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25 February, 07:35

The sides of a triangle are x, x + 1, 2 x - 1 and its area is x root of 10. Find the value of x.

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  1. 25 February, 08:01
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    Ok, I'm going to start off saying there is probably an easier way of doing this that's right in front of my face, but I can't see it so I'm going to use Heron's formula, which is A=√[s (s-a) (s-b) (s-c) ] where A is the area, s is the semiperimeter (half of the perimeter), and a, b, and c are the side lengths.

    Substitute the known values into the formula:

    x√10=√{[ (x+x+1+2x-1) / 2][ ({x+x+1+2x-1}/2) - x][ ({x+x+1+2x-1}/2) - (x+1) ][ ({x+x+1+2x-1}/2) - (2x-1) ]}

    Simplify:

    x√10=√{[4x/2][ (4x/2) - x][ (4x/2) - (x+1) ][ (4x/2) - (2x-1) ]}

    x√10=√[2x (2x-x) (2x-x-1) (2x-2x+1) ]

    x√10=√[2x (x) (x-1) (1) ]

    x√10=√[2x² (x-1) ]

    x√10=√ (2x³-2x²)

    10x²=2x³-2x²

    2x³-12x²=0

    2x² (x-6) = 0

    2x²=0 or x-6=0

    x=0 or x=6

    Therefore, x=6 (you can't have a length of 0).
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