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IMO A,

Lets extend the rectangle then the area of rectangle is 60-(1/2*10*2+1/2*6*4)= 38
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Attachment:
Rectangle.jpg
Rectangle.jpg [ 131.52 KiB | Viewed 1790 times ]

Let the given figure be extended to form a rectangle as shown above. Now it is easy to calculate the are of the given figure.

Area of quadrilateral ABCD = Area of rectangle ABEF - Area of triangle BCE - Area of triangle CDF

Area of rectangle ABEF = 10*6 = 60

Area of triangle BCE = \(\frac{1}{2} * 2 * 10\) = 10

Area of triangle CDF = \(\frac{1}{2} * 6 * 4\) = 12

Area of quadrilateral ABCD = 60 - 10 - 12 = 38

OPTION: A
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Bunuel

What is the area of quadrilateral ABCD?

A. 38
B. 40
C. 42
D. 44
E. 45


Attachment:
#GREpracticequerstion What is the area of quadrilateral ABC.jpg

IMO A.

Draw the line BD, you will get two triangles ABD and BDC.
Area of quadrilateral = A(ABD)+A(BDC)

A(ABD) = 1/2*6*4=12.

By distance formula, BD = \(2\sqrt13\) and CD = \(2\sqrt13\) Therefore, area of BDC = 26.

And A(quadrilateral) = 26+12=38
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Bunuel

What is the area of quadrilateral ABCD?

A. 38
B. 40
C. 42
D. 44
E. 45


Attachment:
#GREpracticequerstion What is the area of quadrilateral ABC.jpg


Area of ABD = 12; Using discriminant method
Area of BCD = 26; Using discriminant method
Area od ABCD = 12 + 26 = 38

IMO A
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