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What is the area of a triangle with the following vertices [#permalink]
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12 Jan 2012, 06:55
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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 M0609
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Last edited by Bunuel on 28 Jun 2014, 03:38, edited 2 times in total.
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Re: What is the area of a triangle with the following vertices [#permalink]
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18 Jan 2012, 23:45
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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]
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19 Jan 2012, 04:31
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NM=sqroot((53)^2+(15)^2)=2sqroot5 LN=sqroot((13)^2+(35)^2)=2sqroot2 LM=sqroot((51)^2+(13)^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]
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19 Jan 2012, 05:27
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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 [ 14.56 KiB  Viewed 24306 times ]
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.
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Re: What is the area of a triangle with the following vertices [#permalink]
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01 Mar 2012, 13:41
Hey Buunel what is the formula for the area of the Triangle Based on the Vertices?



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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: \(area=\frac{a_x(b_yc_y)+b_x(c_ya_y)+c_x(a_yb_y)}{2}\). So for: \(L(1, 3)\), \(M(5, 1)\), and \(N(3, 5)\) the area will be: \(area=\frac{1(15)+5(53)+3(31)}{2}=6\).
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Re: What is the area of a triangle with the following vertices [#permalink]
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01 Mar 2012, 22:03
THank you for the equation!!



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Re: What is the area of a triangle with the following vertices [#permalink]
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30 Mar 2012, 20:34
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: \(area=\frac{a_x(b_yc_y)+b_x(c_ya_y)+c_x(a_yb_y)}{2}\). So for: \(L(1, 3)\), \(M(5, 1)\), and \(N(3, 5)\) the area will be: \(area=\frac{1(15)+5(53)+3(31)}{2}=6\). Formula is good. +1
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Re: What is the area of a triangle with the following vertices [#permalink]
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06 Mar 2013, 10:29
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]
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06 Mar 2013, 11:46
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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]
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07 Mar 2013, 08:46
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]
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07 Mar 2013, 10:13
Hello Sharmila, To find whether the angles in a triangle are 454590 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]
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07 Mar 2013, 11:38
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]
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27 Jun 2014, 08:40
Thanks for the formula. I guess its the best way to solve this type of problem. Any suggestions?
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What is the area of a triangle with the following vertices [#permalink]
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22 Jul 2015, 08:14
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 coordinates 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]
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18 Aug 2016, 05:22
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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