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01 Jun 2017, 00:23
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D
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In a rectangular coordinate plane, points A(3,4), B(6,-5), C(-4,-3) and D(-2,2) are joined to form a quadrilateral. What is the area, in square units, of quadrilateral ABCD?
A. 35
B. 37.5
C. 45
D. 52.5
E. 60
A. 35
B. 37.5
C. 45
D. 52.5
E. 60
GMATAspirer09
Well for co=ordinate geometry questions where we have to find area, its better to divide the quadrilateral in rectangles/triangles and find total area.
But in some cases when they are not straight forward and you know the area formula its better to go with formula instead of trying to divide figure in different parts.
In given figure, it's hard to divide quadrilateral internally. In this case we can surround it with a rectangle and subtract the extra part area from total rectangle area.
Please find attached figure for details solution.
Here we have surrounded quadrilateral ABCD externally by rectangle PQRS.
Found total area of PQRS. And subtracted the area of triangles( AQR, CSR, XDC and YDA) and square (PYXD)for formed outside it.
Answer: D
Well for co=ordinate geometry questions where we have to find area, its better to divide the quadrilateral in rectangles/triangles and find total area.
But in some cases when they are not straight forward and you know the area formula its better to go with formula instead of trying to divide figure in different parts.
In given figure, it's hard to divide quadrilateral internally. In this case we can surround it with a rectangle and subtract the extra part area from total rectangle area.
Please find attached figure for details solution.
Here we have surrounded quadrilateral ABCD externally by rectangle PQRS.
Found total area of PQRS. And subtracted the area of triangles( AQR, CSR, XDC and YDA) and square (PYXD)for formed outside it.
Answer: D
Attachments
Area of Quadrilateral.jpg [ 1.73 MiB | Viewed 10686 times ]
01 Jun 2017, 03:51
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Given data
x1=3,y1=4
x2=6,y2=-5
x3=-4,y3=-3
x4=-2,y4=2
Given the vertices of the quadrilateral
Area = \(\frac{1}{2} *| (x1y2 - y1x2) + (x2y3 - y2x3) + (x3y4 - y3x4) + (x4y1 - y4x1) |\)
Substituting these values,
Area = \(\frac{1}{2} * | -38 -39 -14 -14|\) = \(\frac{1}{2} * | -105|\) = 52.5(since absolute value of -105 is 105) (Option D)
x1=3,y1=4
x2=6,y2=-5
x3=-4,y3=-3
x4=-2,y4=2
Given the vertices of the quadrilateral
Area = \(\frac{1}{2} *| (x1y2 - y1x2) + (x2y3 - y2x3) + (x3y4 - y3x4) + (x4y1 - y4x1) |\)
Substituting these values,
Area = \(\frac{1}{2} * | -38 -39 -14 -14|\) = \(\frac{1}{2} * | -105|\) = 52.5(since absolute value of -105 is 105) (Option D)
