Question 697554
Let x = the shorter leg. Then the longer leg would be (x + 17) and the hypotenuse would be (3x + 4). In order for this to be a right triangle these sides must fit the Pythagorean equation:
{{{(x)^2 + (x+17)^2 = (3x+4)^2}}}
Now we solve for x. First we simplify. (Be sure to use FOIL or the {{{(a+b)^2 = a^2+2ab+b^2}}} pattern to square the x+17 and the 3x+4.)
{{{x^2 + x^2+2*x*17+17^2 = (3x)^2 + 2(3x)(4) +4^2}}}
{{{x^2 + x^2+34x+289 = 9x^2 + 24x + 16}}}
{{{2x^2+34x+289 = 9x^2 + 24x + 16}}}
Since this is a quadratic equation we want one side to be zero. Subtracting the entire left side from each side we get:
{{{0 = 7x^2-10x-273}}}
This will factor, but not easily. So you may prefer to use the Quadratic Formula instead.
{{{0 = (7x+39)(x-7)}}}
From the Zero Product Property:
7x + 39 = 0 or x - 7 = 0
Solving these we get:
x = -39/7 or x = 7
We will reject the first solution not because it is a fraction but because it is negative. x is the length of the shorter leg and we cannot have negative lengths.<br>
So the shorter leg is 7 cm.
The longer leg would be x + 17 = 7 + 17 = 24 cm
And the hypotenuse would be 3x + 4 = 3(7) + 4 = 21 + 4 = 25 cm