length of triangle sides

matteric92

New member
Joined
Sep 1, 2006
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2
Ok, here's the question:

"The length of the hypotenuse of a right triangle is 4 cm more than the longer leg. The length of the longer leg is 14 cm more than the length of the shorter leg. Find the number of centimeters in the length of each side of the right triangle."

I know the answers are 16, 30, 34, but I have to show my work, and when I use the "a^2 + b^2 = c^2" formula, I always get the answer "sqrt[128]". What am I doing wrong?

This is what I do:

. . .x^2 + (x + 14)^2 = (x + 18)^2
. . .2x^2 + 196 = x^2 + 324
. . .x^2 = 128
. . .x = ????
 

stapel

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Feb 4, 2004
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Hint: (a + b)<sup>2</sup> = (a + b)(a + b) = a<sup>2</sup> + 2ab + b<sup>2</sup>. It is not equal to a<sup>2</sup> + b<sup>2</sup>.

Eliz.
 

TchrWill

Full Member
Joined
Jul 7, 2005
Messages
856
matteric92 said:
"The length of the hypotenuse of a right triangle is 4 cm more than the longer leg. The length of the longer leg is 14 cm more than the length of the shorter leg. Find the number of centimeters in the length of each side of the right triangle."

I know the answers are 16, 30, 34, but I have to show my work, and when I use the "a^2 + b^2 = c^2" formula, I always get the answer "sqrt[128]". What am I doing wrong?

This is what I do:

. . .x^2 + (x + 14)^2 = (x + 18)^2
. . .2x^2 + 196 = x^2 + 324
. . .x^2 = 128
You started out right with x^2 + (x + 14)^2 = (x + 18)^2
Expanding
x^2 + x^2 + 28x += 196 = x^2 + 36x + 324

Collecting terms;
x^2 - 8x - 128 = 0

From the quadratic formula
x = {8+/-sqrt(64 + 512)}/2
x = [8+/-sqrt(576)}/2
x = (8 + 24)/2 = 16 making y = 30 and z = 34

Alternatively, 128 = 2^7
The factors of 128 are therefore 1, 2, 4, 8, 16, 32, 64 and 128
Which pair have a difference of 8 and a product of 128?
8 and 16 or course making x^2 - 8x - 128 = (x + 8)(x - 16)
(x - 16) is the only useful factor making x = 16, y = 30 and z = 34
 
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