Question

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?

Answer 1

a)1 / 6
b)1 / 9
c)0.1
d)0.09

Answer 2

where is the picture ( ͡° ͜ʖ ͡°)

Answer 3

@triciaal

Answer 4

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

Answer 5

area of hexagon length 1/3 to area of hexagon length 1

Answer 6

I got 0.1
i saw that 9 triangles make up the 1 triangle
so i did
1+9=10
1/10=0.1

Answer 7

I'm not sure if that's the right answer though

Answer 8

what do you think @triciaal @Hero

Answer 9

think 1/9

Answer 10

ok thanks

Answer 11

@Hero no comments?

Answer 12

@hero would it be 1/9 or 1/10?

Answer 13

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:
\(6*\dfrac{\sqrt{3}*9a^2}{4}\)

\(\dfrac{\text{Area of Blue Hex}}{\text{Area of Red Triangle}} = \dfrac{\dfrac{3\sqrt{3}a^2}{2}}{\dfrac{\sqrt{3}*54a^2}{4}} = \dfrac{1}{9}\)