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A triangle in the xy-coordinate plane has vertices with coordinates (7

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

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New post Updated on: 17 May 2017, 04:47
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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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New post 03 Nov 2013, 11:01
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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.
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Re: A triangle in the xy-coordinate plane has vertices with coordinates (7  [#permalink]

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New post 15 Oct 2013, 02:27
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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?

OA please


Can please somebody give the answer?


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

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New post 03 Nov 2013, 10:40
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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New post 20 Nov 2014, 09:50
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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.
Answer is 87
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Re: A triangle in the xy-coordinate plane has vertices with coordinates (7  [#permalink]

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New post 07 Mar 2017, 12:03
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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New post 06 Aug 2017, 10:26
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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New post 06 Aug 2017, 19:47
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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
BOXMETHODTRAPEZOID.png [ 15.25 KiB | Viewed 7278 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

Answer C

**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
BOXmethod.png [ 11.59 KiB | Viewed 7278 times ]

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Re: A triangle in the xy-coordinate plane has vertices with coordinates (7 &nbs [#permalink] 06 Aug 2017, 19:47
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