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Parallelogram ABCD lies in the xy-plane, as shown in the figur

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Parallelogram ABCD lies in the xy-plane, as shown in the figur  [#permalink]

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New post 08 May 2018, 04:22
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Parallelogram ABCD lies in the xy-plane, as shown in the figure above. The coordinates of point C are (-3, 4) and the coordinates of point B are (-7, 7). What is the area of the parallelogram ?


A. 1

B. \(2 \sqrt{7}\)

C. 7

D. 8

E. \(7 \sqrt{2}\)


Attachment:
xplane.jpg
xplane.jpg [ 23.49 KiB | Viewed 1256 times ]

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Re: Parallelogram ABCD lies in the xy-plane, as shown in the figur  [#permalink]

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New post 08 May 2018, 04:41
Bunuel wrote:
Image
Parallelogram ABCD lies in the xy-plane, as shown in the figure above. The coordinates of point C are (-3, 4) and the coordinates of point B are (-7, 7). What is the area of the parallelogram ?


A. 1

B. \(2 \sqrt{7}\)

C. 7

D. 8

E. \(7 \sqrt{2}\)


Attachment:
xplane.jpg


\(49-12-12-9-9=7\)

Answer: C
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Re: Parallelogram ABCD lies in the xy-plane, as shown in the figur  [#permalink]

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New post 08 May 2018, 05:13
IMO the answer should be C
D: (0,0)
C:(-3, 4)
B :(-7, 7)
the diagonal of the parallelogram divide it into 2 triangles of equal areas.
we can find the triangle area if we know three vertices.
the formulae is given in : http://demonstrations.wolfram.com/TheAr ... terminant/
I find this very useful.
Area of triangle = 7/2
area of parallelogram = 7
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Re: Parallelogram ABCD lies in the xy-plane, as shown in the figur  [#permalink]

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New post 08 May 2018, 06:52
1
Area of parallelogram is B x H so 7 x 7 = 49.
Area of triangle is 1/2 x B x H
Triangle A-(-4,0)-D = 1/2 x 3 x 4 = 6
Triangle C-(0,4)-(0,0) = 1/2 x 3 x 4 = 6
Triangle A-(-7,3)-B = 1/2 x 3 x 4 = 6
Triangle C-(-3,7)-B = 1/2 x 3 x 4 = 6
Area of square is s^2
Square A-(-4,0)-(-7,0)-(-7,3) = 3^2 = 9
Square C-(0,4)-(0,7)-(-3,7) = 3^2 = 9

Put it all together 49-6-6-6-6-9-9 = 7. Answer Choice C.
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Re: Parallelogram ABCD lies in the xy-plane, as shown in the figur  [#permalink]

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New post 08 May 2018, 08:43
1
OA : C
Attachment:
parallelogram area.PNG
parallelogram area.PNG [ 70.23 KiB | Viewed 997 times ]

Area of \(\triangle\)BDE= Area of \(\triangle\)BCD + Area of \(\triangle\)DCF +Area of trapezium BCFE
\(\frac{1}{2}*7*7= Area of \triangle BCD + \frac{1}{2}*3*4 + \frac{1}{2}*(3+7)*3\)
\(Area of \triangle BCD = \frac{1}{2}*( 49-12-30)\) = \(\frac{7}{2}\)

As We know that Diagonal of parallelogram divides it into 2 triangle of equal area.
So \(Area of parallelogram ABCD = 2 *Area of \triangle BCD\) = \(2*\frac{7}{2} =7\)
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Parallelogram ABCD lies in the xy-plane, as shown in the figur  [#permalink]

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New post 08 May 2018, 10:50
Alternative method is to find the area of the rhombus (yes the figure ABCD is a rhombus since this is a Parallelogram with 4 equal sides)

S= d1*d2/2, where d1 and d2 are diagonals. The formula of distance if you know 2 points is Distance = square root of(x2−x1)^2+(y2−y1)^2 Answer C.
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Parallelogram ABCD lies in the xy-plane, as shown in the figur  [#permalink]

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New post 09 May 2018, 07:05
Why can't we use the formula for the areas of parallelograms here?

I calculated that Point A must be at (-4;3) so I had the coordinates of all four points. I took AC = \(\sqrt{2}\) as the height and DC = 5 as the base. As \(\sqrt{2} =~ 1.4\) I calculated 1.4*5 which is 7. Of course 1.4 is only an approximation so the result is not exact.

I am aware that this solution doesn't work here but I want to understand why.
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Re: Parallelogram ABCD lies in the xy-plane, as shown in the figur  [#permalink]

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New post 09 May 2018, 10:08
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Masterscorp wrote:
Why can't we use the formula for the areas of parallelograms here?

I calculated that Point A must be at (-4;3) so I had the coordinates of all four points. I took AC = \(\sqrt{2}\) as the height and DC = 5 as the base. As \(\sqrt{2} =~ 1.4\) I calculated 1.4*5 which is 7. Of course 1.4 is only an approximation so the result is not exact.

I am aware that this solution doesn't work here but I want to understand why.


Masterscorp
to use the area of parallelogram formula here.
First we have to find the equation of CD,
\(y= \frac{{4-0}}{{-3-0}}x+c\)
As CD passes through (0,0); c would be zero
Equation of CD : \(y = - \frac{4}{3}x\)
\(y+\frac{4}{3}x=0\)

As you have already found out the coordinates of point A(-4,3)
We can find perpendicular distance between CD and Point A using
\(D =\frac{{|ax_1 +by_1+c|}}{{\sqrt{a^2+b^2}}}\)
In this case , \(a=\frac{4}{3},b=1,c=0,x_1=-4,y_1=3\)
Solving for D, we will get \(D = \frac{7}{5}=1.4\)
As length of \(CD =5\)
Area of Parallelogram ABCD = \(5*\frac{7}{5}= 7\)

AC is not perpendicular to CD as you assumed
Line equation of AC will \(y = \frac{{4-3}}{{-3+4}}x+c\)
\(y = x+c\)
putting (-3,4) in above equation, we will get \(c=7\)
So equation of line AC would be \(y = x+7\)

As product of slopes of two non vertical perpendicular line should be -1.
Let us check the product of slope of AC and CD, it is coming out be \(1*\frac{-4}{3}\) = \(\frac{-4}{3}\)
So line AC and CD are not perpendicular
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Re: Parallelogram ABCD lies in the xy-plane, as shown in the figur  [#permalink]

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New post 19 Jul 2018, 00:12
Princ wrote:
Masterscorp wrote:
Why can't we use the formula for the areas of parallelograms here?

I calculated that Point A must be at (-4;3) so I had the coordinates of all four points. I took AC = \(\sqrt{2}\) as the height and DC = 5 as the base. As \(\sqrt{2} =~ 1.4\) I calculated 1.4*5 which is 7. Of course 1.4 is only an approximation so the result is not exact.

I am aware that this solution doesn't work here but I want to understand why.


Masterscorp
to use the area of parallelogram formula here.
First we have to find the equation of CD,
\(y= \frac{{4-0}}{{-3-0}}x+c\)
As CD passes through (0,0); c would be zero
Equation of CD : \(y = - \frac{4}{3}x\)
\(y+\frac{4}{3}x=0\)

As you have already found out the coordinates of point A(-4,3)
We can find perpendicular distance between CD and Point A using
\(D =\frac{{|ax_1 +by_1+c|}}{{\sqrt{a^2+b^2}}}\)
In this case , \(a=\frac{4}{3},b=1,c=0,x_1=-4,y_1=3\)
Solving for D, we will get \(D = \frac{7}{5}=1.4\)
As length of \(CD =5\)
Area of Parallelogram ABCD = \(5*\frac{7}{5}= 7\)

AC is not perpendicular to CD as you assumed
Line equation of AC will \(y = \frac{{4-3}}{{-3+4}}x+c\)
\(y = x+c\)
putting (-3,4) in above equation, we will get \(c=7\)
So equation of line AC would be \(y = x+7\)

As product of slopes of two non vertical perpendicular line should be -1.
Let us check the product of slope of AC and CD, it is coming out be \(1*\frac{-4}{3}\) = \(\frac{-4}{3}\)
So line AC and CD are not perpendicular


Hi how do we find out the coordinates of point A. We can't assume them looking at the figure right. ?
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Re: Parallelogram ABCD lies in the xy-plane, as shown in the figur  [#permalink]

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New post 19 Jul 2018, 00:43
abhishek911 wrote:
Princ wrote:
Masterscorp wrote:
Why can't we use the formula for the areas of parallelograms here?

I calculated that Point A must be at (-4;3) so I had the coordinates of all four points. I took AC = \(\sqrt{2}\) as the height and DC = 5 as the base. As \(\sqrt{2} =~ 1.4\) I calculated 1.4*5 which is 7. Of course 1.4 is only an approximation so the result is not exact.

I am aware that this solution doesn't work here but I want to understand why.


Masterscorp
to use the area of parallelogram formula here.
First we have to find the equation of CD,
\(y= \frac{{4-0}}{{-3-0}}x+c\)
As CD passes through (0,0); c would be zero
Equation of CD : \(y = - \frac{4}{3}x\)
\(y+\frac{4}{3}x=0\)

As you have already found out the coordinates of point A(-4,3)
We can find perpendicular distance between CD and Point A using
\(D =\frac{{|ax_1 +by_1+c|}}{{\sqrt{a^2+b^2}}}\)
In this case , \(a=\frac{4}{3},b=1,c=0,x_1=-4,y_1=3\)
Solving for D, we will get \(D = \frac{7}{5}=1.4\)
As length of \(CD =5\)
Area of Parallelogram ABCD = \(5*\frac{7}{5}= 7\)

AC is not perpendicular to CD as you assumed
Line equation of AC will \(y = \frac{{4-3}}{{-3+4}}x+c\)
\(y = x+c\)
putting (-3,4) in above equation, we will get \(c=7\)
So equation of line AC would be \(y = x+7\)

As product of slopes of two non vertical perpendicular line should be -1.
Let us check the product of slope of AC and CD, it is coming out be \(1*\frac{-4}{3}\) = \(\frac{-4}{3}\)
So line AC and CD are not perpendicular


Hi how do we find out the coordinates of point A. We can't assume them looking at the figure right. ?


Quote:
The diagonals of a parallelogram bisect each other

Mid point of AC will be same as Mid point of BD.
Assume the coordinates of point A be (x,y) Given: D(0,0);C(-3, 4);B(-7,7).

\(\frac{x+(-3)}{2} = \frac{-7+0}{2}\)
\(x=-4\)

\(\frac{y+4}{2} = \frac{7+0}{2}\)
\(y=3\)

So Coordinates of A are \((-4,3)\).
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Re: Parallelogram ABCD lies in the xy-plane, as shown in the figur  [#permalink]

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New post 19 Jul 2018, 07:21
Bunuel wrote:
Image
Parallelogram ABCD lies in the xy-plane, as shown in the figure above. The coordinates of point C are (-3, 4) and the coordinates of point B are (-7, 7). What is the area of the parallelogram ?


A. 1

B. \(2 \sqrt{7}\)

C. 7

D. 8

E. \(7 \sqrt{2}\)


Attachment:
xplane.jpg



Hey Bunuel,

Can we solve using the diagonal formula for area(1/2*d1*d2) for a parallelogram? Is my following approach correct?

From the diagram we can say co-ordinates of A are (-4,3)
So, Diagonal AC= \(\sqrt{2}\) (using co-ordinate geometry formula for length)
And, Diagonal BD=7\(\sqrt{2}\)
So, Area of parallelogram= 1/2*AC*BD= 1/2*\(\sqrt{2}\)*7\(\sqrt{2}\) = 7 (Ans) Option C.
Re: Parallelogram ABCD lies in the xy-plane, as shown in the figur &nbs [#permalink] 19 Jul 2018, 07:21
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