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

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Math Expert
Joined: 02 Sep 2009
Posts: 59623
In the xy-plane, a quadrilateral has vertices at (-9,-2), (-1,-2) ,(-1  [#permalink]

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26 Nov 2018, 00:44
00:00

Difficulty:

15% (low)

Question Stats:

80% (01:21) correct 20% (01:33) wrong based on 74 sessions

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In the xy-plane, a quadrilateral has vertices at (-9,-2), (-1,-2) ,(-1,-4) , and (-9,-4). What is the perimeter of the quadrilateral?

A. 4
B. 10
C. 20
D. 24
E. 32

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

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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.

Intern
Joined: 27 Nov 2018
Posts: 44
GMAT 1: 510 Q33 V28
Re: In the xy-plane, a quadrilateral has vertices at (-9,-2), (-1,-2) ,(-1  [#permalink]

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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)
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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