6.Each side of a regular hexagon is extended by a length equal to its own length. The end points of the new segments are joined to form a new and larger regular hexagon. What fraction of the area of the bigger hexagon does the smaller hexagon occupy? (Hint: the area of an equilateral triangle is )
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You can see that te triangle ABO is an equilateral triangle, so: OA = OB = AB = 1
By using the Pythagorean's theorem, you can see that:
(AB/2)² + h² = OA²
h² = OA² - (AB/2)²
h² = 1 - (1/2)²
h² = 3/4
h = (√3)/2
The surface of the troiangle AbO is:
s₁ = (base * height) / 2 → where: base = AB
s₁ = (AB * height) / 2 → where: height = h
s₁ = (AB * h) / 2 → given that: AB = 1
s₁ = h/2 → recall: h = (√3)/2
s₁ = (√3)/4
The total surface area of the smaller hexagon is:
s₂ = 6.s₁
s₂ = 6 * (√3)/4
s₂ = (3√3)/2
The surface area of the triangle RST is:
s₃ = (base * height) / 2 → where: base = ST
s₃ = (ST * height) / 2 → where: height = RS
s₃ = (ST * RS) / 2 → where: RS = OA = 1
s₃ = ST/2 → where: ST = RT.cos(30)
s₃ = [RT.cos(30)]/2 → where: RT = RA + AT = 1 + 1 = 2
s₃ = [2.cos(30)]/2
s₃ = cos(30)
s₃ = (√3)/2
The total surface area of the larger hexagon is:
s₄ = 6.s₃ + s₂ → we've just seen that: s₃ = (√3)/2
s₄ = [6 * (√3)/2] + s₂ → recall: s₂ = (3√3)/2
s₄ = [6 * (√3)/2] + [(3√3)/2]
s₄ = [(6√3) + (3√3)]/2
s₄ = (9√3)/2
The ratio is:
r = s₄/s₂
r = [(9√3)/2] / [(3√3)/2]
r = (9√3)/(3√3)
r = 9/3
r = 3
With minimal cutting and pasting, you can fit exactly 3 of those large right triangles into the original hexagon. So the 6 large right triangles occupy exactly twice the area of the original hexagon. Therefore, the original hexagon occupies exactly 1/3 of the area of the new, large hexagon.
If you don't like my first sentence, here's some help with it. Flip the "top" right triangle down so that its 1-meter side is the 1-m side at the left of the original hexagon. Flip the "bottom" right triangle up so that its 1-m side is the 1-m side at the right of the original hexagon. The two shallow isosceles triangles that remain uncovered in the original hexagon each consist of two SMALL right triangles, and these four small right right triangles can be rearranged into a rectangle exactly as wide as the hexagon, but only 1/2 of a meter high (so exactly half the area of the central rectangle that got covered by the two "flips").