+26367 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 \(\mathbf{\approx 13.7\ cm^2}\)
+9466 
| 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 |
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| the sides across from the 45° angles = AJ = JH = 2 / √2 = √2 | |
| By the Pythagorean Theorem, |
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| AG2 = AJ2 + JG2 |
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| AG2 = AJ2 + (JH + HG)2 |
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| AG2 = (√2)2 + (√2 + 2)2 |
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| AG2 = 2 + 2 + 4√2 + 4 |
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| AG2 = 8 + 4√2 |
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| AG2 ≈ 13.657 |
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| That is in square cm. |
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+128474 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
