Question

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.

Answer 1

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.