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June 16th, 2010, 06:07 PM  #1 
Newbie Joined: Jun 2010 Posts: 4 Thanks: 0  Quadratic Equations
I need help with this question. The hypotenuse of a right triangle is 20cm. The sum of the lengths of the other two sides is 28cm. Find the lengths of the two sides? I am confused, I have been doing math for 2 hours now can someone please help me with this? Thanks 
June 16th, 2010, 07:16 PM  #2  
Member Joined: Jun 2010 Posts: 36 Thanks: 0  Re: Quadratic Equations Quote:
 
June 17th, 2010, 05:42 AM  #3 
Senior Member Joined: Jan 2009 Posts: 345 Thanks: 3  Re: Quadratic Equations
We can call the adjacent leg b or a (lets call it b). We can also call the opposite leg a or b (let's call it a) Then a+b = 28 b = 28  a b^2 = (28a)^2 a^2= (28b)^2 (20)^2 = a^2+b^2 (20)^2 = (28b)^2+(28a)^2 [if we use b=28a, we could solve only for a] [if we use a=28b, we could solve only for b, doesn't matter which one you use] (20)^2 = (28(28a))^2+(28a)^2 400 = a^2+(28a)^2 roots are; a= 12 a = 16 using both, we could solve for b, b = 2812 = 16 b = 2816 = 12 a combination of the two (12, 16) for a or b work. No same values for a or b work. so if a = 12, b = 16 (b is the adjacent leg and a is the opposite leg) or if a = 16, b = 12 (a is the adjacent leg and b is the opposite leg) Those are your solutions! 

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