A right triangle has both a perimeter and an area of 30. Find the side lengths of the triangle.
So far, I drew a right triangle with legs x and y, and hypotenuse √(x^2+y^2).
The 2 equations I have are : xy = 60 and x+y+√(x^2+y^2) = 30
But I get a really weird answer when solving...
Consider this: The perimeter is a whole number.
That means the triangle is a Pythagorean Triple.
The only one that works is 5, 12, 13. So the legs are 5 and 12.
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+2668 You have the right idea guest, here is how you solve it: \(x+y+\sqrt {x^2+y^2}=30\)
Subtarct \(x+y\) from both sides: \(\sqrt{x^2+y^2}=30-x-y\)
Square both sides: \(x^2+y^2=x^2+y^2+2xy-60x-60y+900\)
Subtract \(x^2+y^2\) from both sides:\(0=2xy-60x-60y+900\)
Subsitute \(xy\) for 60:\(1020-60x-60y=0\)
Rewrite as: \(60x+60y=1020\)
Factor the left side: \(60(x+y)=1020\)
Divide both sides by 60: \(x+y = 17\)
From the other equation, divide by \(y\): \(x = {60\over y}\)
Plug into the other equation: \({60\over y}+y=17\)
Multiply by \(y\): \(60+y^2=17y\)
Subtract \(17y\) from both sides: \(y^2-17y+60=0\)
Use quadratic formula, and you find \(y = 12\)
Subsitute into the other equation, and \(x = 5\)
This means that the third side is \(13\), since the triangle must have a perimeter of 30.
Thus, the side legnths are: \(\color{brown}\boxed{5,12,13}\)
