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Manager  Joined: 06 Feb 2010
Posts: 138
Schools: University of Dhaka - Class of 2010
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In the xy-plane, the vertices of a triangle have coordinates (0,0),(3,  [#permalink]

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Question Stats: 88% (01:40) correct 12% (01:35) wrong based on 269 sessions

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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?

Originally posted by monirjewel on 28 Oct 2010, 22:18.
Last edited by Bunuel on 31 Oct 2016, 23:23, edited 1 time in total.
Renamed the topic.
Math Expert V
Joined: 02 Sep 2009
Posts: 59712
Re: In the xy plane, the vertices  [#permalink]

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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 of a triangle have coordinates (0,0),(3,  [#permalink]

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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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Manager  Joined: 06 Feb 2010
Posts: 138
Schools: University of Dhaka - Class of 2010
GPA: 3.63
In the xy plane, the vertices of a triangle have coordinates  [#permalink]

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3
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
Senior Manager  Joined: 13 Aug 2012
Posts: 397
Concentration: Marketing, Finance
GPA: 3.23
Re: In the xy plane, the vertices  [#permalink]

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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

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

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

Manager  Joined: 05 Sep 2014
Posts: 64
Schools: IIMB
Re: In the xy plane, the vertices of a triangle have coordinates  [#permalink]

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Hi Bunnuel,

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

Regards
Megha
Math Expert V
Joined: 02 Sep 2009
Posts: 59712
Re: In the xy plane, the vertices of a triangle have coordinates  [#permalink]

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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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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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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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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 10596 times ]

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

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_________________ Re: In the xy plane, the vertices of a triangle have coordinates   [#permalink] 15 Oct 2019, 03:06
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