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# M06-09

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Manager
Joined: 06 Jul 2013
Posts: 59
Location: United States
GMAT 1: 710 Q48 V40

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11 Aug 2018, 20:00
Bunuel wrote:
Official Solution:

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

Make a diagram:

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 ($$2*\frac{2}{2}$$, $$4*\frac{2}{2}$$ and $$4*\frac{2}{2}$$). Therefore, the area of a triangle LMN is $$16-(2+4+4)=6$$.

Hi Bunuel,

Why can't the height be 4 and the base be 4 as well? Height being the base of the square to point N, and the base being point M to the left side of the square? So b*h/2 or 4*4/2 = 8
Math Expert
Joined: 02 Sep 2009
Posts: 52902

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12 Aug 2018, 00:57
TippingPoint93 wrote:
Bunuel wrote:
Official Solution:

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

Make a diagram:

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 ($$2*\frac{2}{2}$$, $$4*\frac{2}{2}$$ and $$4*\frac{2}{2}$$). Therefore, the area of a triangle LMN is $$16-(2+4+4)=6$$.

Hi Bunuel,

Why can't the height be 4 and the base be 4 as well? Height being the base of the square to point N, and the base being point M to the left side of the square? So b*h/2 or 4*4/2 = 8

Any side of a triangle can be considered as base and similarly any perpendicular from the opposite vertex to the corresponding base can be considered as height. So, for example, if you consider ML to be base, then the height would be perpendicular from N to ML. None of the sides equals 4 in the triangle MNL and none of the heights is equal to 4. That's why your way is not correct.
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Director
Joined: 30 Jan 2016
Posts: 924
Location: United States (MA)

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Updated on: 08 Sep 2018, 00:36
Another approach is to use Pick's formula:

Area= M/2 + N - 1,

where
M of lattice points on the boundary placed on the polygon's perimeter
N - lattice points in the interior located in the polygon

M - green dots;
N - blue dots

Area = 6/3 +4 - 1 = 3+4-1 = 6
>> !!!

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Non progredi est regredi

Originally posted by Akela on 08 Sep 2018, 00:28.
Last edited by Akela on 08 Sep 2018, 00:36, edited 1 time in total.
Director
Joined: 30 Jan 2016
Posts: 924
Location: United States (MA)

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08 Sep 2018, 00:33
Find the area of the following polygon, if the grid is 1 cm X 1cm
>> !!!

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Non progredi est regredi

Intern
Joined: 02 Feb 2018
Posts: 36

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07 Nov 2018, 06:46
dharan wrote:
The area of triangle when 3 vertices are given by

= 1/2 [ x1(y2-y3) + x2(y3-y1) + x3(y1-y2)]

x1,y1 = L(1,3 )
x2,y2 = M(5,1)
x3,y3 = N(3,5)

Substituting the 3 values in the formula we get

=> 1/2 [ 1 ( 1 -5 ) + 5( 5 - 3 ) + 3 ( 3 - 1 )]
= 1/2 [ -4 + 10 + 6 ] = 12/2 = 6

the tricky here is to remembering the correct sequence of the formula.
my short cut to remember this formula is to do a circular rotate first three vertices ( x1,y2,y3) by 1 to get second one ( x2,y3,y1) and do the rotate again to get the third one ( x3,y1,y2). OR you can form a 3x3 matrix to remember well.
1/2 [1,2,3 + 2,3,1 + 3,1,2].
Not sure if this is the wise approach but I could solve faster.

I know its a bit random but does this formula have a name?
Manager
Joined: 09 Jun 2014
Posts: 239
Location: India
Concentration: General Management, Operations
Schools: Tuck '19

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09 Nov 2018, 00:43
good question..but no where close to GMAT styleoutside GMAT scope..
Intern
Joined: 08 Mar 2018
Posts: 3

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11 Nov 2018, 15:54
As mentioned above: It can be better solved using the determinants. Easy TIA
Re: M06-09   [#permalink] 11 Nov 2018, 15:54

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# M06-09

Moderators: chetan2u, Bunuel

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