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What is the area of a triangle with the following vertices

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manalq8
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What is the area of a triangle with the following vertices [#permalink] New post   Updated on: 28 Jun 2014, 04:38
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What is the area of a triangle with the following vertices L(1, 3), M(5, 1), and N(3, 5) ?

A. 3
B. 4
C. 5
D. 6
E. 7

M06-09

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D

Originally posted by manalq8 on 12 Jan 2012, 07:55.
Last edited by Bunuel on 28 Jun 2014, 04:38, edited 2 times in total.
Edited the question.
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Re: What is the area of a triangle with the following vertices [#permalink] New post   19 Jan 2012, 06:27
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manalq8 wrote:
What is the area of a triangle with the following vertices \(L(1, 3)\) , \(M(5, 1)\) , and \(N(3, 5)\) ?

3
4
5
6
7

can someone please explain this please? I get 4 but I can't figure out why it is wrong


There is a direct formula to calculate the are of a triangle based on coordinates of its vertices and one could use it to solve this problem.

Though if you make a diagram minimum simple calculations will be needed:

Attachment:
graph1.PNG
graph1.PNG [ 14.56 KiB | Viewed 51557 times ]


Take note that the area of the blue square is \(4^2 = 16\). The area of the red triangle is the area of this blue square minus the combined areas of three smaller triangles situated at the corners. These little triangles have areas of \(\frac{1}{2}*2*2\), \(\frac{1}{2}*2*4\), and \(\frac{1}{2}*4*2\). Hence, the area of triangle LMN is calculated as \(16 - (2 + 4 + 4) = 6\).

Answer: D.

Hope it's clear.
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Re: What is the area of a triangle with the following vertices [#permalink] New post   01 Mar 2012, 16:38
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gmatpunjabi wrote:
Hey Buunel what is the formula for the area of the Triangle Based on the Vertices?


I really doubt that you'll need it for the GMAT but here you go:

If the vetices 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:

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

So for: \(L(1, 3)\), \(M(5, 1)\), and \(N(3, 5)\) the area will be: \(area=|\frac{1(1-5)+5(5-3)+3(3-1)}{2}|=6\).
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Re: What is the area of a triangle with the following vertices [#permalink] New post   19 Jan 2012, 00:45
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Plot the 3 points.
The base is distance between A(1,3) and B(3,5) = sqrt(8)
Height is distance between C(5,1) and (2,4). 2,4 is the midpoint. So height is sqrt(18)
Area =1/2*sqrt(8)* sqrt(18)=6
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Re: What is the area of a triangle with the following vertices [#permalink] New post   19 Jan 2012, 05:31
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NM=sqroot((5-3)^2+(1-5)^2)=2sqroot5
LN=sqroot((1-3)^2+(3-5)^2)=2sqroot2
LM=sqroot((5-1)^2+(1-3)^2)=2sqroot5

height of the triangle=Sqroot(LM^2-(LN/2)^2))=3sqroot2

so,area=1/2*LN*height=1/2*2sqroot2*3sqroot2=2*3=6
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Re: What is the area of a triangle with the following vertices [#permalink] New post   01 Mar 2012, 14:41
Hey Buunel what is the formula for the area of the Triangle Based on the Vertices?
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Re: What is the area of a triangle with the following vertices [#permalink] New post   01 Mar 2012, 23:03
THank you for the equation!!
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Re: What is the area of a triangle with the following vertices [#permalink] New post   30 Mar 2012, 21:34
Bunuel wrote:
gmatpunjabi wrote:
Hey Buunel what is the formula for the area of the Triangle Based on the Vertices?


I really doubt that you'll need it for the GMAT but here you go:

If the vetices 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:

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

So for: \(L(1, 3)\), \(M(5, 1)\), and \(N(3, 5)\) the area will be: \(area=|\frac{1(1-5)+5(5-3)+3(3-1)}{2}|=6\).


Formula is good. :) +1
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Re: What is the area of a triangle with the following vertices [#permalink] New post   06 Mar 2013, 11:29
In the above question, two sides are same that is 2root5 and the third side is 2root 2. Can't this be an isosceles triangle? I know that I am missing something, please help
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Re: What is the area of a triangle with the following vertices [#permalink] New post   06 Mar 2013, 12:46
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Hello Sharmila,

Yes the above triangle can is an isolsceles triangle and the area can be calculated using the formula (1/2)*b*h where h=(hypotenuse^2(1/4)(base)^2).

h=(2*sqrt(5))^2-((1/4)(2*sqrt(2))^2

It would be great if you could highlight your doubt here.

sharmila79 wrote:
In the above question, two sides are same that is 2root5 and the third side is 2root 2. Can't this be an isosceles triangle? I know that I am missing something, please help

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Re: What is the area of a triangle with the following vertices [#permalink] New post   07 Mar 2013, 09:46
Kris01 wrote:
Hello Sharmila,

Yes the above triangle can is an isolsceles triangle and the area can be calculated using the formula (1/2)*b*h where h=(hypotenuse^2(1/4)(base)^2).

h=(2*sqrt(5))^2-((1/4)(2*sqrt(2))^2

It would be great if you could highlight your doubt here.

sharmila79 wrote:
In the above question, two sides are same that is 2root5 and the third side is 2root 2. Can't this be an isosceles triangle? I know that I am missing something, please help

Hi Kris,
Thanks for the detailed explanation. As soon as I got the lengths I decided that it is a 45:45:90 right triangle and was stuck with that. I was aware that the short leg and long leg (base and height) are longer than the hypotenuse, but still couldn't bring the diagram (your attachment) into picture. My doubt is cleared. I have one more question, if two sides of a triangle are of same length and that length is shorter than the third side, then it has to be a 45:45:90 isosceles triangle. Right?
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Re: What is the area of a triangle with the following vertices [#permalink] New post   07 Mar 2013, 11:13
Hello Sharmila,

To find whether the angles in a triangle are 45-45-90 you would also need to confirm whether the ratio of the sides is 1:1:sqrt(2).

For example, imagine an isosceles triangle with equal sides of 5 cm in length and forming a 120 degree angle between them. The equal angles would be 30 deg each. The third side will still be longer than the equal sides.

Hope this helps! Let me know if I can help you any further.

sharmila79 wrote:
Kris01 wrote:
Hello Sharmila,

Yes the above triangle can is an isosceles triangle and the area can be calculated using the formula (1/2)*b*h where h=(hypotenuse^2(1/4)(base)^2).

h=(2*sqrt(5))^2-((1/4)(2*sqrt(2))^2

It would be great if you could highlight your doubt here.

sharmila79 wrote:
In the above question, two sides are same that is 2root5 and the third side is 2root 2. Can't this be an isosceles triangle? I know that I am missing something, please help

Hi Kris,
Thanks for the detailed explanation. As soon as I got the lengths I decided that it is a 45:45:90 right triangle and was stuck with that. I was aware that the short leg and long leg (base and height) are longer than the hypotenuse, but still couldn't bring the diagram (your attachment) into picture. My doubt is cleared. I have one more question, if two sides of a triangle are of same length and that length is shorter than the third side, then it has to be a 45:45:90 isosceles triangle. Right?
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Re: What is the area of a triangle with the following vertices [#permalink] New post   07 Mar 2013, 12:38
Thanks Kris! I was actually using that 1:1:root2, but did not make any sense with the values I got.
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Re: What is the area of a triangle with the following vertices [#permalink] New post   27 Jun 2014, 09:40
Thanks for the formula. I guess its the best way to solve this type of problem. Any suggestions?

Posted from my mobile device
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Re: What is the area of a triangle with the following vertices [#permalink] New post   22 Jul 2015, 09:14
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basically vertices are L(1, 3), M(5, 1), and N(3, 5), since only area is asked here we can subtract 1 from each of the x and y co-ordinates which makes the new points as L(0, 2), M(4, 0), and N(2, 4) which makes it a right angled traingle framed on the x and y axes with corner at N(2, 4). i t wouldn't be hard to calculate the area of new traingle as 4sq.units.
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Re: What is the area of a triangle with the following vertices [#permalink] New post   18 Aug 2016, 06:22
Bunuel wrote:
manalq8 wrote:
What is the area of a triangle with the following vertices \(L(1, 3)\) , \(M(5, 1)\) , and \(N(3, 5)\) ?

3
4
5
6
7

can someone please explain this please? I get 4 but I can't figure out why it is wrong


There is a direct formula to calculate the are of a triangle based on coordinates of its vertices and one could use it to solve this problem.

Though if you make a diagram minimum simple calculations will be needed:
Attachment:
graph1.PNG
Notice that the area of the blue square is 4^2=16 and the area of the red triangle is 16 minus the areas of 3 little triangles which are in the corners (2*2/2, 4*2/2 and 4*2/2). Therefore the area of a triangle LMN=16-(2+4+4)=6.

Answer: D.

Hope it's clear.


Nice solution. Thanks.

I started my process by calculating the base as LM and were going to calculate the perpendicular line to LM that went through point N to get the height. But at that point I realized it would be time consuming and I had to be doing something wrong.
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Re: What is the area of a triangle with the following vertices [#permalink] New post   08 Nov 2018, 20:34
manalq8 wrote:
What is the area of a triangle with the following vertices L(1, 3), M(5, 1), and N(3, 5) ?

A. 3
B. 4
C. 5
D. 6
E. 7

M06-09


formula for the triangle,whose vertices are given:half of determinant of the three vertices with 1 added to each set as a third parameter--->1/2*det((1,3,1), (5,1,1), (3,5,1))
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Re: What is the area of a triangle with the following vertices [#permalink] New post   11 Nov 2018, 21:55
A = 1/2 | x1 y1 |
|x2 y2 |
|x3 y3|
|x1 y1|
=> Cross multiply A = 1/2 [(x1y2 - x2y1)+(x2y3- x3y2)+(x3y1-x1y3)]
insert values => 1/2 [(1-15)+(25-3)+(9-5)] = 6 Answer !
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Re: What is the area of a triangle with the following vertices [#permalink] New post   20 Jul 2020, 09:26
manalq8 wrote:
What is the area of a triangle with the following vertices L(1, 3), M(5, 1), and N(3, 5) ?

A. 3
B. 4
C. 5
D. 6
E. 7

M06-09



This question can easily be solved using the Heron's formula --> sqrt(s(s-a)(s-b)(s-c))

s = (a + b + c)/2 ; a, b, c are the distances between any two points say LM, MN and LN. This distance can be calculated using the formula -->

sqrt(x2 - x1)^2 + (y2 - y1)^2)

Once you have each of the values a, b and c, plug it into the Heron's formula to get the area of the triangle
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Re: What is the area of a triangle with the following vertices [#permalink] New post   30 Aug 2021, 02:31
manalq8 wrote:
What is the area of a triangle with the following vertices L(1, 3), M(5, 1), and N(3, 5) ?

A. 3
B. 4
C. 5
D. 6
E. 7

M06-09


IMO D

Area of a triangle when coordinates are given = | x1(y2-y3) + x2(y3-y1) + x3(y1-y2) | / 2

Area = | -4 +10 + 6 | / 2 = 6
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Re: What is the area of a triangle with the following vertices [#permalink]
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