SOLUTION: Find the area of a right triangle whose legs have lengths which differ by 7m and whose hypotenuse is 17m long?

Question 828270: Find the area of a right triangle whose legs have lengths which differ by 7m and whose hypotenuse is 17m long?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Find the area of a right triangle whose legs have lengths which differ by 7m and whose hypotenuse is 17m long?
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leg: x meters
leg: x+7 meters
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Pythagoras:
x^2 + (x^2+14x+49) = 17^2
2x^2 + 14x - 240 = 0
x^2 + 7x - 120 = 0
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(x+15)(x-8) = 0
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x = 8 meters
x+7 = 15 meters
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Area = (1/2)8*15 = 60 sq.meters
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Cheers,
Stan H.
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