# Confusion on Parallelogram [duplicate]

i know it's rude and pretentious and maybe stupid to ask such question but, i would like to know the reason why in geometry the area of a parallelogram follows this formula :

$$Base \times Height$$

and not this one :

$$Base \times Side$$

I saw this method which consists of transforming our figure into a rectangle by moving the black one to the right, and that seems logical to me, but why not adopt the second rule, all the horizontal points multiplied by the vertical ones ?

PS : I found the same Visualizing the Area of a Parallelogram, but was unconvinced by the answers

## marked as duplicate by mrtaurho, Cesareo, user21820, Xander Henderson, José Carlos SantosMar 25 at 14:04

• @kkc, did you see the link ? – Bo Halim Mar 24 at 21:46
• In mathematics, there is nothing rude and pretentious about asking questions – Taladris Mar 25 at 4:45
• @Taladris thank you :) – Bo Halim Mar 25 at 8:42

Sometimes a figure is worth 1000 words:

Very long base and very long side and very small area.

Or...

...each of these parallelograms has the same base and side, but manifestly different areas:

• The best answer ever :) – Bo Halim Mar 25 at 8:44
• EVER? My goodness... thanks. – David G. Stork Mar 25 at 17:57

You can't "adopt the second rule" because it gives the wrong answer. The parallelogram in your picture has all four sides of length $$5$$, There are lots of parallelograms like that, depending on the angle between adjacent sides. The largest is a square. If you make the acute angle smaller and smaller you find parallelograms that are nearly on a line, with tiny areas. But your formula would say they all have area $$25$$.

You have to have some notion of what area must satisfy (otherwise, you could just say that area of everything is 0 and be done with it:). I think a principle with which you'll agree is that if one figure is a subset of another, it cannot have greater area.

Well, in that case, imagine a world that looks like yours, where the area of any paralelogram is calculated according to formula you give---and look at the following picture:

As you see, by Bo's formula, the area of AECF would be 13*5=65, while the area of a containing rectangle (which is also a parallelogram) would be (13+3)*4=64<65. Contradiction.

• +1: This is an excellent refutation. – Cameron Buie Mar 25 at 11:48

As you point out, the first method is logical, as it uses a formula we already know to be true (the familiar area formula for a rectangle), and then transforms our figure without changing the area, so that we can apply the rule.

To see why we can't simply multiply $$5$$ and $$5$$ to get the answer, draw the $$5\times 5$$ square. Your parallelogram can't fill it by cut and paste. Alternatively, you can draw $$25$$ unit squares, and try to cut and paste them into the parallelogram. You'll only be able to make $$20$$ of them fit.

Hint:

Drop a $$\perp$$ to a point say $$F$$ from $$C$$, prove that $$\Delta AED \cong \Delta CFB$$ and now consider the $$\text{ar}(\parallel \text{gm} \ ABCD)$$ as the area of this rectangle because they're both equal and the area of the rectangle is very well known to be $$\text{base}\cdot\text{height}$$.

You may also want to see this fact as a consequence of the $$2$$-D analog of Cavalieri's Principle.