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

Guest Dec 13, 2020

#1**0 **

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**

Guest Dec 13, 2020