Parallelogram area (coordinates)

[Alex126]

Homework Statement

The coordinates of the parallelogram ABCD are:

A (-2; 1)
B (5; 2)
C (6; 5)
D (-1; 4)

We also know that the diagonals intercept in the middle of each other (so if the diagonals are AC and BD, and the intercept in point M, then AM = MC, and BM = MD). Not sure if this information is useful or not though.

Homework Equations

Area = base * height

The Attempt at a Solution

So...yea, no clue actually.

I know how to calculate AB = CD = (...) = 5√2
Likewise, BC = AD = (...) = √10

For what they're worth:
AC = 4√5
BD = 2√10

I tried googling around for an answer, and everyone seems to solve it with these steps:
- Calculate the line passing through A and B
- Calculate the distance between point C (or D) and the line passing through AB; this gives you the height
- Do AB*height

However, we didn't do lines equations at school, so this assigned problem should probably be done in some other way (Pythagorean theorem, distance between two points, middle point of a segment).

 

[stevendaryl]

I think the equation "Area = base times height" is difficult to apply in this case, because there is no easy way to compute the height.

I'm not sure if whatever class you're in uses vectors? In terms of vectors, the area is given by the cross-product:

|\vec{AB} \times \vec{AD}|

 

[Alex126]

Nope, no vectors either.

 

[stevendaryl]

Well, there is a very tedious way to compute the height:

I rotated your parallelogram just to make things visually easier, but nothing is changed in the following:

• Drop a perpendicular line from D down to the line AB. Call the point where it hits the line D'.
• Let w = length of AD', x = length of AD, y = length of CD and z = length of BD.
• Pythagoras tells us:

1. h² + w² = x²
2. h² + (y-w)² = z²

So we have two equations and two unknowns, h and w (you know how to calcuate x, y, and z). So we can solve equation 1. for h², and plug that into equation 2. Then solve that for w. Then go back to equation 1. to find h.

 

[Alex126]

That's it, brilliant! Thanks a lot.

 

[scottdave]

Another way you could do it. Make a rectangle which starts at A, goes straight up to be level with C, then across to C, then straight down to the level of A, then straight back over to A. This rectangle is 4 units tall and 8 units wide. Now just subtract off areas which are not part of the parallelogram. So you have the two unit squares at opposite corners. Then you have two right triangles 1x3 each, and two right triangles 7x1 each. Subtract these areas from the bigger rectangle, and you have the parallelogram area. Since it seems as it has already been solved a different way, then I don't think I'm giving too much help.

 

[Alex126]

Since it seems as it has already been solved a different way, then I don't think I'm giving too much help.

Not very helpful for the purpose of solving the assignment on time maybe, but I actually really appreciated your "graphic" method. And yes, it gives the right result.

 

[scottdave]

Not very helpful for the purpose of solving the assignment on time maybe, but I actually really appreciated your "graphic" method. And yes, it gives the right result.

Too bad I had not seen this question, earlier. I'm glad I was able to provide another method for you to learn.

 

[Alex126]

Actually, believe it or not, we discussed the graphic method a few hours after my previous reply lol The only difference is that we discussed subtracting the areas of four triangles instead of four triangles and two squares (calling E the point E (-2; +5), two triangles are AED and CDE; the other two are the symmetrical ones on the lower-right; the height of either one is = 1, and the bases are also easily calculated). Anyways, cheers.