In the rectangular coordinate system above the area of : GMAT Problem Solving (PS)
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# In the rectangular coordinate system above the area of

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22 Nov 2010, 05:27
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In the rectangular coordinate system above the area of triangle PQR is what fraction of the area triangle LMN?

A. 1/9
B. 1/8
C. 1/6
D. 1/5
E. 1/3
[Reveal] Spoiler: OA

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22 Nov 2010, 06:43
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In the rectangular coordinate system above the area of triangle PQR is what fraction of the area triangle LMN?

A. 1/9
B. 1/8
C. 1/6
D. 1/5
E. 1/3

The area of a triangle equals to $$area=\frac{1}{2}*base*height$$.

Now the base of triangle PQR (or simply the length of line segment PR) is 4 units (the difference between x-coordinates of the points P and R), and the height is also 4 units (the altitude from the point Q to the x-axis equals to 4 units). So $$area_{PQR}=\frac{1}{2}*4*4=8$$;

Similarly the base of triangle LMN is 12 units (the difference between x-coordinates of the points L and N), and the height is also 12 units (the altitude from the point M to the x-axis equals to 12 units). So $$area_{LMN}=\frac{1}{2}*12*12=72$$;

Thus $$\frac{area_{PQR}}{area_{LMN}}=\frac{8}{72}=\frac{1}{9}$$.

As for the difficulty level, I'd say around 600.

Hope it's clear.
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Re: In the rectangular coordinate system above the area of [#permalink]

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29 Jul 2013, 02:32
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monirjewel wrote:
Attachment:
Coordinate.jpg
In the rectangular coordinate system above the area of triangle PQR is what fraction of the area triangle LMN?

A. 1/9
B. 1/8
C. 1/6
D. 1/5
E. 1/3

Understanding the question:
Two triangles are given. They look proportional/similar but this should be confirmed. Also, means are provided to calculate length of base and height.

Facts to refer:
If 2 triangles are similar, then the ratio of their area= (ratio of the sides)^2

What's given in the question and what it implies (noted as =>):
Coordinates for L, P, R, N are given => Base of smaller triangle = 4 and that of larger triangle =12
Coordinate for M and Q are given => Height of smaller triangle = 4 and that of larger triangle =12

Area of PQR/Area of LMN => Ratio of the areas

Solution:
Since the base and height of the triangles are of the same ratio, the 2 triangles are similar. (Since the base and height are of equal ratio, the other 2 sides will also be of the same ratio.) Hence the fact given in "Facts to refer" can be used.
Ratio of sides = 4/12 =1/3
(Ratio of areas) = (1/3)^2 = 1/9
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Re: In the rectangular coordinate system above the area of [#permalink]

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21 Jul 2014, 09:19
Bunuel wrote:
In the rectangular coordinate system above the area of triangle PQR is what fraction of the area triangle LMN?

A. 1/9
B. 1/8
C. 1/6
D. 1/5
E. 1/3

The area of a triangle equals to $$area=\frac{1}{2}*base*height$$.

Now the base of triangle PQR (or simply the length of line segment PQ) is 4 units (the difference between x-coordinates of the points P and Q), and the height is also 4 units (the altitude from the point Q to the x-axis equals to 4 units). So $$area_{PQR}=\frac{1}{2}*4*4=8$$;

Similarly the base of triangle LMN is 12 units (the difference between x-coordinates of the points L and N), and the height is also 12 units (the altitude from the point M to the x-axis equals to 12 units). So $$area_{LMN}=\frac{1}{2}*12*12=72$$;

Thus $$\frac{area_{PQR}}{area_{LMN}}=\frac{8}{72}=\frac{1}{9}$$.

As for the difficulty level, I'd say around 600.

Hope it's clear.

bunuel : don't you mean P and R because 10 - 6 = 4?
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Re: In the rectangular coordinate system above the area of [#permalink]

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21 Jul 2014, 09:22
Bunuel wrote:
In the rectangular coordinate system above the area of triangle PQR is what fraction of the area triangle LMN?

A. 1/9
B. 1/8
C. 1/6
D. 1/5
E. 1/3

The area of a triangle equals to $$area=\frac{1}{2}*base*height$$.

Now the base of triangle PQR (or simply the length of line segment PQ) is 4 units (the difference between x-coordinates of the points P and Q), and the height is also 4 units (the altitude from the point Q to the x-axis equals to 4 units). So $$area_{PQR}=\frac{1}{2}*4*4=8$$;

Similarly the base of triangle LMN is 12 units (the difference between x-coordinates of the points L and N), and the height is also 12 units (the altitude from the point M to the x-axis equals to 12 units). So $$area_{LMN}=\frac{1}{2}*12*12=72$$;

Thus $$\frac{area_{PQR}}{area_{LMN}}=\frac{8}{72}=\frac{1}{9}$$.

As for the difficulty level, I'd say around 600.

Hope it's clear.

also for altitude are we subtracting the y coordinate of R from y coordiante of Q? (4 - 0) ?
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Re: In the rectangular coordinate system above the area of [#permalink]

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21 Jul 2014, 09:32
sagnik2422 wrote:
Bunuel wrote:
In the rectangular coordinate system above the area of triangle PQR is what fraction of the area triangle LMN?

A. 1/9
B. 1/8
C. 1/6
D. 1/5
E. 1/3

The area of a triangle equals to $$area=\frac{1}{2}*base*height$$.

Now the base of triangle PQR (or simply the length of line segment PQ) is 4 units (the difference between x-coordinates of the points P and Q), and the height is also 4 units (the altitude from the point Q to the x-axis equals to 4 units). So $$area_{PQR}=\frac{1}{2}*4*4=8$$;

Similarly the base of triangle LMN is 12 units (the difference between x-coordinates of the points L and N), and the height is also 12 units (the altitude from the point M to the x-axis equals to 12 units). So $$area_{LMN}=\frac{1}{2}*12*12=72$$;

Thus $$\frac{area_{PQR}}{area_{LMN}}=\frac{8}{72}=\frac{1}{9}$$.

As for the difficulty level, I'd say around 600.

Hope it's clear.

bunuel : don't you mean P and R because 10 - 6 = 4?

_____________
Yes, edited.
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Re: In the rectangular coordinate system above the area of [#permalink]

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21 Jul 2014, 09:34
sagnik2422 wrote:
Bunuel wrote:
In the rectangular coordinate system above the area of triangle PQR is what fraction of the area triangle LMN?

A. 1/9
B. 1/8
C. 1/6
D. 1/5
E. 1/3

The area of a triangle equals to $$area=\frac{1}{2}*base*height$$.

Now the base of triangle PQR (or simply the length of line segment PQ) is 4 units (the difference between x-coordinates of the points P and Q), and the height is also 4 units (the altitude from the point Q to the x-axis equals to 4 units). So $$area_{PQR}=\frac{1}{2}*4*4=8$$;

Similarly the base of triangle LMN is 12 units (the difference between x-coordinates of the points L and N), and the height is also 12 units (the altitude from the point M to the x-axis equals to 12 units). So $$area_{LMN}=\frac{1}{2}*12*12=72$$;

Thus $$\frac{area_{PQR}}{area_{LMN}}=\frac{8}{72}=\frac{1}{9}$$.

As for the difficulty level, I'd say around 600.

Hope it's clear.

also for altitude are we subtracting the y coordinate of R from y coordiante of Q? (4 - 0) ?

No. The coordinates of Q are (8, 4). So, Q is 4 units above 0.
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Re: In the rectangular coordinate system above the area of [#permalink]

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09 Aug 2015, 00:21
For Area PQR,

Base = 10-6 = 4 (x axis points P & R)
Height = 4 (y axis point Q)

Area of PQR = 1/2 * b * h =1/2 * 4 * 4 = 8

Also for Area LMN, Same procedure as above

Base = 14-2 = 12
Height = 12-0 = 12

Area of LMN = 1/2 * b * h =1/2 * 12 * 12 = 72

Area PQR/ Area LMN = 8/72 = 1/9
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Re: In the rectangular coordinate system above the area of [#permalink]

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14 Apr 2016, 22:10
Jaisri wrote:
monirjewel wrote:
Attachment:
Coordinate.jpg
In the rectangular coordinate system above the area of triangle PQR is what fraction of the area triangle LMN?

A. 1/9
B. 1/8
C. 1/6
D. 1/5
E. 1/3

Understanding the question:
Two triangles are given. They look proportional/similar but this should be confirmed. Also, means are provided to calculate length of base and height.

Facts to refer:
If 2 triangles are similar, then the ratio of their area= (ratio of the sides)^2

What's given in the question and what it implies (noted as =>):
Coordinates for L, P, R, N are given => Base of smaller triangle = 4 and that of larger triangle =12
Coordinate for M and Q are given => Height of smaller triangle = 4 and that of larger triangle =12

Area of PQR/Area of LMN => Ratio of the areas

Solution:
Since the base and height of the triangles are of the same ratio, the 2 triangles are similar. (Since the base and height are of equal ratio, the other 2 sides will also be of the same ratio.) Hence the fact given in "Facts to refer" can be used.
Ratio of sides = 4/12 =1/3
(Ratio of areas) = (1/3)^2 = 1/9

do we really need to confirm the similarity of the 2 triangles for PS question? if I am not mistaken in PS questions the figures are drawn to scale if not stated otherwise (in DS questions figures are NOT drawn to scale if not stated otherwise). Given that the base of the larger triangle relates to the base of the smaller one as 12/4 => 1/3 we can conclude that the areas should follow the ratio (1/3)^2 = 1/9
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Re: In the rectangular coordinate system above the area of   [#permalink] 14 Apr 2016, 22:10
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