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Intern  Joined: 15 Jul 2012
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A triangle in the xy-coordinate plane has vertices with coordinates (7  [#permalink]

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Difficulty:   45% (medium)

Question Stats: 67% (02:33) correct 33% (02:37) wrong based on 357 sessions

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A triangle in the xy-coordinate plane has vertices with coordinates (7, 0), (0, 8), and (20, 10). What is the area of this triangle?

A. 72
B. 80
C. 87
D. 96
E. 100

Originally posted by anu1706 on 14 Oct 2013, 11:03.
Last edited by Bunuel on 17 May 2017, 04:47, edited 2 times in total.
Renamed the topic, edited the question and added the OA.
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Re: A triangle in the xy-coordinate plane has vertices with coordinates (7  [#permalink]

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Quote:
A triangle in the xy-coordinate plane has vertices with coordinates (7, 0), (0, 8), and (20, 10). What is the area of this triangle?

A. 72
B. 80
C. 87
D. 96
E. 100

acacia wrote:
JackReacher wrote:
Area of trapezoid formed by points (0,0) (8,0) (20,0) (20,10) = ((8+13)/2)*20 = 180

Area of Triangle 1 with points (7,0) (20,0) and (20,10) = 1/2 * 13*10 = 65

Area of Triangle 2 with points (0,0) (7,0) and (8,0) = 1/2 * 7*8 = 28

Required Area = Area of Trapezoid minus Area of Triangle 1 minus Area of triangle 2 = 180-65-28 = 87

I hope it's not confusing without an image.

I'm actually quite confused. These points (0,0) (8,0) (20,0) (20,10) are forming a triangle with points (0,0) (8,0) (20,0) in one line. And from definition, trapezoid is a quadrilateral with 2 parallel lines. Can you please explain again.

Attachment: BOXMETHODTRAPEZOID.png [ 15.25 KiB | Viewed 13385 times ]

acacia , you are correct about those points. They make a straight line.

I think JackReacher got the coordinates mixed up. Reverse (8,0) --> (0,8). The figure is indeed a trapezoid. Look at the figure above. Trapezoid area - green triangles' total area = original triangle.

JackReacher decided to use a trapezoid instead of a rectangle to form right triangles around the original triangle.

And I decided to draw the picture because for me this method of finding the area of an off-kilter triangle is much faster than the formula from hell.

The trapezoid is a clever variation on an easy way to solve this problem called the "Box Method," (which @mynhausen sketched in 2013 and I diagram below):

Typically, draw a rectangle around the triangle.

You find the area of the box you have drawn, and subtract the area of right-angled triangles that the box creates, leaving you with the area of the off-kilter triangle. Here that original triangle is ABC.

1. Draw the smallest possible "box" around the triangle ABC. In the diagram, that is rectangle JKBM

2. Use the given vertices, and stay parallel to the axes. List the coordinates of the rectangle's vertices.

Here those coordinates are J (0,0), K (0,10), B (20, 10), M (20,0). You will have created a rectangular shape with right-angled triangles.

3. Find the area of the triangles that surround the original. In the diagram, those triangles are pastel yellow, blue, and purple.

To find the area of the right triangles, you need the length of their legs. EASY to find. Segment JC, for example, has x-coordinates of 7 and 0. Voila, the length is 7.

4. The area of right triangles is easy to find, and is always (leg length * leg length * 1/2).

PURPLE triangle area is $$\frac{(7 * 8)}{2}$$ = 28
YELLOW triangle area is $$\frac{(2 * 20)}{2}$$ = 20
BLUE triangle area is $$\frac{(13*10)}{2}$$ = 65

5. (PURPLE + YELLOW + BLUE) + WHITE = AREA OF RECTANGLE. Isolate the white triangle ABC.

Add the three triangles whose areas are known: 28 + 20 + 65 = 113

Area of rectangle is 20 * 10 = 200

200 - 113 = 87

6. Area of original triangle ABC is 87

**In coordinate geometry, when you need a triangle's area but the triangle's sides are NOT parallel to the x- and y-axes, one method you can use is the box method because it creates right triangles whose areas are easy to find.

Hope it helps.
Attachment: BOXmethod.png [ 11.59 KiB | Viewed 13386 times ]

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Intern  Joined: 15 Dec 2007
Posts: 12
Re: A triangle in the xy-coordinate plane has vertices with coordinates (7  [#permalink]

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just draw a rectangle. Like on the picture. Then substract 3 triangles from big rectangle.
area of rectangle: 20x10 = 200
area of 1: 8*7/2=28
area of 2: 13*10/2=65
area of 3: 2*20/2=20

Answer: 200 - (28+65+20)= 87.
Attachments trwe.jpg [ 191.43 KiB | Viewed 24949 times ]

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Re: A triangle in the xy-coordinate plane has vertices with coordinates (7  [#permalink]

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anu1706 wrote:
anu1706 wrote:
A triangle in the xy-coordinate plane has vertices with coordinates (7, 0), (0, 8), and (20, 10). What is the area of this triangle?

http://www.mathopenref.com/coordtrianglearea.html

this might help. Note that this method is more helpful here because one of the x-co-ordinates here is 0. The answer is 87 sq units.
There is more than one way of doing this problem.

Hope this helps.
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GMAT 1: 670 Q39 V41 GMAT 2: 730 Q49 V41 Re: A triangle in the xy-coordinate plane has vertices with coordinates (7  [#permalink]

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bump for this one, great method from deadmau!

Can anyone else think of a way that's quick? If you have to figure out the distance between each set of points here that would take forever! Must be some other quicker way, beside's deadmau's
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Re: A triangle in the xy-coordinate plane has vertices with coordinates (7  [#permalink]

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anu1706 wrote:
A triangle in the xy-coordinate plane has vertices with coordinates (7, 0), (0, 8), and (20, 10). What is the area of this triangle?

OA:
87

The formula for solving area of triangle when co-ordinates are given is as follows :

Area of Triangle = ½ [{(X1 –X2)*(Y2 – Y3)} – {(Y1 – Y2)*(X2 – X3)}]

Let,
X1 = 7 , Y1 = 0
X2 = 0, Y2 = 8
X3 = 20, Y3 = 10

Area of Triangle = ½ [{(7-0)*(8-10)} – {(0-8)*(0-20)}]
Area of triangle = ½ * (-174)
Area of triangle = -87
Area cannot be negative, hence drop the negative sign.
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GMAT 1: 730 Q49 V40 Re: A triangle in the xy-coordinate plane has vertices with coordinates (7  [#permalink]

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Area of trapezoid formed by points (0,0) (8,0) (20,0) (20,10) = ((8+13)/2)*20 = 180

Area of Triangle 1 with points (7,0) (20,0) and (20,10) = 1/2 * 13*10 = 65

Area of Triangle 2 with points (0,0) (7,0) and (8,0) = 1/2 * 7*8 = 28

Required Area = Area of Trapezoid minus Area of Triangle 1 minus Area of triangle 2 = 180-65-28 = 87

I hope it's not confusing without an image.
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Re: A triangle in the xy-coordinate plane has vertices with coordinates (7  [#permalink]

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JackReacher wrote:
Area of trapezoid formed by points (0,0) (8,0) (20,0) (20,10) = ((8+13)/2)*20 = 180

Area of Triangle 1 with points (7,0) (20,0) and (20,10) = 1/2 * 13*10 = 65

Area of Triangle 2 with points (0,0) (7,0) and (8,0) = 1/2 * 7*8 = 28

Required Area = Area of Trapezoid minus Area of Triangle 1 minus Area of triangle 2 = 180-65-28 = 87

I hope it's not confusing without an image.

I'm actually quite confused. These points (0,0) (8,0) (20,0) (20,10) are forming a triangle with points (0,0) (8,0) (20,0) in one line. And from definition, trapezoid is a quadrilateral with 2 parallel lines. Can you please explain again.
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Re: A triangle in the xy-coordinate plane has vertices with coordinates (7  [#permalink]

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i was trying to think of the following

1. one side is $$7^2$$+$$8^2$$=$$\sqrt{113}$$. That is the base.

2. The height is 10

Area should be ($$\sqrt{113}$$*10)/2

Any idea why this is incorrect?
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Re: A triangle in the xy-coordinate plane has vertices with coordinates (7  [#permalink]

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nausherwan wrote:
i was trying to think of the following

1. one side is $$7^2$$+$$8^2$$=$$\sqrt{113}$$. That is the base.

2. The height is 10

Area should be ($$\sqrt{113}$$*10)/2

Any idea why this is incorrect?

Remember that the height of a triangle must be perpendicular to the base that you choose. So if you take as the base of this triangle the side connecting (0, 8) and (7, 0), then to find the height, you'd need to draw a line from that base, at 90 degrees, to the opposite point at (20, 10). And that isn't straightforward, because it's not clear where that height will even meet the base. There are ways to figure that out, of course, but they aren't easy. The method in mynhauzen's post above (drawing a rectangle completely surrounding the triangle) is usually the easiest way to solve these kinds of questions.
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A triangle in the xy-coordinate plane has vertices with coordinates (7  [#permalink]

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Super simple question of GMAT Coordinate Geometry Just apply the formula and we can easily find the area of the triangle 174/2 = 87. A triangle in the xy-coordinate plane has vertices with coordinates (7   [#permalink] 13 Feb 2019, 16:14
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