varunmaheshwari wrote:

Attachment:

InscribedTraingles.JPG

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 (i.e. the 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)

PS: I am sorry for the formatting of the answer choices like this.

We can arrive at the answer quickly by estimating. The outermost triangle is split into 4 equal triangles 3 of which are completely shaded. Hence the fraction shaded is more than 3/4. So we are probably looking at options (C), (D) and (E).

We are left with the center triangle which occupies 1/4th of the area. 3/4th of this 1/4th is not shaded while a part of 1/4th of 1/4th is shaded. How much of this is completely shaded? 3/4th since the 1/4th of 1/4th triangle is also split into 4 equal areas and 3 of those areas are shaded. Hence the area of the shaded region would be something like this:

3/4 + (3/4)*(1/4)*(1/4) + ... = (3/4)(1 + 1/16 + ...)

We see that this matches option (C) which is 3/4(1 + 1/16 + ...)

Answer (C)

Note that none of the other options can give an area close to this one.

D. 3\4(2^0 + 2^-2 + 2^-4 + 2^-6) = (3\4)(1 + 1/4 + ...) Too much

E. 3\4(2^0 + 2^-1 + 2^-2 + 2^-3) = (3\4)(1 + 1/2 + ...) Too much again

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Karishma

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