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GMAT Club 12 Days of Christmas is a 4th Annual GMAT Club Winter Competition based on solving questions. This is the Winter GMAT competition on GMAT Club with an amazing opportunity to win over $40,000 worth of prizes!
08 Dec 2018, 10:08
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
28% (02:50) correct
72%
(02:49)
wrong
based on 32
sessions
History
Date
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Result
Not Attempted Yet
If the given Quadrilateral is a rectangle, find the area of the shaded region.
A) 23/5
B) 18/5
C) 21/4
D) 13/2
E) 8
Attachment:
Figure.jpg [ 10.87 KiB | Viewed 2404 times ]
08 Dec 2018, 23:28
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The point of intersection at c is the center of the rectangle..
look at the attached figure, we are looking for area of abc, which is equal to Area of adc-area of bdc..
(I) area of adc = \(\frac{1}{2}*ad*dc=\frac{4*2.5}{2}=5\)
(II) area of bdc = \(\frac{1}{2}*bd*dc=\frac{4*(2.5-1)}{2}=2*1.5=3\)
thus area of abc = 5-3=2..
similarly, if we drop a perpendicular from the bigger length, the area of adjoining shaded portion next to abc will be \(\frac{1}{2}*4*2.5-\frac{3*2.5}{2}=5-7.5/2=2.5/2=5/4\)
Thus, area of shaded reion on one part is 2+5/4=13/4..
But we have two such portions therefore 2+13/4=13/2
Note..
The question does not seem to be from a proper source as the questions from reliable sources will always have the choices in order.
look at the attached figure, we are looking for area of abc, which is equal to Area of adc-area of bdc..
(I) area of adc = \(\frac{1}{2}*ad*dc=\frac{4*2.5}{2}=5\)
(II) area of bdc = \(\frac{1}{2}*bd*dc=\frac{4*(2.5-1)}{2}=2*1.5=3\)
thus area of abc = 5-3=2..
similarly, if we drop a perpendicular from the bigger length, the area of adjoining shaded portion next to abc will be \(\frac{1}{2}*4*2.5-\frac{3*2.5}{2}=5-7.5/2=2.5/2=5/4\)
Thus, area of shaded reion on one part is 2+5/4=13/4..
But we have two such portions therefore 2+13/4=13/2
Note..
The question does not seem to be from a proper source as the questions from reliable sources will always have the choices in order.
Attachments
0001.png [ 24.66 KiB | Viewed 2185 times ]
08 Dec 2018, 23:57
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This question is a little extra tricky. The shaded part is not completely symmetrical but symmetrical across the diagonal going through the shaded region and mirror reflected across the point C.
So, the area of triangle ABC will be the area of triangle ADC - the area of the triangle BDC.
Area of ADC = 0.5*2.5*4 = 2.5*2 = 5
Area of BDC = 0.5*(2.5-1)*4 = 1.5*2 = 3
Hence Area of ABC will be 5-3 = 2
Similarly we need to calculate the area of triangle AFC by subtracting area of triangle FEC from area of triangle AEC.
Area of AEC = 0.5*4*2.5 = 5
Area of FEC = 0.5*(4-1)*2.5 = 1.5*2.5 = 3.75
Hence Area of AFC will be 5 - 3.75 = 1.25
Area of the quadrilateral ABCF = 2+1.25 = 3.25
Hence area of shaded part will be 2* area of quadirlateral ABCF As symmetry is through point C and mirror reflected across the diagonal.
Hence Area of shaded is 2*3.25 = 6.5 or 13/2
Option (D) is correct.
Best,
Gladi
So, the area of triangle ABC will be the area of triangle ADC - the area of the triangle BDC.
Area of ADC = 0.5*2.5*4 = 2.5*2 = 5
Area of BDC = 0.5*(2.5-1)*4 = 1.5*2 = 3
Hence Area of ABC will be 5-3 = 2
Similarly we need to calculate the area of triangle AFC by subtracting area of triangle FEC from area of triangle AEC.
Area of AEC = 0.5*4*2.5 = 5
Area of FEC = 0.5*(4-1)*2.5 = 1.5*2.5 = 3.75
Hence Area of AFC will be 5 - 3.75 = 1.25
Area of the quadrilateral ABCF = 2+1.25 = 3.25
Hence area of shaded part will be 2* area of quadirlateral ABCF As symmetry is through point C and mirror reflected across the diagonal.
Hence Area of shaded is 2*3.25 = 6.5 or 13/2
Option (D) is correct.
Best,
Gladi
Attachments
fig1.PNG [ 39.33 KiB | Viewed 2154 times ]
.... anyway will solve it to the end 
