Basic geometry, don't understand

[Sparklez]

Ok so this might seem silly or stupid, but I just "learned" about the area of a parallelogram. I am so confused because it's basically a rectangle, tilted. Why on earth can't I just find the area by multiplying the dimensions of the sides? Why do I have to do this whole base x height thing? The dimensions stay the same, so I'm so confused. If someone could explain why just multiplying the dimensions doesn't work I would really appreciate it. :3

 

[Harry_the_cat]

Ok so this might seem silly or stupid, but I just "learned" about the area of a parallelogram. I am so confused because it's basically a rectangle, tilted. Why on earth can't I just find the area by multiplying the dimensions of the sides? Why do I have to do this whole base x height thing? The dimensions stay the same, so I'm so confused. If someone could explain why just multiplying the dimensions doesn't work I would really appreciate it. :3

Look at the diagram above. ABCD is a parallelogram. If you chop off the triangle from one end and stick it on the other, you get the rectangle ABFE.

Area of this parallelogram = area of this rectangle (because you just cut and pasted one end to the other).

Now area of rectangle = base x height = EF x AE.

So the area of the parallelogram also = EF x AE.

Now notice that EF = DC .

So the area of parallelogram = EF x AE = DC x AE which is the base x perpendicular height NOT base x side.

Hope that makes it clear!

 

[stapel]

Ok so this might seem silly or stupid, but I just "learned" about the area of a parallelogram. I am so confused because it's basically a rectangle, tilted. Why on earth can't I just find the area by multiplying the dimensions of the sides?

Think of a rectangle (say, a cardboard box from which you've removed the top and the bottom). Tip the rectangle (the box) over a bit, so now it's a parallelogram (when you look through the opening). Now continue tipping, until the figure is crushed flat.

The dimensions of the sides are still the same. But is the enclosed area the same? ;-)