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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:


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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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?


https://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]
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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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]
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]
Super simple question of GMAT Coordinate Geometry :)

Just apply the formula and we can easily find the area of the triangle :)

174/2 = 87.

Answer C.
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Re: A triangle in the xy-coordinate plane has vertices with coordinates (7 [#permalink]
The formula when three coordinates of vertices of triangle are given:

|Ax(By-Cy) + Bx(Cy-Ay) + Cx(Ay-By)| / 2

Put values in formula and you will get the answer
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Re: A triangle in the xy-coordinate plane has vertices with coordinates (7 [#permalink]
Official Explanation:


It’s not immediately possible to use the standard formula A = bh / 2 for the area of this triangle, because none of the triangle’s sides is horizontal or vertical. However, if the triangle is circumscribed by a rectangle, as depicted below, the areas of the three surrounding triangles can readily be found and then subtracted from the rectangle’s area to yield the desired result.



First, the triangle to the lower left has b = 7 and h = 8, so its area is (7)(8) / 2 = 28.
Second, the triangle to the lower right has b = 13 and h = 10, so its area is (13)(10) / 2 = 65.
Third, the topmost triangle has b = 20 and h = 2, so its area is (20)(2) / 2 = 20.

The total area of the surrounding triangles—that is, the triangles that are not part of the desired area—is 28 + 65 + 20 = 113 square units. The total area of the rectangle is 20 x 10 = 200 square units, so the triangle’s area is 200 – 113 = 87 square units.

The correct answer is C.
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Re: A triangle in the xy-coordinate plane has vertices with coordinates (7 [#permalink]
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Re: A triangle in the xy-coordinate plane has vertices with coordinates (7 [#permalink]
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