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a triangle have coordinates

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a triangle have coordinates

by sanju09 » Fri Oct 08, 2010 3:37 am
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) √43
(D) 7 + 6 √2
(E) 12 + 3 √2


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by Rahul@gurome » Fri Oct 08, 2010 4:11 am
Distance between (0, 0) and (3, 3) = √[(3-0)^2 + (3-0)^2] = √18
Distance between (0, 0) and (7, 0) = √[(7-0)^2 + (0-0)^2] = √49 = 7
Distance between (3, 3) and (7, 0) = √[(7-3)^2 + (0-3)^2] = √(16+9) = 5
Perimeter of triangle = 5 + 7 + √18 = 12 + 3√2

[spoiler]The correct answer is (E).[/spoiler]
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by GMATGuruNY » Fri Oct 08, 2010 4:54 am
sanju09 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) √43
(D) 7 + 6 √2
(E) 12 + 3 √2


[spoiler]Source: https://www.gmatcram.com[/spoiler]
Distance from (0,0) to (7,0) = 7.
(3,3) and (7,0) form a side of 3 and a side of 4 in a 3:4:5 right triangle, so the distance between them = 5.
7+5 = 12, which is only in answer choice E. (Since the 3rd side is longer than 1, we know that 12+1=13 in answer choice A cannot be correct.)

The correct answer is E.

For the curious, (0,0) to (3,3) from two sides of 3 in a 45:45:90 triangle -- in which the sides are proportioned s:s:s√2 -- so the distance between them is 3√2.
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