SOLUTION: 1. The length of one leg of a right triangle is 6 cm more than the other. If the length of the hypotenuse is 18 cm, what are the lengths of the two legs? Please give the answer in

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: 1. The length of one leg of a right triangle is 6 cm more than the other. If the length of the hypotenuse is 18 cm, what are the lengths of the two legs? Please give the answer in       Log On


   

Question 93459: 1. The length of one leg of a right triangle is 6 cm more than the other. If the length of the hypotenuse is 18 cm, what are the lengths of the two legs? Please give the answer in ascending order rounded to the nearest tenth of a cm.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

Let x=leg #1, y=leg #2

Since the length of one leg of a right triangle is 6 cm more than the other, this means we can write

y=x%2B6

Now lets use Pythagorean's theorem to set up the problem

x%5E2%2By%5E2=h%5E2

x%5E2%2B%28x%2B6%29%5E2=18%5E2 Plug in the hypotenuse h=18 and y=x+6


x%5E2%2Bx%5E2%2B12x%2B36=18%5E2 Foil


2x%5E2%2B12x%2B36=324 Square 18

2x%5E2%2B12x%2B36-324=0 Subtract 324 from both sides

2x%5E2%2B12x-288=0 Combine like terms


Let's use the quadratic formula to solve for x:


Starting with the general quadratic

ax%5E2%2Bbx%2Bc=0

the general solution using the quadratic equation is:

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29

So lets solve 2%2Ax%5E2%2B12%2Ax-288=0 ( notice a=2, b=12, and c=-288)

x+=+%28-12+%2B-+sqrt%28+%2812%29%5E2-4%2A2%2A-288+%29%29%2F%282%2A2%29 Plug in a=2, b=12, and c=-288



x+=+%28-12+%2B-+sqrt%28+144-4%2A2%2A-288+%29%29%2F%282%2A2%29 Square 12 to get 144



x+=+%28-12+%2B-+sqrt%28+144%2B2304+%29%29%2F%282%2A2%29 Multiply -4%2A-288%2A2 to get 2304



x+=+%28-12+%2B-+sqrt%28+2448+%29%29%2F%282%2A2%29 Combine like terms in the radicand (everything under the square root)



x+=+%28-12+%2B-+12%2Asqrt%2817%29%29%2F%282%2A2%29 Simplify the square root (note: If you need help with simplifying the square root, check out this solver)



x+=+%28-12+%2B-+12%2Asqrt%2817%29%29%2F4 Multiply 2 and 2 to get 4

So now the expression breaks down into two parts

x+=+%28-12+%2B+12%2Asqrt%2817%29%29%2F4 or x+=+%28-12+-+12%2Asqrt%2817%29%29%2F4


Now break up the fraction


x=-12%2F4%2B12%2Asqrt%2817%29%2F4 or x=-12%2F4-12%2Asqrt%2817%29%2F4


Simplify


x=-3+%2B3%2Asqrt%2817%29 or x=-3-3%2Asqrt%2817%29


So these expressions approximate to

x=9.36931687685298 or x=-15.369316876853


So our possible solutions are:
x=9.36931687685298 or x=-15.369316876853


Since a negative length doesn't make sense, our only solution is x=9.36931687685298


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Check:

First find y

y=9.369%2B6=15.369

Now plug in the triangle's dimensions

9.369%5E2%2B15.369%5E2=18%5E2

323.984322=324 Simplify. Since we rounded, this is as close as it gets. So our answer is verified.