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Each side of a regular dodecagon A1 A2 A3 ... A12 has length 2. What is the area of the pentagon A1 A2 A3 A4 A5?

 Nov 27, 2019
 #1
avatar+23575 
+2

Each side of a regular dodecagon A1 A2 A3 ... A12 has length 2. What is the area of the pentagon A1 A2 A3 A4 A5?

 

\(\text{Let side of a regular dodecagon $=s $} \\ \text{Let radius of the circle $=r $} \\ \text{Let area of the dodecagon $=A_{\text{dodecagon}} $} \\ \text{Let area of the pentagon $=A_{\text{pentagon}} $} \\ \text{Let area of the triangle in the center $=A_{\triangle} $} \)

 

\(\begin{array}{|rcll|} \hline A_{\text{dodecagon}} &=& 3\times A_{\text{pentagon}} + A_{\triangle} \\ \mathbf{A_{\text{pentagon}}} &=& \mathbf{\dfrac{1}{3} \left( A_{\text{dodecagon}} - A_{\triangle}\right)} \quad | \quad \boxed{A_{\text{dodecagon}}= 3(2+\sqrt{3})s^2} \\ \hline \end{array} \)

 

\(\mathbf{r=\ ?}\)

\(\begin{array}{|rcll|} \hline \sin(15^\circ) &=& \dfrac{\dfrac{s}{2}}{r} \\ 2r &=& \dfrac{s}{\sin(15^\circ)} \quad | \quad \sin(15^\circ) = \dfrac{1}{4}\sqrt{2}(\sqrt{3}-1) \\ \mathbf{2r} &=& \mathbf{\dfrac{4s}{\sqrt{2}(\sqrt{3}-1)}} \\ \hline \end{array} \)

 

\(\mathbf{a=\ ?}\)

\(\begin{array}{|rcll|} \hline \sin(60^\circ) &=& \dfrac{\dfrac{a}{2}}{r} \\ a &=& 2r\sin(60^\circ) \quad | \quad \sin(60^\circ) = \dfrac{\sqrt{3}}{2} \\ a &=& 2r\dfrac{\sqrt{3}}{2} \\ \mathbf{a} &=& \mathbf{\dfrac{4s}{\sqrt{2}(\sqrt{3}-1)}\dfrac{\sqrt{3}}{2}} \\ \hline \end{array}\)

 

\(\mathbf{h=\ ?}\)

\(\begin{array}{|rcll|} \hline \cos(30^\circ) &=& \dfrac{h}{a} \\ h &=& a\cos(30^\circ) \quad | \quad \cos(30^\circ) = \dfrac{\sqrt{3}}{2} \\ h &=& a\dfrac{\sqrt{3}}{2} \\ h &=& \dfrac{4s}{\sqrt{2}(\sqrt{3}-1)}\dfrac{\sqrt{3}}{2}\dfrac{\sqrt{3}}{2} \\ h &=& \dfrac{4s}{\sqrt{2}(\sqrt{3}-1)}\dfrac{3}{4} \\ \mathbf{h} &=& \mathbf{\dfrac{3s}{\sqrt{2}(\sqrt{3}-1)}} \\ \hline \end{array}\)

 

\(\mathbf{A_{\triangle}=\ ?}\)

\(\begin{array}{|rcll|} \hline A_{\triangle} &=& \dfrac{ah}{2} \\ &=& \dfrac{1}{2}\dfrac{4s}{\sqrt{2}(\sqrt{3}-1)}\dfrac{\sqrt{3}}{2}\dfrac{3s}{\sqrt{2}(\sqrt{3}-1)} \\ &=& \dfrac{3\sqrt{3}}{2\left(\sqrt{3}-1 \right)^2}s^2 \quad | \quad \left(\sqrt{3}-1 \right)^2 =2(2-\sqrt{3}) \\ &=& \dfrac{3\sqrt{3}}{2*2(2-\sqrt{3})}s^2 \\ \mathbf{A_{\triangle}} &=& \mathbf{ \dfrac{3\sqrt{3}}{4(2-\sqrt{3})}s^2 } \\ \hline \end{array} \)

 

\(\mathbf{A_{\text{pentagon}}=\ ?}\)

\(\begin{array}{|rcll|} \hline \mathbf{A_{\text{pentagon}}} &=& \mathbf{\dfrac{1}{3} \left( A_{\text{dodecagon}} - A_{\triangle}\right)} \quad | \quad \boxed{A_{\text{dodecagon}}= 3(2+\sqrt{3})s^2} \\\\ &=& \dfrac{1}{3} \left( 3(2+\sqrt{3})s^2 - \dfrac{3\sqrt{3}}{4(2-\sqrt{3})}s^2\right) \\ \\ &=& \dfrac{3}{3} \left((2+\sqrt{3}) - \dfrac{\sqrt{3}}{4(2-\sqrt{3})} \right)s^2 \\ \\ &=& \left(\dfrac{4(2-\sqrt{3})(2+\sqrt{3})-\sqrt{3}}{4(2-\sqrt{3})} \right)s^2 \\ \\ &=& \left(\dfrac{4(4-3)-\sqrt{3}}{4(2-\sqrt{3})} \right)s^2 \\ \\ &=& \left(\dfrac{4 -\sqrt{3}}{4(2-\sqrt{3})} \right)s^2 \\ \\ &=& \left(\dfrac{4 -\sqrt{3}}{4(2-\sqrt{3})}\right)\left(\dfrac{2+\sqrt{3}}{2+\sqrt{3}} \right)s^2 \\ \\ &=& \dfrac{ (4 -\sqrt{3}) (2+\sqrt{3}) }{4(2-\sqrt{3})(2+\sqrt{3})} s^2 \\ \\ &=& \dfrac{ (5+2\sqrt{3}) }{4(4-3)} s^2 \\ \\ \mathbf{A_{\text{pentagon}}}&=& \mathbf{ \dfrac{ (5+2\sqrt{3}) }{4 } s^2 } \quad | \quad s = 2 \\ \mathbf{A_{\text{pentagon}}}&=& \mathbf{ 5+2\sqrt{3} } \\ \hline \end{array}\)

 

laugh

 Nov 28, 2019
 #2
avatar+105989 
+1

That looks very impressive Heureka !

Melody  Nov 28, 2019
 #3
avatar+23575 
+2

Thank you, Melody !

 

laugh

heureka  Nov 29, 2019

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