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Re: What is the area of a triangle with the following vertices [#permalink]
18 Jan 2012, 23:45

1

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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

Re: What is the area of a triangle with the following vertices [#permalink]
19 Jan 2012, 05:27

10

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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 5143 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.

Re: What is the area of a triangle with the following vertices [#permalink]
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

Re: What is the area of a triangle with the following vertices [#permalink]
06 Mar 2013, 11:46

1

This post received KUDOS

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

Re: What is the area of a triangle with the following vertices [#permalink]
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?

Re: What is the area of a triangle with the following vertices [#permalink]
07 Mar 2013, 10:13

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?

Re: What is the area of a triangle with the following vertices [#permalink]
27 Jun 2014, 08:17

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