The picture shows six equilateral triangles. The sides of the triangles are three times longer than are the side of the regular hexagon. What fraction of the whole shape is blue?

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The picture shows six equilateral triangles. The sides of the triangles are three times longer than are the side of the regular hexagon. What fraction of the whole shape is blue?

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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a)1 / 6 b)1 / 9 c)0.1 d)0.09
where is the picture ( ͡° ͜ʖ ͡°)
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challenging might be the area of the blue hexagon to the area of (6*area of triangle) the 6 triangles will form a regular red hexagon
area of hexagon length 1/3 to area of hexagon length 1
I got 0.1 i saw that 9 triangles make up the 1 triangle so i did 1+9=10 1/10=0.1
I'm not sure if that's the right answer though
what do you think @triciaal @Hero
think 1/9
ok thanks
@Hero no comments?
@hero would it be 1/9 or 1/10?
There's a way to know for sure if you use actual numbers for the sides of the hexagon and triangle. Then use the formulas for area of a triangle and hexagon. Area of a regular hexagon = 6*Area of A Triangle Area of a regular hexagon = \(\dfrac{3\sqrt{3}a^2}{2}\) Area of a regular triangle = \(\dfrac{\sqrt{3}b^2}{4}\) In this case b = 3a so: Area of a regular triangle = \(\dfrac{\sqrt{3}(3a)^2}{4}\) = \(\dfrac{\sqrt{3}9a^2}{4}\) But there are six triangles, so the total area of triangles in this case is: \(6*\dfrac{\sqrt{3}*9a^2}{4}\) \(\dfrac{\text{Area of Blue Hex}}{\text{Area of Red Triangle}} = \dfrac{\dfrac{3\sqrt{3}a^2}{2}}{\dfrac{\sqrt{3}*54a^2}{4}} = \dfrac{1}{9}\)

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