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Given area of rectangle = 24 = DC*AD Given DC = 6, so AD = 4 DC = AB = 6, P is the midpoint of AB, so AP = 3 AP = 3, AD = 4, so DP = 5 Area of square = Square of side = 5*5 = 25

Re: In the figure above, P is the midpoint of side AB. If the area of the [#permalink]

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12 Oct 2017, 13:34

I'm wondering the same thing as KS15. I saw this as an easy question inside a bunch of fluff.

Since 6 is the diagonal of the square, you should be able to just solve for one of the sides of the square which comes out to S=6/sqrt{2} which when you square it comes out to 36/2 = 18. (choice B)

Then I looked at how others did it and wonder why my way doesn't work. If we find out that AD = 4 and that we have a 3,4,5 triangle. That means that angle APD is around 53.13. Since BC is also 4, that means that BPC also around 53.13. If that is the case, angle DPC would have to be around 74 degrees which would mean shape PCQD is a rhombus and not a square.

Can someone else take a look and tell me if I'm running my calculations wrong or missing something here?

Re: In the figure above, P is the midpoint of side AB. If the area of the [#permalink]

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13 Oct 2017, 09:23

Hey Everyone, this one is driving me nuts. I am trying to see if my logic is incorrect somewhere. Can we have a couple other eyes look this over and give their thoughts on my comments above?

In looking at it more, it seems to me that if P is the midpoint of AB which is shared as a corner of the square PCQD, and both the rectangle ABCD and square PCQD share points D and C which also has to be the diagonal of PCQD. Given just this information and nothing on the area of the rectangle, we can still assume that angles APD and BPC are both 45 degrees since angle DPC has to be 90 (because PCQD is a square). With just this information, we can assume AD and BC are 3 because triangles APD and BPC are 45,45,90. That would mean the area of the rectangle is 18 (3*6) which is the same as the square.

If we do not assume that PCQD is a square (even though we are told so) and we know that the area of the rectangle is 24, then we can look at the third solution Aditya provided and find that the two triangles add up to 24 also, so the area of PCQD is 24 which is actually just a rhombus and not a square.

Does this logic make sense and can somebody else through in their two cents to make sure I'm not missing some bigger point here?

I believe that there is a problem with the question. I made a couple comments about it already and it's driving me nuts. The only way that PCQD can be a square considering that it shares points C and D with rectangle ABCD and also has a corner P touching the midpoint of AB is for triangles APD and BPC to be 90,45,45 triangles. This would mean that sides AD and BC would have to be 3 (not 4). But since the question tells us the area of ABCD is 24 and we are given that one side is 6, AD and BC have to be 4. If this is the case, PCQD is a rhombus (not a square) and the only way to find the area of PCQD with the information we have is to subtract the area of triangles APD and BPC from the area of the rectangle. This is the right hand method in adityagmatclub's image and the way jajanb solved the problem.

I personally don't believe there is a correct solution with the given information, but I can certainly be wrong and completely missing something. I would love to hear other people's thoughts!