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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
avatar+23082 
+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 \(\mathbf{\approx 13.7\ cm^2}\)

 

laugh

 Aug 2, 2019
 #2
avatar+8725 
+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
avatar+103148 
+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

 

 

 

cool cool cool

 Aug 2, 2019

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