Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

1

This post received KUDOS

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

9

This post received KUDOS

Expert's post

5

This post was BOOKMARKED

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 6030 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

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Type of Visa: You will be applying for a Non-Immigrant F-1 (Student) US Visa. Applying for a Visa: Create an account on: https://cgifederal.secure.force.com/?language=Englishcountry=India Complete...

I started running back in 2005. I finally conquered what seemed impossible. Not sure when I would be able to do full marathon, but this will do for now...