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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?

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  1. anonymous
    • one year ago
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    a)1 / 6 b)1 / 9 c)0.1 d)0.09

  2. imqwerty
    • one year ago
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    where is the picture ( ͡° ͜ʖ ͡°)

  3. anonymous
    • one year ago
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  4. anonymous
    • one year ago
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    @triciaal

  5. triciaal
    • one year ago
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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

  6. triciaal
    • one year ago
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    area of hexagon length 1/3 to area of hexagon length 1

  7. anonymous
    • one year ago
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    I got 0.1 i saw that 9 triangles make up the 1 triangle so i did 1+9=10 1/10=0.1

  8. anonymous
    • one year ago
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    I'm not sure if that's the right answer though

  9. anonymous
    • one year ago
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    what do you think @triciaal @Hero

  10. triciaal
    • one year ago
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    think 1/9

  11. anonymous
    • one year ago
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    ok thanks

  12. triciaal
    • one year ago
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    @Hero no comments?

  13. triciaal
    • one year ago
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    @hero would it be 1/9 or 1/10?

  14. Hero
    • one year ago
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    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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