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A regular octagon has side length 2√2 inches. What is the sum of squares of its diagonals with different lengths? Express your answer in simplest radical form.

 Jun 5, 2024
 #1
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Let  ABCDEFGH  be the octagon  and let it be inscribed in a circle

We can  find   the radius of the circle as follows

 

(2sqrt 2)^2  = 2r^2 - 2r^2* (1/sqrt2)

8 = r^2 (2 - 2/sqrt 2)

8 = r^2 (2 -sqrt (2))

8 / (2 -sqrt 2) = r^2

8 (2 + sqrt 2) / (4 -2)  = r^2

4 (2 + sqrt 2) = r^2

r = 2sqrt [ 2 +sqrt 2] =  sqrt [ 8 + sqrt 32]

 

Let A=   (0 , sqrt [8  + sqrt 32[ )

Let E = ( 0 . -sqrt [8 +sqrt 32] )

Let C = (sqrt [8 + sqrt 32] , 0]

Let D =  ( sqrt [8 + sqrt32] /sqrt 2 , -sqrt[8-sqrt32] /sqrt 2)  = (sqrt [4 + sqrt 8] , -sqrt [4 +sqrt 8] )

 

 

There are only four diagonals of differing lengths

The easiest one  to  calculate   is AE  = 2r =   2sqrt [ 8 +  sqrt32]   = 2sqrt [8 + 4sqrt 2 ]

AD =  sqrt [  ( sqrt [4 + sqrt 8[)^2 +  ( sqrt [8 +sqrt 32) ]  + sqrt [4 +sqrt 8] ^2 ] = sqrt [ 24 + 24sqrt 2]

AB = sqrt (8) 

AC =  sqrt [  ( sqrt [8 + sqrt 32 ] )^2 +  ( sqrt [8 +sqrt 32 ] )^2 ]= sqrt [ 16 + 8sqrt 2 ]

 

Sum of squares of differing length diagonals

 

AE^2 + AD^2 + AB^2 + AC^2  = 

 

4 [8 + 4sqrt 2] + [24 + 24sqrt 2]  + 8  + 16 + 8sqrt 2 = 80 + 48sqrt 2  

 

cool cool cool

 Jun 6, 2024

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