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# 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...

Feb 14, 2022

#1
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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

.

Feb 14, 2022
#2
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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: $$\color{brown}\boxed{5,12,13}$$

Feb 15, 2022