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triangle in xy plane

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triangle in xy plane

by alex.gellatly » Tue Jun 05, 2012 6:19 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. srt34
C srt43
D 7+6srt2
E 12+3srt2

[spoiler]I got 10srt58, which is not in the ans choice. I got this by noticing it's a right triangle with side a=7, b=3, and then finding side c = srt58. Is this a bad question or am I way off track?[/spoiler]

Thanks...
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by Jim@StratusPrep » Tue Jun 05, 2012 9:12 am
This is not a right triangle. Let's give each point a label

A = (0,0)
B = (3,3)
C = (7,0)

AB = 3sqrt2 found via a 45-45-90 triangle
AC = 7 simply the length on the X-axis
BC = 5 If you drop an altitude from (3,3) you will for a 3-4-5 right triangle with the hypotenuse equal to BC

Answer = 12 + 3sqrt2

Also, once you get the 3sqrt2 there is only 1 answer that includes it --- D
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by Anurag@Gurome » Tue Jun 05, 2012 7:45 pm
alex.gellatly 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. srt34
C srt43
D 7+6srt2
E 12+3srt2

[spoiler]I got 10srt58, which is not in the ans choice. I got this by noticing it's a right triangle with side a=7, b=3, and then finding side c = srt58. Is this a bad question or am I way off track?[/spoiler]

Thanks...
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 = [spoiler]12 + 3√2[/spoiler]

The correct answer is E.
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by Matt@VeritasPrep » Thu Mar 17, 2016 9:11 pm
Jim@StratusPrep wrote: Also, once you get the 3sqrt2 there is only 1 answer that includes it --- D
There are two (D and E), and D is wrong.

Another approach to the problem is here.
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