Question Number 179811 by Acem last updated on 02/Nov/22
Commented by Acem last updated on 02/Nov/22
$$\:\ast\:\ell\:{is}\:{an}\:{arc} \\ $$$$\:\ast\:{we}\:{can}\:{get}\:\mathrm{2}\:{equilateral}\:{triangles}\:{formed}\:{by} \\ $$$$\:\:\:\:\:{it}\:{vertices} \\ $$$$ \\ $$
Commented by Acem last updated on 02/Nov/22
$${Find}\:{the}\:{area}\:{of}\:{the}\:{equilateral}\:{triangles} \\ $$$$\:{formed}\:{by}\:{the}\:{vertices}\:{of}\:{this}\:{hexagon} \\ $$
Commented by mr W last updated on 02/Nov/22
Commented by mr W last updated on 02/Nov/22
$${area}\:{of}\:{equilateral}\:\left({red}\right)\:{is}\:{half}\:{of} \\ $$$${the}\:{area}\:{of}\:{the}\:{hexagon}: \\ $$$$\frac{\mathrm{6}}{\mathrm{2}}×\frac{\sqrt{\mathrm{3}}{l}^{\mathrm{2}} }{\mathrm{4}}=\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{4}}\left(\frac{\pi}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} =\frac{\sqrt{\mathrm{3}}\pi^{\mathrm{2}} }{\mathrm{4}} \\ $$
Answered by Acem last updated on 05/Nov/22
$${We}\:{will}\:{solve}\:{it}\:{in}\:{two}\:{cases}:\: \\ $$$$\left.{A}\right)\:\mathrm{2}\:{seperated}\:{right}\:{triangles} \\ $$$$\left.{B}\right)\:{Area}\:{of}\:{union}\:{of}\:\mathrm{2}\:{triangles}\:“{hexagonal}\:{star}'' \\ $$$${L}_{{H}} \:=\:{r}\:=\:\frac{\ell}{\theta}\:;\:\theta=\:\frac{\pi}{\mathrm{3}}\:\Rightarrow\:{L}_{{H}} =\:\sqrt{\mathrm{3}} \\ $$$$\boldsymbol{{A}}\bullet\:{Area}_{\mathrm{2}\:{triang}.} \:=\:{Area}_{{Hexag}.} \:=\:\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}\:{L}_{{H}} ^{\mathrm{2}} \:=\:\frac{\mathrm{9}\sqrt{\mathrm{3}}}{\mathrm{2}}\:{un}^{\mathrm{2}} \\ $$$$ \\ $$$$\boldsymbol{{B}}\bullet\:{Area}_{{hex}.\:{star}} =\:{Area}_{\mathrm{2}\:{triang}.} −\:{Area}_{{Small}\:{hexagon}} …\left(\mathrm{1}\right) \\ $$$$\:;\:{Small}\:{hexagon}\:=\:{common}\:{area}\:{of}\:\mathrm{2}\:{right}\:{trian}. \\ $$$$\: \\ $$$$\:{L}_{{small}\:{hex}.} =\:\frac{\mathrm{1}}{\mathrm{3}}\:{L}_{{H}} \:\Rightarrow\:{Area}_{{sm}.\:{hex}} =\:\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}\:{un}^{\mathrm{2}} \\ $$$$\:\Rightarrow\:{Area}_{{hex}.\:{star}} =\:\mathrm{3}\:\sqrt{\mathrm{3}}\:{un}^{\mathrm{2}} \\ $$$$\: \\ $$$$\boldsymbol{{Notes}}: \\ $$$$\mathrm{1}\bullet\:{Area}_{{Hexagon}} =\:\mathrm{2}×\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{L}+{L}\right){L}\:\mathrm{cos}\:\mathrm{30}=\:\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}\:{L}^{\mathrm{2}} \\ $$$$\mathrm{2}\bullet\:{Side}\:{length}_{{equilat}.\:{triang}} =\:\mathrm{2}×{L}\:\mathrm{cos}\:\mathrm{30}=\:\sqrt{\mathrm{3}}\:{L} \\ $$$$\mathrm{3}\bullet\:{Area}_{{equilat}.\:{triang}} =\:\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{3}{L}^{\mathrm{2}} \:\mathrm{sin}\:\mathrm{60}=\:\frac{\mathrm{3}\:\sqrt{\mathrm{3}}}{\mathrm{4}}\:{L}^{\mathrm{2}} \\ $$