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In the xy-plane, a quadrilateral has vertices at (-9,-2), (-1,-2) ,(-1

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Math Expert
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Joined: 02 Sep 2009
Posts: 60647
In the xy-plane, a quadrilateral has vertices at (-9,-2), (-1,-2) ,(-1  [#permalink]

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New post 26 Nov 2018, 00:44
00:00
A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

81% (01:22) correct 19% (01:32) wrong based on 84 sessions

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V
Joined: 18 Jul 2018
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Location: India
Concentration: Finance, Marketing
GMAT 1: 590 Q46 V25
GMAT 2: 690 Q49 V34
WE: Engineering (Energy and Utilities)
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Re: In the xy-plane, a quadrilateral has vertices at (-9,-2), (-1,-2) ,(-1  [#permalink]

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New post 26 Nov 2018, 00:53
Applying the distance between the points formula.

\(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

The distance between the points (-9,-2) & (-1,-2) is \(\sqrt{(-1+9)^2+(-2+2)^2}\) = 8.

Similarly, for other points, we get the distance as 2,8 and 2.

Hence the perimeter is 8+2+8+2 = 20.

C is the answer.
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Re: In the xy-plane, a quadrilateral has vertices at (-9,-2), (-1,-2) ,(-1  [#permalink]

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New post 24 Nov 2019, 07:21
It can solved faster by drawing points in co ordinate plane. By drawing, we will come to know that it is a rectangle with one side length=8 units, second side length=2 units.
Perimeter of rectangle= 2 x (l +b) = 2 x 10 = 20 which is answer C (Hit Kudos if my stated procedure helped you)
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Re: In the xy-plane, a quadrilateral has vertices at (-9,-2), (-1,-2) ,(-1   [#permalink] 24 Nov 2019, 07:21
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In the xy-plane, a quadrilateral has vertices at (-9,-2), (-1,-2) ,(-1

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