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Re: If a rectangle WXYZ has 240 area and 68 perimeter and A, B, C, D are m [#permalink]
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If a rectangle WXYZ has 240 area and 68 perimeter and A, B, C, D are midpoints, what is the perimeter of a rhombus ABCD?

A. 48
B. 50
C. 52
D. 54
E. 56


--> In order to satisfy the question that the area is 240 and the perimeter is 68, WX=ZY=24, WZ=XY=10. Then, AX=WA=ZD=DY=12 and XC=CY=WB=BZ=5. AC=CD=DB=BA=13 is derived. So, the perimeter of the rhombus ABCD is 13*4=52.
Therefore, the answer is C.
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Re: If a rectangle WXYZ has 240 area and 68 perimeter and A, B, C, D are m [#permalink]
You can save time if you notice that Area = 240 ends in a 0, which means that there is either a 0 or 5 as the units digit of one of the factors. Since we also know that the 2 factors L+W = 34(Even number), we can't have a single or 2 odd numbers (can only be Even+Even=Even) so one factor must be a multiple of 10.

If you try 240/10, it's pretty easy to see the other is 24 and that 24+10 = 34.

Since we have midpoints, half of these is 5 and 12, which we should recall as the 5-12-13 pythagorean triple. Thus P rhombus is 4*13 = 52, C.
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Re: If a rectangle WXYZ has 240 area and 68 perimeter and A, B, C, D are m [#permalink]
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