Question

A city lot has the shape of a right triangle whose hypotenuse is 2 ft longer than one of the other sides. the perimeter of the lot is 364 ft. how long is each side of the lot?

Answer 1

Suppose that, in the triangle ABC, the hypothenuse is AC. Let's call its length [tex ] \overline{AC} = x [/tex].

The hypothenuse is 2 ft longer than one of the legs (say AB, for example). This means that $\overline{AB} = x-2$

Finally, we can find the other leg with the pythagorean theorem:

$\overline{BC} = \sqrt{\overline{AC}^2 - \overline{AB}^2} = \sqrt{x^2 - (x-2)^2} = \sqrt{x^2 - x^2 + 4x - 4} = \sqrt{4x-4} = \sqrt{4(x-1)} = 2\sqrt{x-1}$

So, the perimeter (i.e. the sum of the sides) is given by

$x + (x-2) + 2\sqrt{x-1} = 364 \iff 2x - 2 + 2\sqrt{x-1} = 364$

Isolate the square root to get

$2\sqrt{x-1} = 2 - 2x + 364 = 366 - 2x$

Divide all sides by 2:

$\sqrt{x-1} = 183 - x$

Square both sides:

$x-1 = x^2 - 366 x + 33489 \iff x^2 - 367x + 33490 = 0$

This equation has solutions $x = 170$ or $x=197$. These solutions lead to, in the first case,

$\overline{AC} = x = 170,\quad \overline{AB} = x-2 = 168,\quad \overline{BC} = 2\sqrt{x-1} = 2\sqrt{169} = 2\cdot 13 = 26$

In the second case, you have

$\overline{AC} = x = 197,\quad \overline{AB} = x-2 = 195,\quad \overline{BC} = 2\sqrt{x-1} = 2\sqrt{196} = 2\cdot 14 = 28$