides of the equation gets
solving for either x or y gets
-------------------
pythagorean formula states that
this becomes
squaring both sides of the equation, we get (***** this five star equation used below)
substituting (27-x) for y gets
this becomes
this becomes
i could not find any common multiples so i used the quadratic equation to solve.
the quadratic equation is and
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
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