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anonymous
 one year ago
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?
anonymous
 one year ago
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0a)1 / 6 b)1 / 9 c)0.1 d)0.09

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.0where is the picture ( ͡° ͜ʖ ͡°)

triciaal
 one year ago
Best ResponseYou've already chosen the best response.1challenging 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

triciaal
 one year ago
Best ResponseYou've already chosen the best response.1area of hexagon length 1/3 to area of hexagon length 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got 0.1 i saw that 9 triangles make up the 1 triangle so i did 1+9=10 1/10=0.1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm not sure if that's the right answer though

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what do you think @triciaal @Hero

triciaal
 one year ago
Best ResponseYou've already chosen the best response.1@hero would it be 1/9 or 1/10?

Hero
 one year ago
Best ResponseYou've already chosen the best response.0There'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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