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For what value of a is there a right triangle with sides a+1, 6a, and 2a+1?

 Dec 13, 2020
 #1
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Solve for a:
36 a^2 = (a + 1)^2 + (2 a + 1)^2

Expand out terms of the right hand side:
36 a^2 = 5 a^2 + 6 a + 2

Subtract 5 a^2 + 6 a + 2 from both sides:
31 a^2 - 6 a - 2 = 0

Divide both sides by 31:
a^2 - (6 a)/31 - 2/31 = 0

Add 2/31 to both sides:
a^2 - (6 a)/31 = 2/31

Add 9/961 to both sides:
a^2 - (6 a)/31 + 9/961 = 71/961

Write the left hand side as a square:
(a - 3/31)^2 = 71/961

Take the square root of both sides:
a - 3/31 = sqrt(71)/31 or a - 3/31 = -sqrt(71)/31

Add 3/31 to both sides:
a = 3/31 + sqrt(71)/31 or a - 3/31 = -sqrt(71)/31

Add 3/31 to both sides:

 a = 3/31 + sqrt(71)/31 ==0.36859

 Dec 13, 2020
 #2
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Via Pythagorean Theorem

  (a + 1)2 + (2a + 1)2  = 36 a2


  5a2 + 6a + 2   =  36 a2       re-arrange

-31a2 +6a +2 = 0                 Plug into Quadratic Formula     a = -31    b = 6     c = 2

      to find    a = .368585


 

 Dec 14, 2020

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