The hypotenuse of a right triangle is two more than the length of one of its legs. Find the side lengths of the right triangle given the perimeter= 60 and it's area= 120

9514 1404 393

  10 and 24

Step-by-step explanation:

We know that some of the Pythagorean triples that appear in math problems are (3, 4, 5), (5, 12, 13), (7, 24, 25), (8, 15, 17).

These have (perimeter, area) values of (12, 6), (30, 30), (56, 84), (40, 60).

For some scale factor n, we want (p·n, a·n²) = (60, 120). Of the triangles listed, we see that the (5, 12, 13) triangle scaled by n=2 will satisfy the problem requirements. (30·2, 30·2²) = (60, 120)

The side lengths are 10 and 24.

__

Check

For the side lengths we found, the perimeter is 10+24+26 = 60; the area is 1/2(10)(24) = 120. The hypotenuse is 2×13 = 26 = 24+2.

__

In the attached, one side is x, the other is y. The hypotenuse is (x+2). The square root equation comes from ...

  x² +y² = (x+2)²   ⇒   y² = (x² +4x +4) -x²   ⇒   y² = 4x +4 = 4(x +1)

_____

Additional comment

The graph shows the solution of the various constraints. At least, the combination of constraints will give a quadratic equation in x. They can be combined in a way that gives a cubic equation in x. Either way, we prefer the graphical or "guess and check" approach (above) as being easier to do.

Using the third equation in the attachment to write an expression for y, we have ...

  y = 58 -2x

Substituting that into the second equation gives ...

  (x(58 -2x)/2 = 120

  29x -x² = 120

  x² -29x +120 = 0

  (x -5)(x -24) = 0 . . . . x = 5 or 24.

The root x=5 is a legitimate solution to the pair of equations we chose to solve. The line y=58-2x intersects the hyperbola xy/2 = 120 in two places. However, (x, y) = (5, 48) does not satisfy the hypotenuse requirement that x+2 > y.

i believe the answer is letter c) 8

step-by-step explanation:

move it to the other side

step-by-step explanation: