SOLUTION: Find the area of a right triangle whose legs have lengths which differ by 7 meters and whose hypotenuse is 17 meters long. (Show your solution using Quadratic Equation)

Question 353843: Find the area of a right triangle whose legs have lengths which differ by 7 meters and whose hypotenuse is 17 meters long. (Show your solution using Quadratic Equation)
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
Find the area of a right triangle whose legs have lengths which differ by 7 meters and whose hypotenuse is 17 meters long. (Show your solution using Quadratic Equation)
.
Once you see "right triangle" you can apply the Pythagorean theorem.
.
Let x = shortest leg
then
x+7 = other leg
17 = hypotenuse
.
x^2 + (x+7)^2 = 17^2
x^2 + (x+7)(x+7) = 17^2
x^2 + x^2+14x+49 = 289
2x^2+14x+49 = 289
2x^2+14x-240 = 0
x^2+7x-120 = 0 (your quadratic)
(x+15)(x-8) = 0
x = {-15, 8}
Throw out the negative solution leaving
x = 8 meters (shortest leg)
x+7 = 8+7= 15 meters (other leg)

Area is
(1/2)bh
(1/2)(8)(15)
(4)(15)
60 square meters
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