Posted by purpletree on Sunday, March 29, 2009 at 12:58pm.
The longer leg of a right triangle is 3 ft longer than three times the shorter leg. The hypotenuse is 3 ft shorter than four times the shorter leg. Find the lengths of the three sides of the right triangle. (Please provide steps/explanation to solve this problem.)
- Quadratic Equations - Brianna, Sunday, March 29, 2009 at 2:58pm
shoter leg = x
longer leg = 3x+3
to solve for hypotenuse you use
a^2 + b^2 = c^2, so
(x)^2 + (3x+3)^2 = (4x-3)^2
(x)(x) + (3x+3)(3x+3) = (4x-3)(4x-3)
(x^2) + (9x^2+18x+9) = (16x^2-24x+9)
then combine like terms
10x^2 + 18x + 9 = 16x^2 - 24x + 9
then i subtracted 10x^2 from both sides
18x + 9 = 6x^2 - 24x +9
next, i subtracted 18x from both sides:
9 = 6x^2 - 42x + 9
and then subtracted 9 from both sides
0 = 6x^2 - 42x
then pull out a 6x on the right side of the equation
0 = 6x(x-7)
so x= 0 and 7
so the shorter leg is :
0 units or 7 units
the longer leg is :
3 units or 24 units
and the hypotenuse is :
-3 units or 25
and since you cant have a hypotenuse with a negative value
i think the three lengths of your triangle are
7, 24, and 25
- Quadratic Equations - purpletree, Monday, March 30, 2009 at 9:22am
Thanks for the help. :)
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