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# In the rectangular coordinate system above the area of

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In the rectangular coordinate system above the area of [#permalink]  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
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In the rectangular coordinate system above the area of [#permalink]  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]  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]  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]  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]  21 Jul 2014, 09:32
Expert's post
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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Posts: 29791
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Kudos [?]: 53540 [0], given: 8165

Re: In the rectangular coordinate system above the area of [#permalink]  21 Jul 2014, 09:34
Expert's post
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]  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
Re: In the rectangular coordinate system above the area of   [#permalink] 09 Aug 2015, 00:21
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