A certain right triangle has area 30 in. squared . one leg of the triangle measures 1 in. less than the hypotenuse. let x represent the length of the hypotenuse. a) express the length of the leg mentioned above in terms of x. give the domain of x. b) express the length of the other leg in terms of x. c) write an equation based on the information determined thus far. square both sides and then write the equation with one side as a polynomial with integer coefficients, in descending powers, and the other side equal to 0. divide out any common factors to find the most simplified form. d) solve the equation in part (c) graphically. find the lengths of the three sides of the triangle.

a) The length of the longer leg is x-1

b) Based on the area, the other leg is 2*30/(x -1). Based on the Pythagorean theorem, the other leg is √(x^2 -(x -1)^2).

c) Equating the two expressions for the shorter leg, we have
.. 60/(x -1) = √(2x -1)
.. 3600/(x -1)^2 = (2x -1)
.. (2x -1)(x^2 -2x +1) = 3600
.. 2x^3 -5x^2 +4x -3601 = 0

d) There is one positive real root, at x=13. A graphical solution works well.

The three sides of the triangle are 5 in, 12 in, 13 in.

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5-12-13 is a well-known Pythagorean triple. It is the next smallest one after 3-4-5.