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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 (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?
[Reveal] Spoiler: OA

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Last edited by Bunuel on 31 Oct 2016, 23:23, edited 1 time in total.
Renamed the topic.
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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 ...

 ! Please post PS questions in the PS subforum: gmat-problem-solving-ps-140/Please post DS questions in the DS subforum: gmat-data-sufficiency-ds-141/

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 = $$12 + 3\sqrt{2}$$

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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
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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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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

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

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Re: In the xy plane, the vertices of a triangle have coordinates [#permalink]

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10 Aug 2015, 02:38
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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?

Regards
Megha
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Posts: 39702
Re: In the xy plane, the vertices of a triangle have coordinates [#permalink]

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31 Jul 2016, 02:25
megha_2709 wrote:
Hi Bunnuel,

Is there any formula to calculate area of triangle when all coordinates are known?

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:

$$area=|\frac{a_x(b_y-c_y)+b_x(c_y-a_y)+c_x(a_y-b_y)}{2}|$$.
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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?

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)

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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)
Attachments

Triangle.jpg [ 685.55 KiB | Viewed 1477 times ]

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Re: In the xy plane, the vertices of a triangle have coordinates   [#permalink] 14 Nov 2016, 07:09
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