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Fifteen unit circles are inscribed in an equilateral triangle in such a way that each circle is externally tangent to its neighbors. What is the area of the triangle?

 

 Jun 22, 2020
 #1
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A = 14.22724134 units squared

 Jun 22, 2020
 #2
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Fifteen unit circles are inscribed in an equilateral triangle in such a way that
each circle is externally tangent to its neighbors.
What is the area of the triangle?

 

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

\(\begin{array}{|rcll|} \hline A &=&\rho * s \quad | \quad s=\dfrac{a+a+a}{2},\ \rho = 1 \\ A &=&1 *\dfrac{3a}{2} \\ A &=& \dfrac{3a}{2} \\ && \boxed{ 2A = a^2*\sin{60^\circ},\ \sin{60^\circ} = \dfrac{\sqrt{3}}{2} \\ 2A = a^2*\dfrac{\sqrt{3}}{2} \\ A = a^2*\dfrac{\sqrt{3}}{4} }\\ a^2*\dfrac{\sqrt{3}}{4} &=& \dfrac{3a}{2} \\ a*\dfrac{\sqrt{3}}{4} &=& \dfrac{3}{2} \\ a &=& \dfrac{3}{2} *\dfrac{4}{\sqrt{3}} \\ a &=& \dfrac{6}{\sqrt{3}} * \dfrac{\sqrt{3}}{\sqrt{3}} \\ a &=& \dfrac{6\sqrt{3}}{3} \\ \mathbf{a} &=& \mathbf{2\sqrt{3}} \\ \hline \end{array}\)

 

The area of the triangle:

\(\begin{array}{|rcll|} \hline 2A &=& \left(8+2\sqrt{3} \right)^2 \sin{60^\circ} \quad | \quad \sin{60^\circ} = \dfrac{\sqrt{3}}{2} \\ 2A &=& \left(8+2\sqrt{3} \right)^2 \dfrac{\sqrt{3}}{2} \\ 4A &=& \sqrt{3}\left(8+2\sqrt{3} \right)^2 \\ 4A &=& \sqrt{3}(64+32\sqrt{3}+12) \\ 4A &=& \sqrt{3}(76+32\sqrt{3}) \\ 4A &=& 96+76\sqrt{3} \quad | \quad : 4 \\ \mathbf{A} &=& \mathbf{24+19\sqrt{3}} \\ \mathbf{A} &=& \mathbf{56.9089653438} \\ \hline \end{array}\)

 

The area of the triangle is \(\approx \mathbf{56.9}\)

 

laugh

 Jun 22, 2020

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