# solving triangles

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• October 10th 2009, 02:25 PM
samtheman17
solving triangles
The hypotenuse of a right triangle is 8 more than the shorter leg. The longer leg is 4 more than the shorter leg. Find the length of the shorter leg.

i know your supposed to set up the equation like:

(x+8)^2 = x^2 + (x+4)^2 and then solve for x

however, i keep getting that x = sqrt(48)..which is wrong.
• October 10th 2009, 03:13 PM
skeeter
Quote:

Originally Posted by samtheman17
The hypotenuse of a right triangle is 8 more than the shorter leg. The longer leg is 4 more than the shorter leg. Find the length of the shorter leg.

i know your supposed to set up the equation like:

(x+8)^2 = x^2 + (x+4)^2 and then solve for x

however, i keep getting that x = sqrt(48)..which is wrong.

your equation is correct ... show how you calculated $x = \sqrt{48}$
• October 10th 2009, 03:22 PM
aidan
Quote:

Originally Posted by samtheman17
The hypotenuse of a right triangle is 8 more than the shorter leg. The longer leg is 4 more than the shorter leg. Find the length of the shorter leg.

i know your supposed to set up the equation like:

(x+8)^2 = x^2 + (x+4)^2 and then solve for x

however, i keep getting that x = sqrt(48)..which is wrong.

The set up is correct.
Go ahead & expand the expressions and simplify.
You will get a simple quadratic expression.
Show your work & we'll see where the differences occur.

.
• October 10th 2009, 05:04 PM
samtheman17
well when i expand and simplify i get x^2 - 48 = 0

therefore, x^2 = 48 and x = sqrt(48)
• October 10th 2009, 05:09 PM
skeeter
Quote:

Originally Posted by samtheman17
well when i expand and simplify i get x^2 - 48 = 0

therefore, x^2 = 48 and x = sqrt(48)

show your expansion ...
• October 10th 2009, 05:16 PM
samtheman17
x^2 + (x+4)^2 = (x+8)^2

=> X^2 +x^2 + 16 = x^2 + 64

=> x^2 + 16 = 64

=> x^2 = 48

=> x = sqrt(48)
• October 10th 2009, 05:24 PM
skeeter
Quote:

Originally Posted by samtheman17
x^2 + (x+4)^2 = (x+8)^2

=> X^2 + x^2 + 16 = x^2 + 64 here are your mistakes

=> x^2 + 16 = 64

=> x^2 = 48

=> x = sqrt(48)

$(x+4)^2 \ne x^2 + 16$ and $(x+8)^2 \ne x^2 + 64$

this is the pattern for squaring a binomial ...

$(a+b)^2 = a^2 + 2ab + b^2$

try again.
• October 10th 2009, 05:36 PM
samtheman17
ahhh
thank you so much, i finally figured it out :)