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June 16th, 2010, 06:07 PM   #1
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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
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June 16th, 2010, 07:16 PM   #2
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Re: Quadratic Equations

Quote:
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?
sa-ri-ga-ma is offline  
June 17th, 2010, 05:42 AM   #3
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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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