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ps -triangle.

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ps -triangle. [#permalink]

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New post 13 Sep 2008, 13:55
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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)?

3
4
5
6
7
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Re: ps -triangle. [#permalink]

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New post 13 Sep 2008, 16:55
gmatnub wrote:
x2suresh 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


6


would like to see your approach.
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New post 13 Sep 2008, 17:22
Here is how I would approach this.
(1) Use the distance formula and find the length of the sides.
(2) Use the triangle formula (A=b*h) to calculate the area. Note that the height of the triangle is found using the right angle perpendicular.

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New post 13 Sep 2008, 17:53
Yes. it should be 6.. D
This question can not be solved in 2 mins.. I guess. I used the following way :

1 ) distance between the points ( you will have to do this verbally as we wont have much time ) L(1, 3), M(5, 1), and N(3, 5) are 2v5 , 2v5 and 2v2 .. so its an isosceles triangele

2 ) now considering 2v2 as the base, the height will be = v[ (2v5)^2 - (v2)^2 ] { V2- as it is the half of the base }
= 3v2

3 ) area= 1/2*h*b = 1/2* 2v2 * 3v2 = 6

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New post 13 Sep 2008, 17:55
Method I:
When the points are plotted, and when the triangle is inscribed in a square, the side of the square is 4.

Area of required triangle = Area of sq- (area of 3 rt. triangles)
= 16- (4+2+4)
= 6

Method 2:

To find the area of the triangle when all the sides are gives as: a,b,c.
Semi-perimeter of the triangle s = (a+b+c)/2
and area A = sqrt[s(s-a)(s-b)(s-c)]

Here a, b, c are 2sqrt(2), 2sqrt(5) and 2sqrt(5).
Area = sqrt[ (2V5+V2) (2V5-V2) (V2) (V2)]
=sqrt(18*2)
=6

Hence D
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Last edited by leonidas on 13 Sep 2008, 18:07, edited 2 times in total.

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New post 13 Sep 2008, 18:03
Wao man.. you are really bringing up coordinate geometry here .. neways, I tried your formula and it works..

A = sqrt[s(s-a)(s-b)(s-c)]

s=(a+b+c)/2 = (2v5 +2v5 +2v2)/2= 2v5 +v2

now s(s-a)(s-b)(s-c) = (2v5 +v2) ( v2)(v2)(2v5-v2) = (20-2)* 2= 36

so area= v36 = 6

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New post 13 Sep 2008, 18:09
ssandeepan wrote:
Wao man.. you are really bringing up coordinate geometry here .. neways, I tried your formula and it works..

A = sqrt[s(s-a)(s-b)(s-c)]

s=(a+b+c)/2 = (2v5 +2v5 +2v2)/2= 2v5 +v2

now s(s-a)(s-b)(s-c) = (2v5 +v2) ( v2)(v2)(2v5-v2) = (20-2)* 2= 36

so area= v36 = 6


ssandeepan.... Thanks....I made a mistake earlier- used 2V2 instead of 2V5.... It will suck if I made these kind of mistakes in the real test.
I corrected that and re-posted.
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Re: ps -triangle. [#permalink]

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New post 13 Sep 2008, 20:29
leonidas' Approach #1 is the best way.

Find the area of the square and then subrtract out the area of the 3 right triangles with the remainder being the area of the triangle in question.

I found the area of the square [yellow + orange] and then subtracted out the individual yellow triangles. Much easier since all 3 yellow tiangles are right triangles.

Attachment:
AreaOfTriangle.jpg
AreaOfTriangle.jpg [ 56.59 KiB | Viewed 849 times ]

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Re: ps -triangle. [#permalink]

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New post 13 Sep 2008, 21:05
jallenmorris wrote:
leonidas' Approach #1 is the best way.

Find the area of the square and then subrtract out the area of the 3 right triangles with the remainder being the area of the triangle in question.

I found the area of the square [yellow + orange] and then subtracted out the individual yellow triangles. Much easier since all 3 yellow tiangles are right triangles.

Attachment:
AreaOfTriangle.jpg


Nice diagram!!!!!
I guess I was too lazy to draw one.......
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Re: ps -triangle. [#permalink]

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New post 13 Sep 2008, 21:13
you're not lazy. I already had photoshop open for a different project so I did it.
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Re: ps -triangle. [#permalink]

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New post 13 Sep 2008, 23:31
x2suresh 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


IMO D

find the sides first : MN=ML=20^1/2
NL=8^1/2

hence find the height by joining mid point of NL to M ,P is the midpoint (2,4) .
hence height is 3 * 2^1/2

hence area=1/2 * NL* MP=6
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Re: ps -triangle.   [#permalink] 13 Sep 2008, 23:31
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