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# If the given Quadrilateral is a rectangle, find the area of the shaded

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Retired Moderator
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08 Dec 2018, 09:08
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30% (02:41) correct 70% (02:43) wrong based on 20 sessions

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GMATbuster's Weekly Quant Quiz#12 Ques #5

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

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Figure.jpg [ 10.87 KiB | Viewed 911 times ]

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08 Dec 2018, 22:28
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.
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08 Dec 2018, 22:57
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,
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09 Dec 2018, 00:19
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Official Explanation

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09 Dec 2018, 03:01
:D at least tried to solve it
we have two isosceles triangles (45 45 90) with two sides equal 1, so to find altitude of these triangles i need to to draw the line that bisects this tringle let height be x then x^2+(sqrt2)^2=1^2 ----> x^2= 1^2-(sqrt2)^2 ---> x^2 = -1 so altitude is 1. Area of triangle = base *height ---> 1*1 =1

i also need to find areas of two longer triangles.

Diaogonal of rectangle is sqroot W^2+l^2 ----> 5^2+8^2 = 89 i will round it to 90

so diagonal of rectangle is 90

Area of rhombus is length * width

Width is 1 and length is of each rhombus is 90/2 i.e. 45

so Area of Rhombus is 1*45/2 = 22.5

got a "bit" confused .... anyway will solve it to the end

so if area of one rhombus is 22.5 then area of two rhombus is 90

so area of shaded region is 90

and why hell i was calculating area of triangle
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09 Dec 2018, 04:51
IMO D.

Both triangles are similar. Find the area of one shaded region and multiply it by 2.

13/4 is area of one region and when multiplied by 2 gives 13/2 .
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09 Dec 2018, 06:18
Area of the rectangle = 8*5 = 40
Diagonal AC divides it in half , 20 each.
DO divides area AOC into 2 parts. 10 each
KO divides AOD into further 2 parts 5 each.

Area of AOK = 5 when base is 2.5.
when base is 1 , area = 2

area of OGC = 5 when base is 4
when base is 1 , area = 5/4 = 1.25

Area of shaded region = 2(2+1.25) = 6.5
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09 Dec 2018, 08:22
If we divide the rectangle into four equal parts and then analyse 1st part APOS
Area of this part = 4*5/2 = 10
There will be two unshaded triangles.
Area of 1st unshaded triangle = 1/2*3*5/2 = 15/4
Area of 2nd unshaded triangle = 1/2*3/2*3 = 9/4
So area of shaded part in the 1st part = 10-15/4-9/4 = 10-6 = 4
2nd part BPOQ has no shaded part
3rd part CQOR has shaded part and its area will also be =4
4th part DROS has no shaded part.

Total area of shaded part =4+4 = 8

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02 Apr 2020, 07:01
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Re: If the given Quadrilateral is a rectangle, find the area of the shaded   [#permalink] 02 Apr 2020, 07:01