# Solving GMAT Geometry Problems That Involve Infinite Figures

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).

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

# Quarter Wit, Quarter Wisdom: Beware of Sneaky Answer Choices on the GMAT!

Test-takers often ask for tips and short cuts to cut down the amount of work necessary to solve a GMAT problem. As such, the Testmaker might want to award the test-taker who pays attention to detail and puts in the required effort.

Today, we will look at an example of this concept – if it seems to be too easy, it is a trap!

In the figure given above, the area of the equilateral triangle is 48. If the other three figures are squares, what is the perimeter, approximately, of the nine-sided shape they form?

(A) 8√(2)
(B) 24√(3)
(C) 72√(2)
(D) 144√(2)
(E) 384

The first thing I notice about this question is that we have an equilateral triangle. So I am thinking, the area = s^2 * √(3)/4 and/or the altitude = s*√(3)/2.

The irrational number in play is √(3). There is only one answer choice with √(3) in it, so will this be the answer?

Now, it actually makes me uncomfortable that  there is only one option with √(3). At first glance, it seems that the answer has been served to us on a silver plate. But the question format doesn’t seem very easy – it links two geometrical figures together. So I doubt very much that the correct answer would be that obvious.

The next step will be to think a bit harder:

The area of the triangle has √(3) in it, so the side would be a further square root of √(3). This means the actual irrational number would be the fourth root of 3, but we don’t have any answer choice that has the fourth root of 3 in it.

Let’s go deeper now and actually solve the question.

The area of the equilateral triangle = Side^2 * (√(3)/4) = 48

Side^2 = 48*4/√(3)
Side^2 = 4*4*4*3/√(3)
Side = 8*FourthRoot(3)

Now note that the side of the equilateral triangle is the same length as the sides of the squares, too. Hence, all sides of the three squares will be of length 8*FourthRoot(3).

All nine sides of the figure are the sides of squares. Hence:

The perimeter of the nine sided figure = 9*8*FourthRoot(3)
The perimeter of the nine sided figure =72*FourthRoot(3)

Now look at the answer choices. We have an option that is 72√(2). The other answer choices are either much smaller or much greater than that.

Think about it – the fourth root of 3 = √(√(3)) = √(1.732), which is actually very similar to √(2). Number properties will help you figure this out. Squares of smaller numbers (that are still greater than 1) are only a bit larger than the numbers themselves. For example:

(1.1)^2 = 1.21
(1.2)^2 = 1.44
(1.3)^2 = 1.69
(1.414)^2 = 2

Since 1.732 is close to 1.69, the √(1.732) will be close to the √(1.69), i.e. 1.3. Also, √(2) = 1.414. The two values are quite close, therefore, the perimeter is approximately 72√(2). This is the reason the question specifically requests the “approximate” perimeter.

We hope you see how the Testmaker could sneak in a tempting answer choice – beware the “easiest” option!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

# Quarter Wit, Quarter Wisdom: Squares and Square Roots on the GMAT

In today’s post, we will try to clear up your doubts regarding positive and negative solutions in the case of squares and square roots. We will explain the reasons behind each case, which will help you recall the fundamentals when you need to use them. While preparing for the GMAT, you have probably come across a discussion that says x^2 = 4 has two roots, 2 and -2, while √4 has only one value, 2.

Now, let’s try to understand why this is so:

1) x^2 = 4
Basic algebra tells us that quadratics have two roots. Here, x can be either 2 or -2; each, when squared, will give you 4.

x^2 – 4 = 0 and (x + 2)*(x – 2) = 0 when x equals -2 or 2.

2) √x is positive, only
Now this is odd, right? √4 must be 2. Why is that? Shouldn’t it be 2 or -2. After all, when we square both 2 and -2, we get 4 (as discussed above). So, √4 should be 2 or -2.

Here is the concept: √x denotes only the principal square root. x has two square roots – the positive square root (or principal square root) written as √x and the negative square root written as -√x. Therefore, when you take the square root of 4, you get two roots: √4 and -√4, which  is 2 and -2 respectively.

On a GMAT question, when you see √x, this is specifically referring to the positive square root of the number. So √4 is 2, only.

3) (√x)^2 = x
This is fairly straightforward – since x has a square root, it must be non-negative. When you square it, just the square root sign vanishes and you are left with x.

4) √(x^2) = |x|
Now this isn’t intuitive either. √(x^2) should simply be x – why do we have absolute value of x, then? Again, this has to do with the principal square root concept. First you will square x, and then when you write √, it is by default just the principal square root. The negative square root will be written as -√(x^2). So, irrespective of whether x was positive or negative initially, √(x^2) will definitely be positive x. Therefore, we will need to take the absolute value of x.

Here’s a quick recap with some examples:

• √9 = 3
• x^2 = 16 means x is either 4 or -4
• √(5^2) = 5
• √(-5^2) = 5
• (√16)^2 = 16
• √100 = 10

To see this concept in action, let’s take a look at a very simple official problem:

If x is not 0, then √(x^2)/x =

(A) -1
(B) 0
(C) 1
(D) x
(E) |x|/x

We know that √(x^2) is not simply x, but rather |x|. So, √(x^2)/x = |x|/x.

Depending on whether x is positive or negative, |x|/x will be 1 or -1 – we can’t say which one. Hence, there is no further simplification that we can do, and our answer must be E.

Now that you are all warmed up, let’s examine a higher-level question:

Is √[(x – 3)^2] = (3 – x)?

Statement 1: x is not 3
Statement 2: -x * |x| > 0

We know that √(x^2) = |x|, so √[(x – 3)^2] = |x – 3|.

This means that our question is basically:

Is |x – 3| = 3 – x?

Note that 3 – x can also be written as -(x – 3).

Is |x – 3| = -(x – 3)?

Recall the definition of absolute values: |a| = a if a is greater than or equal to 0, and -a if a < 0.

So, “Is |x – 3| = -(x – 3)?” depends on whether (x – 3) is positive or negative. If (x – 3) is negative (or 0), then |x – 3| is equal to -(x – 3).

So our question now boils down to:

Is (x – 3) negative (or 0)?

Statement 1: x is not 3

This means we know that (x – 3) is not 0, but we still don’t know whether it is negative or positive. This statement is not sufficient.

Statement 2: -x * |x| > 0

|x| is always non-negative, so for the product to be positive, “-x” must also be positive. This means x must be negative. If x is negative, x – 3 must be negative, too.

If (x – 3) is negative, |x – 3| is equal to -(x – 3). Hence, this statement alone is sufficient, and our answer is B.