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# In the xy plane, the vertices of a triangle have coordinates

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In the xy plane, the vertices of a triangle have coordinates [#permalink]  05 Nov 2010, 19:31
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Difficulty:

5% (low)

Question Stats:

85% (02:02) correct 15% (00:41) wrong based on 108 sessions
In the xy plane, the vertices of a triangle have coordinates(0,0), (3,3), and (7,0). What is the perimeter of the triangle?

(A) 13
(B) 34
(C) root 43
(D) 7+root3
(E) 12 +3root2
[Reveal] Spoiler: OA

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Re: In the xy plane, the vertices [#permalink]  05 Nov 2010, 20:08
Expert's post
monirjewel wrote:
In the xy plane, the vertices of a triangle have coordinates(0,0), (3,3), and (7,0). What is the perimeter of the triangle?
(A) 13
(B) 34
(C) root 43
(d) 7+root3
(E) 12 +3root2

The formula to calculate the distance between two points $$(x_1,y_1)$$ and $$(x_2,y_2)$$ is $$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$$.

Distance between (0, 0) and (7, 0) is $$d=7$$;
Distance between (0, 0) and (3, 3) is $$d=\sqrt{(0-3)^2+(0-3)^2}=\sqrt{18}=3\sqrt{2}$$;
Distance between (3, 3) and (7, 0) is $$d=\sqrt{(3-7)^2+(3-0)^2}=5$$;

$$P=7+3\sqrt{2}+5=12+3\sqrt{2}$$.

For more check Coordinate Geometry chapter of Math Book: math-coordinate-geometry-87652.html

Hope it helps.
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Kudos [?]: 275 [0], given: 11

Re: In the xy plane, the vertices [#permalink]  03 Jan 2013, 03:10
monirjewel wrote:
In the xy plane, the vertices of a triangle have coordinates(0,0), (3,3), and (7,0). What is the perimeter of the triangle?
(A) 13
(B) 34
(C) root 43
(d) 7+root3
(E) 12 +3root2

Get distance between (0,0) and (3,3): $$\sqrt{(3 - 0)^2 + (3-0)^2} = \sqrt{(18)} = 3 \sqrt{(2)}$$
Get distance between (3,3) and (7,0): $$\sqrt{(7-3)^2 + (3-0)^2} = \sqrt{16 + 9}= \sqrt{25}=5$$
Get distance between (0,0) and (7,0): 7

$$P = 7 + 5 + 3\sqrt{2} = 12 + \sqrt{2}$$

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Re: In the xy plane, the vertices   [#permalink] 03 Jan 2013, 03:10
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