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This is from Australian Maths 2013.

In a regular hexagon,the midpoints of the sides are joined to form he shaded regular hexagon.What fraction of the larger hexagon is shaded?

Since the larger hexagon can be divided into 6 triangles. Let the height of the triangle be $h$ and base be $b$.

Area of larger hexagon=6 triangles= $6(\frac 12)(bh)=3bh$

Then,I know that since the midpoints of the sides are used to form another hexagon,there are another 6 triangles within the shaded hexagon.However I don't know the height and base,so I can't find the area.

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Hint: Divide the larger hexagon into $24$ congruent triangles as shown below:

Can you tell what fraction of the larger hexagon's area is taken up by the smaller hexagon?

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  • $\begingroup$ Yes.I can see the answer.But how do you know to divide into 3 congruent triangles for each bigger triangle? $\endgroup$ – Arc Neoepi Jul 9 '16 at 5:58
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HINT: Those triangles are equilateral, so $h=\frac{\sqrt3}2b$. The new triangles are also equilateral, and they have side $h$.

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  • $\begingroup$ So I use Pythagoras'theorem to find the height of the new triangles? $\endgroup$ – Arc Neoepi Jul 9 '16 at 5:20
  • $\begingroup$ @ArcNeoepi: You can, or you can use the fact that that their height is also $\frac{\sqrt3}2$ times their side. Or you can work directly from the fact that the smaller hexagon is similar to the original one, and you know the similarity factor. $\endgroup$ – Brian M. Scott Jul 9 '16 at 5:23
  • $\begingroup$ Thanks.I get it now!But how do you know they are equilateral.I thought they were isosceles triangles. $\endgroup$ – Arc Neoepi Jul 9 '16 at 5:25
  • $\begingroup$ @ArcNeoepi: There are six of them, so the angle at the centre of the hexagon is $\frac{360}{6}=60$ degrees. Since they are, as you say, isosceles, all three angles must be $60$ degrees. You're welcome! $\endgroup$ – Brian M. Scott Jul 9 '16 at 5:28

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