Question-189790

Question Number 189790 by normans last updated on 21/Mar/23

Answered by mr W last updated on 22/Mar/23

Commented by mr W last updated on 22/Mar/23

B E = a \tan α

B D = \sqrt{3} a

\Rightarrow E D = (\sqrt{3} - \tan α) a

\frac{F D}{A D} = \frac{\sin (60 ° - α)}{\sin (60 ° + α)} = \frac{\sqrt{3} - \tan α}{\sqrt{3} + \tan α}

\Rightarrow F D = \frac{2 a (\sqrt{3} - \tan α)}{\sqrt{3} + \tan α}

A_{1} = \frac{a^{2} \tan α}{2}

\Rightarrow \tan α = \frac{2 A_{1}}{a^{2}}

A_{2} = \frac{F D \times E D \sin 30 °}{2} = \frac{a^{2} {(\sqrt{3} - \tan α)}^{2}}{2 (\sqrt{3} + \tan α)}

A_{2} = \frac{a^{2} {(\sqrt{3} - \frac{2 A_{1}}{a^{2}})}^{2}}{2 (\sqrt{3} + \frac{2 A_{1}}{a^{2}})}

3 a^{4} - 2 \sqrt{3} (2 A_{1} + A_{2}) a^{2} + 4 A_{1} (A_{1} - A_{2}) = 0

\Rightarrow a^{2} = \frac{2 A_{1} + A_{2} + \sqrt{A_{2} (8 A_{1} + A_{2})}}{\sqrt{3}}

\Rightarrow a^{2} = \frac{2 \times 15 + 8 + \sqrt{8 (8 \times 15 + 8)}}{\sqrt{3}} = \frac{70}{\sqrt{3}}

a r e a o f h e x a g o n :

6 \times \frac{\sqrt{3} a^{2}}{4} = \frac{6 \sqrt{3}}{4} \times \frac{70}{\sqrt{3}} = 105 ✓

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