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In the xy-plane, the vertices of a triangle have coordinates (0,0),(3, [#permalink]
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 28 Oct 2010, 22:18
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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) (14) (C) (16) (D) (17) (E) 12+3 root 2 How to draw graph on this?
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Last edited by Bunuel on 31 Oct 2016, 23:23, edited 1 time in total.
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Re: In the xy-plane, the vertices of a triangle have coordinates (0,0),(3, [#permalink]
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29 Oct 2010, 00:27
Moving to PS sub-forum ... To answer this question ... all you need is distance between vertices : (0,0) & (7,0) : 7 (0,0) & (3,3) : \(3\sqrt{2}\) (7,0) & (3,3) : \(\sqrt{16+9}=5\) Perimeter = sum of these three sides = \(12 + 3\sqrt{2}\) Answer : e
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In the xy plane, the vertices of a triangle have coordinates [#permalink]
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05 Nov 2010, 20:31
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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
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Re: In the xy plane, the vertices [#permalink]
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05 Nov 2010, 21:08
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 \(d=\sqrt{(3-7)^2+(3-0)^2}=5\); \(P=7+3\sqrt{2}+5=12+3\sqrt{2}\). Answer: E. For more check Coordinate Geometry chapter of Math Book: math-coordinate-geometry-87652.htmlHope it helps.
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Re: In the xy plane, the vertices [#permalink]
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03 Jan 2013, 04:10
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 \(P = 7 + 5 + 3\sqrt{2} = 12 + \sqrt{2}\) Answer: E
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Re: In the xy plane, the vertices of a triangle have coordinates [#permalink]
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31 Jul 2016, 02:15
Hi Bunnuel,
Is there any formula to calculate area of triangle when all coordinates are known?
Please help. Many thanks in advance for your kind response
Regards
Megha
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Re: In the xy plane, the vertices of a triangle have coordinates [#permalink]
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31 Jul 2016, 02:25
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Re: In the xy plane, the vertices of a triangle have coordinates [#permalink]
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01 Aug 2016, 13:42
Bunuel wrote: megha_2709 wrote: Hi Bunnuel,
Is there any formula to calculate area of triangle when all coordinates are known?
Please help. Many thanks in advance for your kind response
Regards
Megha I really doubt that you'll need it for the GMAT but here you go: If the vertices of a triangle are: \(A(a_x, a_y)\), \(B(b_x, b_y)\) and \(C(c_x,c_y)\) then the area of ABC is: \(area=|\frac{a_x(b_y-c_y)+b_x(c_y-a_y)+c_x(a_y-b_y)}{2}|\). Ohh, A day before I saw question like that on gmat club. Anyways many thanks for your response and telling me its highly unlikely to come.  Regards Megha
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Re: In the xy-plane, the vertices of a triangle have coordinates (0,0),(3, [#permalink]
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31 Oct 2016, 17:09
Relatively straight-forward problem, though I'd suggest plotting it so you can visualize the sides and what two side lengths you will need to calculate.
Base length is 7
(1) Distance between (0,0) & (3,3) --> sqrt[(3^2)+(3^2)] = 3sqrt(2)
(2) Distance between (3,3) & (7,0) --> sqrt[(3-0)^2 + (3-7)^2] = sqrt(9+16) = sqrt(25) =5
Add them all up --> 12+ 3sqrt(2)
E
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Re: In the xy plane, the vertices of a triangle have coordinates [#permalink]
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14 Nov 2016, 07:09
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) (14)
(C) (16)
(D) (17)
(E) 12+3 root 2
How to draw graph on this? Plot the graph and you will get two triangles one is pythagorean triples(3, 4, 5) and 3, 3, and 3sqrt(2) by this perimeter is 12+3sqrt(2)
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Triangle.jpg [ 685.55 KiB | Viewed 1465 times ]
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Re: In the xy plane, the vertices of a triangle have coordinates
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