varunmaheshwari wrote:
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. \(\frac{1}{4}(2^0 + 2^{-4} + 2^{-8} + 2^{-12})\)
B. \(\frac{1}{4}(2^0 + 2^{-2} + 2^{-4} + 2^{-6})\)
C. \(\frac{3}{4}(2^0 + 2^{-4} + 2^{-8} + 2^{-12})\)
D. \(\frac{3}{4}(2^0 + 2^{-2} + 2^{-4} + 2^{-6})\)
E. \(\frac{3}{4}(2^0 + 2^{-1} + 2^{-2} + 2^{-3})\)
Attachment:
InscribedTraingles.JPG
Though i marked the wrong answer, Lets try to solve easy way..
1/128 = 1/2^7 . So center triangle will go upto 8th triangle in the center and will be unshaded.
Let A be the area of bigger triangle. Lets number biggest triangle as 1 and then smaller center triangles as 2 and next smaller center triangle as 3 and so on.
So, First focusing on 1st bigger triangle no. 1..
Shaded region = 3/4 A
Now , In 2nd smaller triangle no. 2 , we have only shaded region from 1/4th of the part i.e from triangle no.3. So we will consider it in the next calculation in triangle no.3
In triangle 3,
Shaded region = 3/4 * triangle 3 = 3/4 *(1/4)^2 A
In triangle no. 4 , we have only shaded region from 1/4th of the part i.e from triangle no.5. So we will consider it in the next calculation in triangle no.5
In triangle 5,
Shaded region = 3/4 * triangle 5 = 3/4* 1/4 *1/4 * triangle 3 = 3/4 *(1/4)^2 * (1/4)^2 A
In triangle 6, we have only shaded region from 1/4th of the part i.e from triangle no.7. So we will consider it in the next calculation in triangle no.7
Shaded region = 3/4 * (1/4)^2 * (1/4)^2 * (1/4)^2 A
Sum of shaded regions = 3/4 A + 3/4 *(1/4)^2 A + 3/4 *(1/4)^2 * (1/4)^2 A + 3/4 * (1/4)^2 * (1/4)^2 * (1/4)^2 A
= 3/4 (1/2^0 + 1/2^4 + 1/2^8 + 1/2^12) A
Answer C