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.

The correct answerC

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,

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

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!

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!

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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