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Bunuel
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Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
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Hello guys,

I am following the logic of this for the most part. But if the adjacent side lengths are the same, isn't it a square then, and not a rectangle? That confused me about the question.
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Raben
Hello guys,

I am following the logic of this for the most part. But if the adjacent side lengths are the same, isn't it a square then, and not a rectangle? That confused me about the question.


Hey, fair question. Yes you are right, if the adjacent sides are the same then it is indeed a square. But it's important to know that a square is a rectangle, it is a special case of a rectangle where all sides are the same length, hence, all squares are rectangles but not all rectangles are squares. The same goes for parallelograms too, not all parallelograms are rectangles but all rectangles are parallelograms.
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Hello guys,

I am following the logic of this for the most part. But if the adjacent side lengths are the same, isn't it a square then, and not a rectangle? That confused me about the question.


Hey, fair question. Yes you are right, if the adjacent sides are the same then it is indeed a square. But it's important to know that a square is a rectangle, it is a special case of a rectangle where all sides are the same length, hence, all squares are rectangles but not all rectangles are squares. The same goes for parallelograms too, not all parallelograms are rectangles but all rectangles are parallelograms.


Thank you very much for that thorough explanation!

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Yes, a square is a special rectangle!

(1) See light orange in the pic. Each of the things in light orange are figured out by the first piece of info. It is NOT Sufficient to calculate the entire perimeter of the pic.

(2) See light blue in the pic. Now we know the "rectangle" is a special rectangle. It is a square. Everything in light blue is now known... BUT, we know NOTHING about the parts in light orange. This data on its own is NOT Sufficient to calculate the perimeter of the entire object.

TOGETHER: 4(2√2)+√2 π = 8√2+√2 π

SUFFICIENT!!
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