A regular hexagon is inscribed in an equilateral triangle, as shown. If the hexagon has an area of $12$ , find the area of this triangle.

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Solution

$△ A O B$ is an equilateral triangle
$∠ A O B = \frac{360}{60} = 60^{o}$ and $O A = O B$
Area of hexagon $= 6 \times$ area of OAB
$12 = 6 \times \frac{\sqrt{3}}{4} a^{2}$
$a^{2} = \frac{12 \times 4}{6 \sqrt{3}}$
$a = \sqrt{\frac{8}{\sqrt{3}}}$
Sides of triangle $= 3 a$
$= 3 \sqrt{\frac{8}{\sqrt{3}}}$
Area of triangle $= \frac{\sqrt{3}}{4} a^{2}$
$= \frac{\sqrt{3}}{4} \times 9 \times \frac{8}{\sqrt{3}}$
$= 18$ sq. units

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A regular hexagon is inscribed in an equilateral triangle, as shown. If the hexagon has an area of 12, find the area of this triangle.

△AOB is an equilateral triangle∠AOB=36060=60o and OA=OBArea of hexagon =6×area of OAB12=6×√34a2a2=12×46√3a=√8√3Sides of triangle =3a=3√8√3Area of triangle =√34a2=√34×9×8√3=18 sq. units

A regular hexagon is inscribed in an equilateral triangle, as shown. If the hexagon has an area of $12$ , find the area of this triangle.