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Geometry - Infinite Figures Inscribed

Sometimes, we come across GMAT geometry questions that involve figures inscribed inside other figures. One shape inside of another shape may not be difficult to work with, but how do we handle problems that involve infinite figures inscribed inside one another? Such questions can unsettle even the most seasoned test takers. Let’s take a look at one of them today:


A square is drawn by joining the midpoints of the sides of a given square. A third square is drawn inside the second square in this way and this process is continued indefinitely. If a side of the first square is 4 cm. Determine the sum of areas of all squares?

(A) 18
(B) 32
(C) 36
(D) 64
(E) None


Now the first thing that might come to our mind is this – how do we mathematically, in the time limit of approximately 2 minutes, calculate areas of infinite squares?

There has to be a formula for this. Recall that we do, in fact, have a formula that calculates the sum of infinite terms – the geometric progression formula! Let’s see if we can use that to find the areas of the squares mentioned in this problem.

First, we’ll see if we can find a pattern in the areas of the squares:


Say the side of the outermost square is “s“. The area of the outermost square will be s^2 and half of the side will be s/2. The side of the next square inside this outermost square (the second square) forms the hypotenuse of a right triangle with legs of length s/2 each. Using the Pythagorean Theorem:


Hypotenuse^2 = (s/2)^2 + (s/2)^2 = s^2/2
Hypotenuse = s/√(2)


So now we know the sides of the second square will each equal s/√(2), and the area of the second square will be s^2/2.


Our calculations will be far easier if we note that the diagonal of the second square will be the same length as the side of the outer square. We know that area of a square given diagonal d is d^2/2, so that would directly bring us to s^2/2 as the area of the second square.

The second square and the square inscribed further inside it (the third square) will have the same relation. The area of the third square will be (s^2/2)*(1/2) = s^2/4.

Now we know the area of every subsequent square will be half the area of the outside square. So the total area of all squares = s^2 + s^2/2 + s^2/4 + s^2/8 + …Each term is half the previous term.


Therefore, the sum of an infinite Geometric Progression where the common ratio is less than 1 is:

Total Sum = a/(1-r)
a: First Term
r: Common Ratio


Sum of areas of all squares = s^2 + s^2/2 + s^2/4 + s^2/8 + …
Sum of areas of all squares = s^2/(1 – 1/2)
Sum of areas of all squares = 2s^2


Since s is the length of the side of the outermost square, and s = 4 (this fact is given to us by the questions stem), the sum of the areas of all the squares = 2*4^2 = 32 cm^2. Therefore, our answer is B.


We hope you understand how we have used the geometric progression formula to get to our answer. To recap, the sum of an infinite geometric progression is a/(1 – r).


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