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18 Feb 2016, 18:58
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C
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:
76% (02:50) correct
24%
(02:44)
wrong
based on 80
sessions
History
Date
Time
Result
Not Attempted Yet
Attachment:
rectangle WXYZ.jpg [ 7.75 KiB | Viewed 3658 times ]
A. 48
B. 50
C. 52
D. 54
E. 56
* A solution will be posted in two days.
19 Feb 2016, 07:42
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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
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
19 Feb 2016, 10:41
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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
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
