In the figure, the 6 small squares are identical, each with sides of

Look at the figure ...
Don't get confused with the start points of the corner squares..

How many lines moving left to right - 4
How many moving down to up - 2
So perimeter = 2(4+2)=2*6=12

The small square on the left has an outer perimeter of 4 - 1 = 3. The larger square (which is composed 4 smaller squares) in the middle has an outer perimeter of 8 - 2 = 6. The small square on the right has an outer perimeter of 4 - 1 = 3. Thus, the outer perimeter of the entire figure is 3 + 6 + 3 = 12.

Answer: B

We know that all squares have a length of 1 and computing the outer perimeter goes like this :

A. Top and Bottom of the four adjacent squares = 2 + 2 = 4
B. Side square = 2 (four adjacent square side length) - 1 (side square bottom) + 1 + 1 + 1 (side square top left right) = 4
C. We have 2 side squares which therefore 2 * 4 = 8

Add up A + C = 4 + 8 = 12

Therefore B is the answer

Hello, adkikani. I understand how, intuitively, it might seem as though you cannot come up with a measure for the vertical segments, and that is partially correct: you will never know the exact measure of either the top or bottom segments on either the left or right side of the figure in question. However, since you know that the six small squares are identical, and that they each have a side length of 1, you can picture sliding those two outer squares vertically to align with one of the horizontal segments at the top or bottom, and then you can appreciate how, even without knowing the lengths of the individual vertical segments in question, you can tell with certainty what they must equal together. The "broken up" side must still equal 1 on both the left and right.

I hope that helps clarify the issue. For a tougher problem dealing with a similar concept, see this official question. Good luck.

- Andrew

Video solution from Quant Reasoning:

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Can you please elaborate the highlighted part? How does one account for vertical lines since four horizontal lines are complete and each measure as 1 units?

Answer: Option B

Video solution by GMATinsight



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I'm surprised nobody has mentioned this yet, since it's a useful approach to some geometry problems to switch between 2D and 3D.

Let's imagine the figure as a 2D rendering of the aerial view of a building.
Now, imagine standing in front of the building (you can pick whichever side you want as the front...I'll go with the bottom of the figure).

How many units wide is the building? Four.
Now go to the right side of the building and look at it. How many units is it from front to back? Two.
How about the back side of the building? Four.
How about the left side? Two.

4+2+4+2 = 12

Answer choice B.

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