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December 20, 2014

December 20, 2014

Posted by **ann** on Wednesday, April 4, 2012 at 2:00pm.

when given long leg b is 1/2x +11

when given hypotenuse c is 2x + 1

i put (-1/2x + 11) +(2x + 1) = x and came out with 20 but that did not work out when squared A + b = C.

I estimated x/a = 8, then b = 15 and c = 17 and that works 8^2 + 15^2 = 17^2 or 64 + 225 = 289 for A^2 +b^2 = c^2

could you please show me how to do it the right way?

thanks

- algebra -
**Reiny**, Wednesday, April 4, 2012 at 2:23pmYou are not even using the correct formula.

it is A^2 + B^2 = C^2 , where C is the hypotenuse, so

x^2 + ((1/2)x + 11)^2 = (2x+1)^2

x^2 + (1/4)x^2 + 11x + 121 = 4x^2 + 4x + 1

times 4 to get rid of my fraction

4x^2 + x^2 + 44x + 484 = 16x^2 + 16x + 4

-11x^2 + 28x + 480 = 0

11x^2 - 28x - 480 = 0

(x-8)(11x + 60) = 0

x = 8 or x = -60/11 , but x can't be negative

so x = 8

one side is 8

the other is (1/2)x+11 = 15

and the hypotenuse is 2x+1 = 17

- algebra -
**ann**, Thursday, April 5, 2012 at 1:04pmReiny, thank you so much, I really needed that help.

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