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In a rectangular coordinate plane, points A(3,4), B(6,-5), C(-4,-3) an

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In a rectangular coordinate plane, points A(3,4), B(6,-5), C(-4,-3) an  [#permalink]

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New post 01 Jun 2017, 00:23
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A
B
C
D
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Question Stats:

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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
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In a rectangular coordinate plane, points A(3,4), B(6,-5), C(-4,-3) an  [#permalink]

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New post 01 Jun 2017, 03:51
1
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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)
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Re: In a rectangular coordinate plane, points A(3,4), B(6,-5), C(-4,-3) an  [#permalink]

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New post 28 Aug 2017, 12:19
1
pushpitkc wrote:
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 = \(\frac{1}{2} *| (x1y2 - y1x2) + (x2y3 - y2x3) + (x3y4 - y3x4) + (x4y1 - y4x1) |\)
Substiuting 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)


Formulas are a big NO NO for GMAT.
pushpitkc can you please suggest some other method to do this question? :|
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In a rectangular coordinate plane, points A(3,4), B(6,-5), C(-4,-3) an  [#permalink]

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New post Updated on: 26 Nov 2017, 00:08
9
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
Attachments

Area of Quadrilateral.jpg
Area of Quadrilateral.jpg [ 1.73 MiB | Viewed 3278 times ]


Originally posted by Nikkb on 04 Sep 2017, 09:13.
Last edited by Nikkb on 26 Nov 2017, 00:08, edited 1 time in total.
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Re: In a rectangular coordinate plane, points A(3,4), B(6,-5), C(-4,-3) an  [#permalink]

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New post 07 Sep 2017, 18:37
pushpitkc wrote:
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 = \(\frac{1}{2} *| (x1y2 - y1x2) + (x2y3 - y2x3) + (x3y4 - y3x4) + (x4y1 - y4x1) |\)
Substiuting 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)


Hi pushpitkc

can you please elaborate on the what principle you have used to solve this question, I havent come across the formula used by you
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Re: In a rectangular coordinate plane, points A(3,4), B(6,-5), C(-4,-3) an  [#permalink]

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New post 25 Nov 2017, 13:34
Nikkb wrote:
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

PS: Took help from someone to solve it but not sure of his gmatclub id to tag.

+1 kudos if you like the solution.


hi, my solution is exact xerox copy of yours, kudos to you!!

Thanks pushpitkc for formula, saves lot of time.
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Re: In a rectangular coordinate plane, points A(3,4), B(6,-5), C(-4,-3) an  [#permalink]

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New post 15 Jan 2018, 12:53
hellosanthosh2k2 wrote:
Nikkb wrote:
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

PS: Took help from someone to solve it but not sure of his gmatclub id to tag.

+1 kudos if you like the solution.


hi, my solution is exact xerox copy of yours, kudos to you!!

Thanks pushpitkc for formula, saves lot of time.



Thanks for the diagram - any way to do this in under 3mins? I see avg time of 17 mins..
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In a rectangular coordinate plane, points A(3,4), B(6,-5), C(-4,-3) an  [#permalink]

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New post 18 Jan 2018, 10:17
pushpitkc wrote:
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)


Did some googling, and found out that the name of the formula used was the "shoelace formula". Basically, its a formula that allows you to calculate the area of a quadrilateral if you have all the vertices.

An easier version to remember of this formula would be
Area = \(\frac{1}{2} *| x1y2 + x2y3 + x3y4 + x4y1 - y1x2 - y2x3 - y3x4 - y4x1 |\)
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Re: In a rectangular coordinate plane, points A(3,4), B(6,-5), C(-4,-3) an  [#permalink]

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New post 19 Jan 2018, 11:20
Thanks for helping people with the name, blazeowlss

I had studied it in school and didn't really remember the name :)
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Re: In a rectangular coordinate plane, points A(3,4), B(6,-5), C(-4,-3) an  [#permalink]

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New post 19 Jan 2018, 22:06
(d1xd2)/2
d1 is diagonal AC
d2 is diagonal BD

Ans. is "D".
Please correct if I am wrong.
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In a rectangular coordinate plane, points A(3,4), B(6,-5), C(-4,-3) an  [#permalink]

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New post 25 Jan 2018, 13:02
niteshwaghray wrote:
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


The Area of Triangle with \(A(x1,y1) B(x2,y2) C(x3,y3)\) = \(\frac{1}{2}|x1(y2-y3)+x2(y3-y1)+x3(y1-y2)\)|. Now, Divide the Quadrilateral in two triangles and find the area using the formula. Hence, D.
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