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Question 1200812: the hypotenuse of a right triangle is 3 in. longer than the longer. the shorter leg is 3 in. shorter than the longer leg. find the lengths of the side of the triangle.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! b is the longer leg
a is the shorter leg
c is the hypotenuse
c = b + 3
a = b - 3
c^2 = a^2 + b^2 becomes (b+3)^2 = (b-3)^2 + b^2
subtract (b-3)^2 + b^2 from both sides of the equation to get:
(b+3)^2 - (b-3)^2 - b^2 = 0
simplify to get:
b^2 + 6b + 9 - (b^2 - 6b + 9) - b^2 = 0
simplify further to get:
b^2 + 6b + 9 - b^2 + 6b - 9 - b^2 = 0
combine like terms to gtet:
-b^2 + 12b = 0
multiply both sides of this equation by -1 to get:
b^2 - 12b = 0
factor out the b to get:
b * (b-12) = 0
you get:
b = 0 or b = 12.
b can't be equal to 0, so go with b = 12 and see what happens.
(b+3)^2 - (b-3)^2 - b^2 = 0 becomes:
(12+3)^2 - (12-3)^2 - 12^2 = 0 which becomes:
15^2 - 9^2 - 12^2 = 0 which becomes 0 = 0, confirming that b = 12 is good.
your triangle has a hypotenuse of 15 and a shorter leg of 9 and a longer leg of 12, making it a 9-12-15 triangle, which is similar to a 3-4-5 triangle.
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