A
B
C
D
E
Consider the following case:The figure is made up of a series of inscribed equilateral triangles.If the pattern continues until the length of a side of the largest triangle( entire figure) is exactly 128 times that of the smallest triangle, what fraction of the total figure will be shaded.
A. (1\4)(2^0 + 2^-4 + 2^-8 + 2^-12)
B. (1\4)(2^0 + 2^-2 + 2^-4 + 2^-6)
C. (3\4)(2^0 + 2^-4 + 2^-8 + 2^-12)
D. (3\4)(2^0 + 2^-2 + 2^-4 + 2^-6)
E. (3\4)(2^0 + 2^-1 + 2^-2 + 2^-3)
Thank You Mitch. This explanation was very easy to understand.GMATGuruNY wrote:Here is the complete problem, with answer choices:
Consider the following case:The figure is made up of a series of inscribed equilateral triangles.If the pattern continues until the length of a side of the largest triangle( entire figure) is exactly 128 times that of the smallest triangle, what fraction of the total figure will be shaded.
A. (1\4)(2^0 + 2^-4 + 2^-8 + 2^-12)
B. (1\4)(2^0 + 2^-2 + 2^-4 + 2^-6)
C. (3\4)(2^0 + 2^-4 + 2^-8 + 2^-12)
D. (3\4)(2^0 + 2^-2 + 2^-4 + 2^-6)
E. (3\4)(2^0 + 2^-1 + 2^-2 + 2^-3)
In the figure above, each side of the large outer triangle (2) is DOUBLE each side of the four small interior triangles (1).
Implication:
As the triangles increase in size, each successive triangle must have a side that is DOUBLE that of the next smallest triangle.
Thus, from smallest to biggest, the lengths of the sides must be proportioned as follows:
2�, 2¹, 2². 2³, 2�, 2�, 2�, 2�.
Because all of the triangles have the same combination of angles -- 60, 60, 60 -- they are all SIMILAR.
RULE:
If two triangles are similar, and each side of the larger triangle is X TIMES the corresponding side in the smaller triangle, then the AREA of the larger triangle is X² TIMES the area of the smaller triangle.
In the problem here, since the sides from smallest to biggest keep DOUBLING, the areas from smallest to biggest must be QUADRUPLING.
Thus, from smallest to biggest, the areas must be proportioned as follows:
2�, 2², 2�, 2�, 2�, 2¹�, 2¹², 2¹�.
The figure above indicates that -- for each successive pair of triangles -- 3/4 of the larger triangle will be shaded.
The blue values in the list above represent the larger triangle in each successive pair.
Since 3/4 of these areas will be shaded, we get:
Total shaded = (3/4)(2² + 2� + 2¹� + 2¹�).
The total AREA is equal to the area of the largest triangle, represented by the greatest blue value (2¹�).
Thus:
(total shaded)/(total area) = [(3/4)(2² + 2� + 2¹� + 2¹�)]/2¹� = (3/4)(2¯¹² + 2¯� + 2¯� + 2�).
The correct answer is C.
