anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
a)1 / 6 b)1 / 9 c)0.1 d)0.09
imqwerty
  • imqwerty
where is the picture ( ͡° ͜ʖ ͡°)
anonymous
  • anonymous
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anonymous
  • anonymous
@triciaal
triciaal
  • triciaal
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
triciaal
  • triciaal
area of hexagon length 1/3 to area of hexagon length 1
anonymous
  • anonymous
I got 0.1 i saw that 9 triangles make up the 1 triangle so i did 1+9=10 1/10=0.1
anonymous
  • anonymous
I'm not sure if that's the right answer though
anonymous
  • anonymous
what do you think @triciaal @Hero
triciaal
  • triciaal
think 1/9
anonymous
  • anonymous
ok thanks
triciaal
  • triciaal
@Hero no comments?
triciaal
  • triciaal
@hero would it be 1/9 or 1/10?
Hero
  • Hero
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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