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Parallelogram ABCD lies in the xyplane, as shown in the figur
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08 May 2018, 03:22
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49% (03:09) correct 51% (02:52) wrong based on 77 sessions
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Re: Parallelogram ABCD lies in the xyplane, as shown in the figur
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08 May 2018, 03:41
Bunuel wrote: Parallelogram ABCD lies in the xyplane, 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. \(7 \sqrt{2}\) \(49121299=7\) Answer: C
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Re: Parallelogram ABCD lies in the xyplane, as shown in the figur
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08 May 2018, 04:13
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 xyplane, as shown in the figur
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08 May 2018, 05:52
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 49666699 = 7. Answer Choice C.



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Re: Parallelogram ABCD lies in the xyplane, as shown in the figur
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08 May 2018, 07:43
OA : C Attachment:
parallelogram area.PNG [ 70.23 KiB  Viewed 1132 times ]
Area of \(\triangle\)BDE= Area of \(\triangle\)BCD + Area of \(\triangle\)DCF +Area of trapezium BCFE \(\frac{1}{2}*7*7= Area of \triangle BCD + \frac{1}{2}*3*4 + \frac{1}{2}*(3+7)*3\) \(Area of \triangle BCD = \frac{1}{2}*( 491230)\) = \(\frac{7}{2}\) 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 xyplane, as shown in the figur
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08 May 2018, 09:50
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 xyplane, as shown in the figur
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09 May 2018, 06:05
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 xyplane, as shown in the figur
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09 May 2018, 09:08
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. Masterscorpto use the area of parallelogram formula here. First we have to find the equation of CD, \(y= \frac{{40}}{{30}}x+c\) 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 \(y = \frac{{43}}{{3+4}}x+c\) \(y = x+c\) 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 xyplane, as shown in the figur
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18 Jul 2018, 23:12
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. Masterscorpto use the area of parallelogram formula here. First we have to find the equation of CD, \(y= \frac{{40}}{{30}}x+c\) 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 \(y = \frac{{43}}{{3+4}}x+c\) \(y = x+c\) 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 xyplane, as shown in the figur
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18 Jul 2018, 23:43
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. Masterscorpto use the area of parallelogram formula here. First we have to find the equation of CD, \(y= \frac{{40}}{{30}}x+c\) 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 \(y = \frac{{43}}{{3+4}}x+c\) \(y = x+c\) 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 xyplane, as shown in the figur
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19 Jul 2018, 06:21
Bunuel wrote: Parallelogram ABCD lies in the xyplane, 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. \(7 \sqrt{2}\) 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 coordinates of A are (4,3) So, Diagonal AC= \(\sqrt{2}\) (using coordinate 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 xyplane, as shown in the figur &nbs
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