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.

\(Area \hspace{3} of \hspace{3} equilateral \hspace{3} triangle = \frac{\sqrt{3}}{4}*(side)^2\)

\(Total \hspace{3} Area = \frac{\sqrt{3}}{4}(128)^2=\frac{\sqrt{3}}{4}(2^7)^2=\frac{\sqrt{3}}{4}*2^{14}\)

\(Shaded \hspace{3} portion \hspace{3} of \hspace{3} outer \hspace{3} most= \frac{3}{4}*\frac{\sqrt{3}}{4}*(2^7)^2\) {:Side=128=2^7}

Skip the next: {:Side=64}

\(Shaded \hspace{3} portion \hspace{3} of 3^{rd}= \frac{3}{4}*\frac{\sqrt{3}}{4}*(2^5)^2\) {:Side=32=2^5}

\(Shaded \hspace{3} portion \hspace{3} of 5^{th}= \frac{3}{4}*\frac{\sqrt{3}}{4}*(2^3)^2\) {:Side=8=2^3}

\(Shaded \hspace{3} portion \hspace{3} of 7^{th}= \frac{3}{4}*\frac{\sqrt{3}}{4}*(2^1)^2\) {:Side=2=2^1}

Take the fraction:

\(\frac{\frac{3}{4}*\frac{\sqrt{3}}{4}(2^{14}+2^{10}+2^{6}+2^{2})}{\frac{\sqrt{3}}{4}*2^{14}}\)

\(=\frac{\frac{3}{4}(2^{14}+2^{10}+2^{6}+2^{2})}{2^{14}}\)

\(=\frac{3}{4}(\frac{2^{14}}{2^{14}}+\frac{2^{10}}{2^{14}}+\frac{2^{6}}{2^{14}}+\frac{2^{2}}{2^{14}})\)

\(=\frac{3}{4}(2^{(14-14)}+2^{(10-14)}+2^{(6-14)}+2^{(2-14)})\)

\(=\frac{3}{4}(2^0+2^{-4}+2^{-8}+2^{-12})\)

Ans: "C"

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~fluke

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