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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
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 Originally Posted by cholride 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
If one side is x, the other side will be (28 - x). What is the relation between the two sides and the hypotenuse of the right angled triangle?

 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 = (28-a)^2 a^2= (28-b)^2 (20)^2 = a^2+b^2 (20)^2 = (28-b)^2+(28-a)^2 [if we use b=28-a, we could solve only for a] [if we use a=28-b, we could solve only for b, doesn't matter which one you use] (20)^2 = (28-(28-a))^2+(28-a)^2 400 = a^2+(28-a)^2 roots are; a= 12 a = 16 using both, we could solve for b, b = 28-12 = 16 b = 28-16 = 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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