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union can be a star

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union can be a star

by sanju09 » Mon Mar 07, 2011 1:19 am
When two congruent equilateral triangles share a common center, then their union can be a star as shown. If their overlap is a regular hexagon of area 60, then what is the area of one of the original equilateral triangles?
(A) 10
(B) 15
(C) 30
(D) 60
(E) 90
The mind is everything. What you think you become. -Lord Buddha



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by Haldiram Bhujiawala » Mon Mar 07, 2011 1:54 am
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by Anurag@Gurome » Mon Mar 07, 2011 2:23 am
sanju09 wrote:When two congruent equilateral triangles share a common center, then their union can be a star as shown. If their overlap is a regular hexagon of area 60, then what is the area of one of the original equilateral triangles?
(A) 10
(B) 15
(C) 30
(D) 60
(E) 90
Easiest Solution:
Area of any of the original triangle must be greater than the overlapping region. Only such option is 90, i.e. option E.

Methodical Tricky Solution:
Join the center of the triangles (also of the hexagon) with the vertices of the hexagon. This will result 6 small equilateral triangles inside the hexagon, each of equal are as shown in the following figure.

Image

Now, each of these six small equilateral triangles are congruent to each of the six equilateral triangles outside the hexagon.

Now, area of hexagon = 6*(Area of an small equilateral triangle) = 60
=> Area of an small equilateral triangle = 60/6 = 10

Area of one of the original equilateral triangle = 9*(Area of an equilateral small triangle) = 90

Correct answer is E.
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by sanju09 » Mon Mar 07, 2011 3:32 am
Anurag@Gurome wrote:
sanju09 wrote:When two congruent equilateral triangles share a common center, then their union can be a star as shown. If their overlap is a regular hexagon of area 60, then what is the area of one of the original equilateral triangles?
(A) 10
(B) 15
(C) 30
(D) 60
(E) 90
Easiest Solution:
Area of any of the original triangle must be greater than the overlapping region. Only such option is 90, i.e. option E.

Methodical Tricky Solution:
Join the center of the triangles (also of the hexagon) with the vertices of the hexagon. This will result 6 small equilateral triangles inside the hexagon, each of equal are as shown in the following figure.

Image

Now, each of these six small equilateral triangles are congruent to each of the six equilateral triangles outside the hexagon.

Now, area of hexagon = 6*(Area of an small equilateral triangle) = 60
=> Area of an small equilateral triangle = 60/6 = 10

Area of one of the original equilateral triangle = 9*(Area of an equilateral small triangle) = 90

Correct answer is E.
your easiest solution is much appreciated because it also serves the purpose of the answer choices for why they are so made.
The mind is everything. What you think you become. -Lord Buddha



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by Haldiram Bhujiawala » Mon Mar 07, 2011 11:47 am
Unfortunately I got the diagram wrong.
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