The three sides of a triangular lot are represented by x, 2x and 3x+2. How do you find each side if the perimeter of the triangular lot is 362 feet?

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Answer 1 · 2017-01-09T03:54:42

The solution is negative in the proximity of $- 28.5$ $-28.5$ ..Illustrative graph for the zero of the resulting cubic is inserted.

Use Heron/s formula.

$s = \frac{a + b + c}{2} = \frac{x + 2 x + 3 x + 2}{2} = 3 x + 1 .$ $s = (a+b+c)/2=(x + 2x +3x+2)/2=3x+1.$

The area

$= \sqrt{s (s - a) (s - b) (c - c)}$ $= sqrt(s(s-a)(s-b)(c-c))$ =362.

Note that s-c = 3x+1-(3x+2)=-1.

Squaring and simplifying by expansion'

6x^3+11x^2+6x+131034=0

All coefficients have the same sign. So, there is no positive solution.

The graph reveals x = -28.5, nearly.

The question has to be reviewed. for proper changes.

graph{6x^3+11x^2+6x+131045 [-44.5, 44.5, -22.26, 22.24]} }