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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! Tags equations, quadratic Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post PlzzHelp Algebra 2 February 5th, 2013 01:23 PM Ter Algebra 1 April 16th, 2012 09:08 AM jarman007 Algebra 4 January 13th, 2010 11:19 AM Debjani Algebra 2 May 3rd, 2009 05:06 AM luke22 Applied Math 0 December 31st, 1969 04:00 PM

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