Question 81476: in a right triangle, one leg is 7 more than the other. the hypotenuse is 17. find both legs. thanks.
Answer by praseenakos@yahoo.com(507)   (Show Source):
You can put this solution on YOUR website! QUESTION:
In a right triangle, one leg is 7 more than the other. the hypotenuse is 17. find both legs.
ANSWER:
Given ,
Hypotenuse = 17
Assume that the third side is 'x'
Then the second side is = x + 7
By pythagorean theorem,
(hypotenuse)^2 = (thrid side)^2 + (second side)^2
==> 17^2 = x^2 + (x+7)^2
==> 289 = x^2 + x^2 + 14x + 49
==> 289 = 2x^2 + 14x + 49
Subtract 289 from both sides
Hypotenuse = 17
Assume that the third side is 'x'
Then the second side is = x + 7
By pythagorean theorem,
(hypotenuse)^2 = (thrid side)^2 + (second side)^2
==> 17^2 = x^2 + (x+7)^2
==> 289 = x^2 + x^2 + 14x + 49
==> 289 - 289 = 2x^2 + 14x + 49 - 289
==> 0 = 2x^2 + 14x- 240
==> 2x^2 + 14x- 240 = 0
This is a quadratic equation and can be solved by quadratic formula,
Comparing with the general quadratic equation, ax^2 + bx + c = 0
we have, a = 2, b = 14 and c = -240
so,
==> x = (-14 + 46) /4 or x = (-14 - 46)/4
==> x = 32/4 or x = -60/4
==> x = 8 or x = -15
Since negative value is not admissible for the side of a triangle,
we can take the thirds side = 8
Then the second side = x + 7 = 8 + 7 = 15
so the sides of the triangle are, 8, 15 and 17.
Hope you found the explanation useful.
Regards.
Praseena.
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