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# Parallelogram ABCD lies in the xy-plane, as shown in

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Parallelogram ABCD lies in the xy-plane, as shown in [#permalink]  06 Aug 2017, 12:28
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53% (01:47) correct 46% (02:19) wrong based on 15 sessions

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#GREpracticequestion Parallelogram ABCD lies in the xy-plane, as shown in the figure above..jpg [ 23.49 KiB | Viewed 206 times ]

Parallelogram ABCD lies in the xy-plane, as shown in the figure above. The coordinates of point C are (-3, 4) and the coordinates of point B are (-7, 7). What is the area of the parallelogram ?

A. 1

B. $$2 \sqrt{7}$$

C. 7

D. 8

E. $$7 \sqrt{2}$$
[Reveal] Spoiler: OA

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Re: Parallelogram ABCD lies in the xy-plane, as shown in [#permalink]  09 Aug 2017, 23:18
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can u explain the solution?
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Re: Parallelogram ABCD lies in the xy-plane, as shown in [#permalink]  31 Aug 2017, 21:20
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The easiest way to solve this problem is to draw a rectangle around the parallelogram, find its area, and substract area of the triangles that emerge around the parallelogram, within the rectangle (but that are not part of the parallelogram).
Since ABCD is a parallelogram, line segments AB and CD have the same length and the same slope. Therefore, in the diagram above, point A is at (-4,3), The square has an area of 7*7=49. By drawing carefully and exploiting similar triangles created by various parallel lines, you can label the height of each triangle 3, and each base 7. Each triangles has area 1/2hb=1/2*3*7=21/2. Therefore, the area of the parallelogram ABCD equals 49-4*(21/2)=49-42=7.
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Kudos [?]: 7 [1] , given: 11

Re: Parallelogram ABCD lies in the xy-plane, as shown in [#permalink]  16 Jan 2018, 05:28
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First find the co-ordinates of point A. For now, lets assume the co-ordinates are (x,y)
Since it is mentioned that ABCD is a parallelogram. So slope of BC must be equal to slope of AD since BC||AD.
Hence we can get the equation 3x=4y=0 -- equation(1) [by equating slope of line BC and AD]

Similarly by equating the lines AB and CD since (AB||CD), we get the equation 4x+3y= -7 -- (2)

Solving for equations (1) and (2) we get x = -4 and y= 3. Thus A is (-4,3).

If we observe now, all sides are equal in length i.e. each side AB=BC=CD=AD=5. Thus ABCD is a rhombus. The area of a rhombus is (product of lengths of diagonals)/2 i.e. in our case (BD*AC)/2.

BD is 7*(2)^(1/2) and AC is 2^(1/2). Thus area of ABCD is 7{sqrt2} * {sqrt2} = 7

option C
Re: Parallelogram ABCD lies in the xy-plane, as shown in   [#permalink] 16 Jan 2018, 05:28
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