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# regular octagon ABCDEFGH has a side length of 2 cm. Find the area of square ACEG.

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## regular octagon ABCDEFGH has a side length of 2 cm. Find the area of square ACEG.

off-topic
Aug 1, 2019

#1
+3

Regular octagon $$ABCDEFGH$$ has a side length of 2 cm.

Find the area of square $$ACEG$$.

$$\text{ Let the side length of the square =s } \\ \text{ Let the area of the square =s^2 } \\ \text{ Let \angle ABC =135^\circ \qquad 135^\circ = 180^\circ-\dfrac{360^\circ}{8} }$$

cosine formula:

$$\begin{array}{|rcll|} \hline s^2 &=& 2^2+2^2-2\cdot 2\cdot 2\cdot \cos(135^\circ) \\ s^2 &=& 8-8\cdot \cos(135^\circ) \\ && \boxed{ \cos(135^\circ) \\ =\cos(180^\circ-45^\circ)\\ = -\cos(45^\circ)\\=-\dfrac{\sqrt{2}}{2} } \\ s^2 &=& 8-8\cdot \left( -\dfrac{\sqrt{2}}{2}\right) \\ \mathbf{s^2} &=& \mathbf{8+4\cdot\sqrt{2}} \\ \mathbf{s^2} &=& \mathbf{13.6568542495\ cm^2} \\ \hline \end{array}$$

The area of square $$ACEG$$ is $$\mathbf{\approx 13.7\ cm^2}$$ Aug 2, 2019
#2
+5 Because the exterior angle of an octagon is 45°, we can extend sides  BA  and  GH  to  point  J  to create 45-45-90 triangle AJH In △AJH, the side across from the 90° angle  =  AH  =  2 the sides across from the 45° angles  =  AJ  =  JH  =  2 / √2  =  √2 By the Pythagorean Theorem, AG2  =  AJ2 + JG2 AG2  =  AJ2 + (JH + HG)2 AG2  =  (√2)2 + (√2 + 2)2 AG2  =  2 + 2 + 4√2 + 4 AG2  =  8 + 4√2 AG2  ≈  13.657 That is in square cm.
Aug 2, 2019
#3
+2

Using the Law of Sines....we have that

(1/2) side length of  the square  / sin (67.5°)   = 2 / sin (90°)          .....multiply both sides by 2.....

side length of the square =  4 sin (67.5°)  =  4 √ [  (1  - cos (135°)  ) / 2 ] =  4 √  [ (1 + √2 /2) / 2 ]  =  4 √  [ (2 + √2)/4 ]

So....the area of the square is the square of this  =   16 (2 + √2)/4  =  4 (2 + √2) units^2   Aug 2, 2019