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What is the area of four sided figure formed by the intersection of  [#permalink]

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Difficulty:   55% (hard)

Question Stats: 68% (02:16) correct 32% (02:20) wrong based on 985 sessions

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What is the area of four sided figure formed by the intersection of the set of lines represented by |2x| + |y| = 6?

A. 18
B. 24
C. 30
D. 36
E. 48
Math Expert V
Joined: 02 Sep 2009
Posts: 57996
Re: What is the area of four sided figure formed by the intersection  [#permalink]

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What is the area of four sided figure formed by the intersection of the set of lines represented by |2x| + |y| = 6?

A. 18
B. 24
C. 30
D. 36
E. 48

Can someone give a diagrammatic answer to this? Would be really helpful to understand what kind of rhombus or square gets created :)

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GMAT 1: 750 Q49 V43 Re: What is the area of four sided figure formed by the intersection of  [#permalink]

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Turkish wrote:
What is the area of four sided figure formed by the intersection of the set of lines represented by |2x|+|y| = 6?

1. 18
2.24
3.30
4.36
5.48

we just need four XY pair to define the four-sided figure.
Lets works with the concept of X and Y intercepts
When |x|=0; then |y|=6 .. right
y=-6 and 6 ; XY pairs are {0,-6} and {0,6}

when |y|=0 then |2x|= 6
x=-3 and 3 ; The XY pairs are (-3,0 and 3,0)

When you plot these XY pairs you will see that you get a diamond shaped figure (an upright RHOMBUS ) which can be divided into 2 equal triangles. the base of these two triangles is the x-axis

Now area of a triangle = 1/2 * base * height = 1/2 * 6*6 (the distance between -3 and 3 is 6) and the height is 6-0 = 6
Therefore Area of the triangle = $$\frac{1}{2}*36$$ =====>18
There are 2 such triangle
Total area = 2*18 =36
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Originally posted by LogicGuru1 on 24 Jul 2016, 23:29.
Last edited by LogicGuru1 on 05 Oct 2016, 13:41, edited 1 time in total.
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Re: What is the area of four sided figure formed by the intersection of  [#permalink]

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Turkish wrote:
What is the area of four sided figure formed by the intersection of the set of lines represented by |2x|+|y| = 6?

1. 18
2.24
3.30
4.36
5.48

took me some time to draw some points...
but i believe the easiest way is:
x=0, then y=6 or -6.
y=0, x=3 or -3.
we get 4 similar 90 degrees triangles, with the 90 degree angle at the origin.
short leg is 3, long leg is 6.
area is leg*leg/2 = 18/2 = 9.
since we have 4 of these triangles, the area is 9*4 = 36.

D
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Re: What is the area of four sided figure formed by the intersection of  [#permalink]

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4
9
We need to find the equation |2X| + |Y| = 6

Equation 1 : 2X + Y = 6 which is Y = -2X + 6 . Points where this lines intersects the X and Y axis ( X = 0 , Y = 6 ) , ( X = 3 , Y = 0 )
Equation 2 : -2X - Y = 6 which is Y = -2X - 6 . Points where this lines intersects the X and Y axis ( X = 0 , Y = -6 ) , ( X = -3 , Y = 0 )

From here we know the intersects of the lines ,
Using area of triangles = 0.5 * X * Y
= 0.5 * 3 * 6
= 9
Using symmetry 4 triangles = 9 * 4
= 36 .
Answer is D . ( Sketch a diagram to visualize it helps !! )
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Re: What is the area of four sided figure formed by the intersection of  [#permalink]

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Is it correct if I assume that the area is the product of the following two equations?

2x+y=6
2x+y=-6
Intern  Joined: 06 Jul 2015
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Re: What is the area of four sided figure formed by the intersection of  [#permalink]

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2
These are four equations. If you carefully observe these equations because of the way mod functions are plotted around y and x axis as opposite images, these four lines form a rhombus with diagonals of lengths as 6 and 12 (if x =0 then possible y values 0,6 and 0,-6) and then (if y =0 x intercepts at 3,0 and -3,0. Area of rhombus is 1/2 * (product of diagonals) = 1/2 * 12 *6 = 36.
Manager  B
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Re: What is the area of four sided figure formed by the intersection of  [#permalink]

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What's the fastest way to solve this?

I took almost 4 mins. First opened mods and found the 4 equations. Then found 2 points per equation. Then found area of one triangle and multiplied it by 4 (for area of the 4 triangles formed)
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Re: What is the area of four sided figure formed by the intersection of  [#permalink]

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Can someone give a diagrammatic answer to this? Would be really helpful to understand what kind of rhombus or square gets created :)
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Re: What is the area of four sided figure formed by the intersection of  [#permalink]

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Please correct me if i am wrong.

|2x|+|y|=6

When, x=2, y=2,-2 Points are (2,2) and (2,-2)
When, x=-2 y=2,-2 Points are (-2,2) and (-2,-2)

I get a square with an area of (4x4) =16.
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Re: What is the area of four sided figure formed by the intersection of  [#permalink]

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Rishovnits wrote:
Please correct me if i am wrong.

|2x|+|y|=6

When, x=2, y=2,-2 Points are (2,2) and (2,-2)
When, x=-2 y=2,-2 Points are (-2,2) and (-2,-2)

I get a square with an area of (4x4) =16.

It's a rhombus. Check here: https://gmatclub.com/forum/what-is-the- ... l#p1910705
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What is the area of four sided figure formed by the intersection of  [#permalink]

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From the equation, we know that for all the four lines:
y-intercept is either +6 or -6
&
x-intercept is either +3 or -3

So, the shape of the quadrilateral is a rhombus with one diagnol length as 12 (from +6 to -6) and other diagnol length as 6 (from +3 to -3)

Therefore, Area of Rhombus= (d1xd2)/2 = (12x6)/2 = 36.

Ans D.

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What is the area of four sided figure formed by the intersection of  [#permalink]

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Turkish wrote:
What is the area of four sided figure formed by the intersection of the set of lines represented by |2x| + |y| = 6?

A. 18
B. 24
C. 30
D. 36
E. 48

By drawing a diagram we see that we have a rhombus .We know that each side of the rhombus is X so $$Area=base*hight=x^2$$ must be a perfect square of a number D. $$36=6^2$$ fits
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Re: What is the area of four sided figure formed by the intersection of  [#permalink]

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Turkish wrote:
What is the area of four sided figure formed by the intersection of the set of lines represented by |2x| + |y| = 6?

A. 18
B. 24
C. 30
D. 36
E. 48

Can't (1,4) (-1,4)
(2,2), (-2,-2) be used to solve the problem? Maybe I didn't understand the question properly... Re: What is the area of four sided figure formed by the intersection of   [#permalink] 16 Oct 2018, 04:37
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