hard area question

Question

A square,s1, has a perimeter of 40 inches. The vertices of a second square,s2, are the midpoints of he sides of s1. The vertices of a third square,s3, are the midpoints of the sides of s2. Assume the process continues indefinitely with the vertices of s(k+1) being the midpoints of the sides of sk for every positive integer k. What is the sum of the areas,in square inches, of s1,s2,s3?

 

A)40/3

B)20

C)70

D)400/3

E)200

 

 

I have no idea how to do this. Shouldn't the answer be infinity because you have infinitely small squares. lol this is me going back to ap calc and chemistry. No but on a serious note shouldn't the answer be infinity??? ANY HELP WILL BE APPRECIATED!!!!!

Answer 1

The sum would only be infinity if the sides kept getting bigger (or stayed the same size). But they get smaller. Any one side of S2 forms a 45-45-right triangle with 2 half sides of S1. So the length of a side of S2 must be 5√2. In the same way, the length of any side of S3 must be (5√2)/2*√2 = 5. Then an S4 side would be 5/2*√2, and so on. So each successive square S(k+1) has a side that is √2/2 *(length of Sk).

 

AREA = (side length)², so we have a geometric series, where r = (√2/2)² = 1/2 and the first term a = 10² = 100

 

Area sum = a/(1 - r) = 100/(1 - 1/2) = 200 ANSWER IS E