Algebra

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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...

Guest Feb 14, 2022

Guest · Answer 1 · 2022-02-14T11:13:53

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.

_.

BuilderBoi · Answer 2 · 2022-02-15T12:15:21

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: $\color{brown}\boxed{5,12,13}$