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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 7599
GMAT 1: 760 Q51 V42 GPA: 3.82
If a rectangle WXYZ has 240 area and 68 perimeter and A, B, C, D are m  [#permalink]

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Difficulty:   35% (medium)

Question Stats: 76% (02:49) correct 24% (02:48) wrong based on 54 sessions

### HideShow timer Statistics Attachment: rectangle WXYZ.jpg [ 7.75 KiB | Viewed 1482 times ]

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

* A solution will be posted in two days.

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Concentration: Strategy, General Management
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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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Area of rectange = WX * WZ = 240
Area of rhombus = 0.5 * AD * BC = 0.5 * WZ * WX = 120

Area of rhombus + Area of 4 triangles = 240
120 + Area of 4 triangles = 240
Area of 4 triangles = 120
Since A, B, C and D are midpoints, each of the 4 triangles must be of equal area. So area of 1 triangle = 30
0.5 * BZ * ZD = 30
BZ * ZD = 60.....(1)

2(WX + WZ) = 68
2(2ZD + 2BZ) = 68
4(BZ + ZD) = 68
BZ + ZD = 17.....(2)

From (1) and (2), 12 and 5 can be the values of BZ and ZD.

In triangle BZD, BZ^2 + ZD^2 = BD^2
BD^2 = 144 + 25 = 169
BD = 13

Perimeter of rhombus = 4*BD = 52

Manager  Joined: 09 Jul 2013
Posts: 109
Re: If a rectangle WXYZ has 240 area and 68 perimeter and A, B, C, D are m  [#permalink]

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An alternate method requiring fewer steps:
We are told that the area of the rectangle is 240. $$l*w=240$$
We are told that the perimeter of the triangle is 68. $$2l+2w=68$$ or $$w=34-l$$
Substituting into the equation for the area
$$l*(34-l)=240$$
$$l^2-34l+240=0$$
Now factor the equation to find the possible values for $$l$$
$$(l-10)(l-24)=0$$
so possible values of $$l$$ are 10 and 24. So length and width of the rectangle are 10 and 24, but we don't care which is which to solve our problem.

What the problem is asking is the perimeter of the rhombus. A rhombus has 4 equal sides, so we need to find the length of one side. Looking at triangle BZD, we need to find BD. But we now have BZ and ZD, since those are half of the length and width.
$$BZ^2+ZD^2 = BD^2$$
$$5^2+12^2 = BD^2$$
$$BD=13$$ (You should recognize the 5-12-13 triangle)

Perimeter of rhombus ABCD is 4*13 = 52

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Dave de Koos
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
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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.
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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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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. Re: If a rectangle WXYZ has 240 area and 68 perimeter and A, B, C, D are m   [#permalink] 25 Feb 2019, 15:37
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