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# f the coordinates of point B are (-3, -4) and

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Manager
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f the coordinates of point B are (-3, -4) and [#permalink]  10 Sep 2018, 16:30
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Question Stats:

80% (00:50) correct 20% (01:56) wrong based on 5 sessions
Attachment:

geometry single 002.JPG [ 13.06 KiB | Viewed 2045 times ]

If the coordinates of point B are (-3, -4) and the coordinates of point C are (-7, -7), what is the area of the parallelogram?

A. 1

B. $$2 \sqrt{7}$$

C. 7

D. 8

E. $$7 \sqrt{2}$$

I do not understand how did they get C as the answer.
[Reveal] Spoiler: OA

Last edited by Carcass on 07 Apr 2019, 01:03, edited 1 time in total.
Edited by Carcass
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Re: Find the area of a parallelogram [#permalink]  11 Sep 2018, 05:56
kruttikaaggarwal wrote:

I do not understand how did they get C as the answer.

Who are they?

Can you plz check if the signs of all the coordinates are correct, as all the values are in 2nd quadrant
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Re: Find the area of a parallelogram [#permalink]  11 Sep 2018, 06:23
[quote="kruttikaaggarwal"]If the coordinates of point B are (-3, -4) and the coordinates of point C are (-7, -7), what is the area of the parallelogram?

Attachment:
geometry single 002.JPG

A. 1
B. 2√7
C. 7
D. 8
E. 7√2

I do not understand how did they get C as the answer.[/quotes]

Really is the figure true? it contradicts with points given.
In gre graphically representation can be taken exactly as in figure unlike geometric figures
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Re: Find the area of a parallelogram [#permalink]  11 Sep 2018, 10:56
pranab01 wrote:
kruttikaaggarwal wrote:

I do not understand how did they get C as the answer.

Who are they?

Can you plz check if the signs of all the coordinates are correct, as all the values are in 2nd quadrant

Hi guys, I did re-check the question and I actually copy-pasted it as is. It is from one of the GRE prep club Quants test.

This is the explanation they gave and I find it bizzare:

Area of parallelogram ABCD = 2×ABD
The coordinates of the vertices of triangle ABD are A(0,0),B(−3,−4),C(−7,−7). If one of the vertices of a triangle is at the origin and other two being (a,b),(c,d), then the area of the triangle can be written as : Area = |(ad−bc)/2|
Applying this formula to the triangle ABD, we get the area as [(−3)∗(−7)−(−4)∗(−7)]/2=7/2
So, the area of the parallelogram is twice the area of triangle ABD or 7/2 ×2
or 7sq.units.

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Re: Find the area of a parallelogram [#permalink]  11 Sep 2018, 18:18
Expert's post
Hi Krutika,

Could you please share the question number so that we can fix the graph.

The are aof a triangle with coordinate A ($$A_x$$,$$A_y$$), B ($$B_x$$,$$B_y$$) and C ($$C_x$$,$$C_y$$) can be written as:

area=$$\mod{\frac{A_x(B_y-C_y)+B_x(C_y-A_y)+C_x(A_y-B_y)}{2}}$$.

Here one of the coordinates is (0,0). Thus the reduced expression in the explanation.

Here is great tool to vizualize the same.

https://www.mathopenref.com/coordtrianglearea.html
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Re: Find the area of a parallelogram [#permalink]  11 Sep 2018, 22:09
sandy wrote:
The are aof a triangle with coordinate A ($$A_x$$,$$A_y$$), B ($$B_x$$,$$B_y$$) and C ($$C_x$$,$$C_y$$) can be written as:

area=$$\mod{\frac{A_x(B_y-C_y)+B_x(C_y-A_y)+C_x(A_y-B_y)}{2}}$$.

Hi Sandy,

Can we accept in GRE the sign of the co-ordintes donot matter,

As point c (-7,-7) and point B (-3, -4). both of these lies in II quadrant as such both y- coordinates should be positive.

Can you plz enlighten me
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Re: Find the area of a parallelogram [#permalink]  12 Sep 2018, 05:20
sandy wrote:
Hi Krutika,

Could you please share the question number so that we can fix the graph.

The are aof a triangle with coordinate A ($$A_x$$,$$A_y$$), B ($$B_x$$,$$B_y$$) and C ($$C_x$$,$$C_y$$) can be written as:

area=$$\mod{\frac{A_x(B_y-C_y)+B_x(C_y-A_y)+C_x(A_y-B_y)}{2}}$$.

Here one of the coordinates is (0,0). Thus the reduced expression in the explanation.

Here is great tool to vizualize the same.

https://www.mathopenref.com/coordtrianglearea.html

Hi Sandy,

It is Question 13 from Quantitative Test 1.

I do not think that just graph is the issue. I think either the coordinates or the options also need to change. Even if we are unable to use the formula you or the explanation gave, we should be able to use Area of a parallelogram = bh and both should give the same result.

Please explain and correct me if my concept is wrong. Thanks!
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Re: Find the area of a parallelogram [#permalink]  12 Sep 2018, 14:17
Expert's post
Area of a parallelogram = bh

Do you mean base $$\times$$ height?
If so then that is not a a paralleogram but a rectangle.
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Re: Find the area of a parallelogram [#permalink]  12 Sep 2018, 14:45
sandy wrote:
Area of a parallelogram = bh

Do you mean base $$\times$$ height?
If so then that is not a a paralleogram but a rectangle.

Yes Area= base X height. It also says this everywhere. I got it from Barron's book. I also just cross checked using Google. All sources says that Area of a Parallelogram= base X height, where height is at 90 degress with the base.
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Re: Find the area of a parallelogram [#permalink]  13 Sep 2018, 00:09
Expert's post
kruttikaaggarwal wrote:
sandy wrote:
Area of a parallelogram = bh

Do you mean base $$\times$$ height?
If so then that is not a a paralleogram but a rectangle.

Yes Area= base X height. It also says this everywhere. I got it from Barron's book. I also just cross checked using Google. All sources says that Area of a Parallelogram= base X height, where height is at 90 degress with the base.

This question is from coordinate geometry as apposed to regular geometry, so here you cannot use base $$\times$$ height. You have to break the paralleogram into two triangle and find the area of each smaller triangle.

This is a tough problem by GRE standards but by no means is it beyond GRE level.
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Re: Find the area of a parallelogram [#permalink]  13 Sep 2018, 06:15
sandy wrote:
kruttikaaggarwal wrote:
sandy wrote:
Area of a parallelogram = bh

Do you mean base $$\times$$ height?
If so then that is not a a paralleogram but a rectangle.

Yes Area= base X height. It also says this everywhere. I got it from Barron's book. I also just cross checked using Google. All sources says that Area of a Parallelogram= base X height, where height is at 90 degress with the base.

This question is from coordinate geometry as apposed to regular geometry, so here you cannot use base $$\times$$ height. You have to break the paralleogram into two triangle and find the area of each smaller triangle.

This is a tough problem by GRE standards but by no means is it beyond GRE level.

Oh, alright! Did not know that. Is this applicable only for parallelograms or for other geometric figures as well?

Thanks again Sandy!
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Re: Find the area of a parallelogram [#permalink]  06 Apr 2019, 20:31
kruttikaaggarwal wrote:
If the coordinates of point B are (-3, -4) and the coordinates of point C are (-7, -7), what is the area of the parallelogram?

Attachment:
The attachment geometry single 002.JPG is no longer available

A. 1
B. 2√7
C. 7
D. 8
E. 7√2

.

Here,

Let join the sides from Point C to point G on Y axis and point F on X axis.

The distance from C to G = distance from C to F = 7

The Area of the Square AFCG = $$7^2$$ = $$49$$

Point D will be the mirror image of point B and co ordinates of Point D = -4,3

Now let consider the $$\triangle FDA$$, $$\triangle FDC$$, $$\triangle CGB$$ and $$\triangle BGA$$

Area of $$\triangle FDA$$ =$$\frac{1}{2}* 7 * 3 = \frac{21}{2}$$

Area of $$\triangle FDC$$ = $$\frac{1}{2}* 7 * 3 = \frac{21}{2}$$
Area of $$\triangle CGB$$ = $$\frac{1}{2} * 7 * 3 = \frac{21}{2}$$

Area of $$\triangle BGA$$ = $$\frac{1}{2} * 7 * 3 = \frac{21}{2}$$

Sum of all the $$\triangle$$ = $$\frac{21}{2} * 4 = 42$$

Therefore the area of parallelogram = Area of the Square - Area of the sum of All triangles = $$49 - 42 = 7$$
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geometry single 003.jpg [ 21.04 KiB | Viewed 751 times ]

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Re: Find the area of a parallelogram   [#permalink] 06 Apr 2019, 20:31
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