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...

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

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.

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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