|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
niteshwaghray
GMAT 1: 700 Q45 V40
Posts: 59
01 Jun 2017, 00:23
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
58% (03:10) correct
42%
(03:04)
wrong
based on 256
sessions
History
Date
Time
Result
Not Attempted Yet
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 10708 times ]
01 Jun 2017, 03:51
Kudos
Bookmarks
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)
