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GMAT 1: 760 Q51 V42
Posts: 10,057
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
77% (02:51) correct
23%
(02:44)
wrong
based on 81
sessions
History
Date
Time
Result
Not Attempted Yet
Attachment:
rectangle WXYZ.jpg [ 7.75 KiB | Viewed 3770 times ]
A. 48
B. 50
C. 52
D. 54
E. 56
* A solution will be posted in two days.
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
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
Posts: 96
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
_________________
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
GMAT Aficionado
GMAT 1: 760 Q51 V42
Posts: 10,057
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.
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.
Posts: 292
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.
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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