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# If a rectangle WXYZ has 240 area and 68 perimeter and A, B, C, D are m

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

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18 Feb 2016, 17:58
1
4
00:00

Difficulty:

35% (medium)

Question Stats:

78% (02:59) correct 22% (02:13) wrong based on 64 sessions

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rectangle WXYZ.jpg [ 7.75 KiB | Viewed 1185 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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MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
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"Only $99 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" SC Moderator Joined: 13 Apr 2015 Posts: 1689 Location: India Concentration: Strategy, General Management GMAT 1: 200 Q1 V1 GPA: 4 WE: Analyst (Retail) Re: If a rectangle WXYZ has 240 area and 68 perimeter and A, B, C, D are m [#permalink] ### Show Tags 19 Feb 2016, 06:42 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 Answer: C 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] ### Show Tags 19 Feb 2016, 09:41 1 1 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 Answer: C _________________ Dave de Koos GMAT aficionado Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6524 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: If a rectangle WXYZ has 240 area and 68 perimeter and A, B, C, D are m [#permalink] ### Show Tags 20 Feb 2016, 22:39 1 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. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$99 for 3 month Online Course"
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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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02 Sep 2018, 23:32
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Re: If a rectangle WXYZ has 240 area and 68 perimeter and A, B, C, D are m &nbs [#permalink] 02 Sep 2018, 23:32
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