Quantitative Section Test1 -> Question number 13

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You have the base and height wrong. In a parallelogram shown in the figure below the base and height are marked. In this case, it would be a very long calculation to calculate the base and height from the given coordinates.



The problem is solved by calculating the area of the triangle ABD.

We can find area of triangle with Shoelace formula:

Area=$\frac{1}{2}(x_1(y_2−y_3)+x_2(y_3−y_1)+x_3(y_1−y_2))$

Where $x_n, y_n$ are coordinates of the vertices of a triangle. Now one vertex is (0,0) so the formula reduces to :

Area=$\frac{1}{2}(x_1(y_2)+x_2(−y_1))$ $x_3=0$ and $y_3=0$.

And the area of the triangle is indeed $\frac{1}{2}$ area of the parallelogram ABCD.

Hence the answer is 7.

Thanks for your response.

I tried to solve by matrix method and got the answer is 7/2 for triangle BCD. I assumed the area of parallelogram is twice the triangle as there are two triangle present.

Is my understanding correct?

Yes, that is correct. Well done, it is a tough question.

Is there any other easy approach?

Unfortunately, I think this might be the easiest method available.

You may look at it this way..
1) measure if diagonals
a) C is (-7,7) so we are looking for hypotenuse with sides 7 each
So $\sqrt{7^2+7^2}=7√2$
b) diagonal BD
Since B is (-3,4), D is mirror image from line AC which is x=y
So D is (-4,3)
Diagonal BD is hypotenuse with sides (-3-(-4) and (4-3) so 1 each..
Diagonal = $\sqrt{1^2+1^2}=√2$


Now area will be half of product of diagonals = 7√2*√2*1/2=7*2/2=7

Hi Sandy,
How we want to find the area of ABD and we are using the coordinate of C?

Hi,
Unfortunately Sandy, is not with us

However, to answer your ques. - Simple trick don't complicate with formulas

Plz see the attach diag.

Let us extended the line from co-ordinate C to point M (-7,0) and to Point O (0,7)
Extend the line from the co-ordinate B (-3,4) to Point O (0,7)

Extend the line from the co-ordinate D (4,-3) to Point M (-7,0) ( co-ordinate of point D = mirror image of the co-ordinate of Point B and hence the sign changes and value gets interchanged)

Now you can see 4 $\triangle$ s

Since we need the Area of the parallelogram ABCD and that can be found = Area of the square AMCO - Sum of the Areas of all the 4 \triangle s

i.e. Area of the parallelogram ABCD = Area of the square AMCO - { Area of $\triangle$ AMD + Area of $\triangle$ DMC + Area of $\triangle$ COB + Area of $\triangle$ OAB }

First :

Area of the square $AMCO = 7 * 7 = 49$ (distance between each point =7)

Now,

Area of $\triangle$ AMD = $\frac{1}{2} * 7 * 3$ = $\frac{21}{2}$( base =7 and height = distance from base to point D = 3)

Area of $\triangle$ DMC = $\frac{1}{2}* 7 * 3$ = $\frac{21}{2}$( base =7 and height = distance from base to point D = 3)

Area of $\triangle$COB = $\frac{1}{2} * 7 * 3$ =$\frac{21}{2}$( base =7 and height = distance from base to point B = 3)

Area of $\triangle$ OAB = $\frac{1}{2} * 7 * 3$ =$\frac{21}{2}$( base =7 and height = distance from base to point B = 3)


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

Therefore Area of the parallelogram ABCD = 49 - 42 =7

It is the best approach! Thanks!

Formula of Area of Parallelogram = 2 * Area of Triangle
Formula of Area of Triangle when 2 coordinate points are given:
|ad-bc| / 2
(a,b) = (-3,4)
(c,d) = (-7,7)

Area of Triangle: (ad - bc) /2 = (-3*7) - (4*-7) / 2 = (-21 + 28) / 2 = 7/2
Area of Parallelogram = 2 * 7/2 = 7 Option C


Please correct me if I am wrong

Hi
This approach is not correct,

when you mention "ad" - it doesn't mean = a * d, infact it is the distance from "a" to "d"

Here is an example from khan academy



Regards

Thanks for correcting me. I saw this explanation which I have written on GREClub test explanations.

No worries,

Would you mind providing the question ID and question set of the GRECLUB test.

Regards

Hi, I tried solving this by finding the distance between the two points. For example, I first found the distance between point B and origin. Then point B and C. I was obviously wrong in this approach, but could you tell me how I was wrong here?

Please Sir,

read the explanations above. They go in depth

Quant Section 1: ID: Q02-51