# Geometry: Area of Parallelograms, etc...

I think I was taught the hard way to find the area of a parallelogram. I was taught, perhaps to apprecaite it, that you have to find the area by partitioning the shape into two right triangles (on either end) and a box in the middle…and then summing the components. But… it appears that b*h would still work because the two triangles are symmetrical, and if you mentally flip one around to join up with the other, you end up with a rectangle. Right? Also, it seems a similar approach would also work for trapeziods…at least trapezoids in text books which can also be parsed into two symmetrical triangles at either end.

Crude sketches:
(a) parallelogram (b) trapezoid
/.-----.–/ /.--------.
/ . . / / . .
/–.-----./ /–.--------.–\

Dots are crude vertical lines showing how I’d parse the shapes. (I need the ability to insert a half-space to align things properly. You have mentally picture flipping one parsed triangle over to match-up with the othe triangle to build rectangles…

Is this a sound method? (Never gave it much thought before…)
Thanks! - Jinx

I tried to fix, but the dang Board doesn’t seem to like all those additional space bar entries to form the shapes I want. Grrr! :mad: Maybe some SDoper is mroe adroit than I with these things!

Sorry for failing to preview earlier! :eek:
Thanks, again! - Jinx

Parallelograms: yes.

Trapezoids: Maybe. If the two triangles are equal, then yes, but don’t forget you can have a trapezoid with 2 right angles. A better formula is (B[sub]1[/sub]+B[sub]2[/sub])h/2, because then you’re just averaging the top and bottom. B[sub]1[/sub] and B[sub]2[/sub] are just the top and bottom sides, respectively.

You can use either of two formulas for a parallelogram:

s1 x h1
(where s1 is the length of one side, and h1 is the vertical distance between the two sides of length h1)

s1 x s2 x sin(a1)
(where s1 and s2 are the lengths of two adjacent sides, and a1 is any of the angles of the parallelogram)

It’s pretty obvious why h1 = s2 x sin(a1)

Use the code tag. Like this:

``````

________
/       /
/       /
/_______/

``````

And preview, the spacing is still a little wonky sometimes.
(Wonders whether the above looks like dung on other people’s browsers?)

Try the code tags:

``````
(a)parallelogram (b)trapezoid
_________      _____
/        /     /     \
/        /     /       \
/        /     /         \
/________/     /___________\

``````

Takes a bit of futzing, but works eventually.

Sorry – I should have said “two sides of length s1” in the 3rd line there.

I just read a great book on this topic yesterday. Yes, your intuition is right about parallelograms. Cut off the end triangle and paste it to the OTHER end, and you have a rectangle. If you believe that the area of the rectangle is Width times Length, or Height times Base (which it is) then you are right to calculate the area that way.
As to trapezoids, you can do the same thing. Duplicate it and flip it over and slide it next to the original. Now you have a parallelogram. So you can figure its area the way you did the last one. EXCEPT, since you doubled the trapezoid to get to the parallelogram, when you get to the final area, you have to divide it by 2. So the formula for the area of a trapezoid is really the bottom base plus the top base times the height - all of which must be divided by two.

``````(a+b) x H  /  2      where a = long base and b = short base
``````

You can even use this thinking (and I know I can’t do it here with my graphic skills) to illustrate to yourself how it is that the area of a circle is calculated by multiplying pi times the radius squared, and that method is essentially a version of “turn it into a parallelogram” amazingly enough.

In general the formula for a trapezoid consists in making another copy of the trapezoid and rotating it halfway around so that it fits onto one side of the first copy. Now you’ve got a parallelogram with double the area, a base the length of the sum of the two parallel sides of the original trapezoid, and the same height as the original. The area of this parallelogram is then h(b[sub]1[/sub]+b[sub]2[/sub]), and the area of the trapezoid is half that.