The positive difference between real numbers c and d are defined to be absolute values of c minus d. The hypotenuse of the right triangle is 5 cm and the perimeter is 11 cm. Find the number of cm in the positive difference between the lengths of the two legs.
What I've done: I tried the 3 4 5 but the perimeter then would be 12, I'm not really sure what this question is asking for.... I've also tried labeling the two legs c and d, c^2 + d^2 = 25, and c+d+5=11 and tried solving systems of equations which did not work...
I would greatly appreciate hints for this problem.
This is not a Pythagorean triple where $1)$ there is a right angle and $2)$ all sides are integers. Instead, it could be anything from an isosceles where the two legs are equal to a near-isosceles where the hypotenuse almost equals one of the legs.
In the former, the difference is zero. In the latter, the hypotenuse is only arbitrarily larger than one leg and the difference approaches $5-1=4$.
In between, there is a right angle between the two legs but their lengths are not integers. The answer(s) can be found by solving the simultaneous equations: $$c^2+d^2=25\qquad h=5\qquad c+d+h=11\\ \implies c^2+d^2=25\qquad c+d=6$$
$$ \implies d=6-c\\ \implies c^2+(6-c)^2\\ =2 c^2 - 12 c + 36=25\\ \implies 2 c^2 - 12 c + 11 =0$$ $$ c=\frac{12\pm\sqrt{12^2-4(2)11}}{2\cdot2}\\ =\frac{12 \pm 2 \sqrt{14}}{4} =3 \pm \sqrt{\frac{7}{2}} $$
This is confirmed by WolframAlpha's solution here where
$$c=\left(3 + \sqrt{\dfrac{7}{2}}\right) \approx 4.87083\\ d=\left(3 - \sqrt{\dfrac{7}{2}}\right)\approx 1.12917$$
or vice versa. We can see that $$\sqrt{\left(3 + \sqrt{\dfrac{7}{2}}\right)^2 + \left(3 - \sqrt{\dfrac{7}{2}}\right)^2}=5$$
and the difference is
$$\left(3 + \sqrt{\dfrac{7}{2}}\right) - \left(3 - \sqrt{\dfrac{7}{2}}\right)= \sqrt{14} \approx 3.74165$$.