I've never been to good with Mathematics... I'm wanting to change that as game development has always been a passion of mine. I started studying math on Khan Academy about a month ago and everything has been making sense up to this point (https://www.khanacademy.org/math/cc-sixth-grade-math/cc-6th-geometry-topic/cc-6th-area/v/area-of-a-parallelogram). From what I understand he's taking the measurement of two triangles areas and getting the area of a parallelogram. I'm having a hard time understanding this due to...

Area of a Parallelogram

How's the height of this parellelogram any different from it's width? I always thought A=lw and A=bh was always the same? In the image, you can see height is 6cm, length is 8cm, and base is 10cm. Why is height not equal to 8cm like it's width?

Please explain this in a way I could possibly understand. Thank you very much for your time.

  • $\begingroup$ A = l*w applies only to rectangles, but not all parallelograms are rectangles. $\endgroup$ – Zack Li Dec 16 '13 at 19:07
  • $\begingroup$ Yes but any parrallelogram can be made into a rectangle? So, A=l*w would apply in that case? $\endgroup$ – W3Geek Dec 16 '13 at 19:15
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    $\begingroup$ Good ol semantics. Height is something that goes up – even though 8 cm side kind of goes up, it's inclined, so real height is only 6. As for the width, it's more informal, since objects have height, width, and sometimes depth. In your 2D toy example, there's already a height, so other one is width. If you want to stretch your mind a bit just flip your parallelogram on its 8 cm side and try to find what's height and width in that case. $\endgroup$ – Kaster Dec 16 '13 at 19:26
  • $\begingroup$ @Kaster wouldn't it be 80cm squared then because height could be measured as the 10cm side now? Hmm but how do you know if the actual height of this parallelogram is 6cm? Did a person make this up or is there some kind of odd calculation to get 6cm from 8cm and 10cm? Thank you. $\endgroup$ – W3Geek Dec 16 '13 at 21:58
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    $\begingroup$ @W3Geek yes, actual height might be anything between 0 and 8, depending on how inclined that side is. In your case it's 6. $\endgroup$ – Kaster Dec 16 '13 at 22:31

Two jokes I tell my students (though I tell them better in person) when width/height confusion strikes (don't feel bad ... it happens often!):

A guy came across his buddy in the middle of a field, desperately struggling to push a collapsible tape measure to the far-out-of-reach top of a pole sticking out of the ground. The guy asks, "If it's that important, then why don't you just knock over the pole and measure it while it lies on the ground?" His buddy responded, "Because ... I want to know how tall it is, not how long it is!"


A pilot and co-pilot found themselves needing to make an emergency landing on an unfamiliar airstrip. "We're coming in too fast!" warned the co-pilot, "We're going to run out of runway!" The pilot adjusted the flaps and the angle of approach to reduce speed and buy some time. "Not enough!" said the co-pilot, "We're still going to run out of runway!" The pilot made more adjustments, and more again, but the co-pilot was still frantic: "We're going to run out of runway!" In a final feat of aerodynamic magic, the masterful pilot managed to get the plane's approach speed down so that the craft touched down ever-so-gently and taxied to a safe and secure stop with its nose just past the end of the runway. "Manoman," sighed the exhausted pilot. "That's the shortest runway I've ever encountered!" "Yeah," agreed the co-pilot, "... but look how wide it is!"

... and that, my friends, is why I always endeavor to phrase the formula for the area of a parallelogram as "base-times-altitude".

Importantly, in the formula "base-time-altitude", the "base" is indeed any side you choose, but "altitude" is the altitude perpendicular to the chosen base.

So, in your problem, while $8$cm is a perfectly-good base measurement, the $6$cm measurement is not perpendicular to that base, so these numbers don't fit together into the area formula. (The altitude for the $8$cm base doesn't appear in the diagram at all.) The $6$cm measure is of the altitude perpendicular to the base of length $10$cm, so that the area is $10\times 6 = 60\text{cm}^2$.


I hope this analogy helps,

the area of all parallelograms is

A = base * height,

where the height is the length of the perpendicular from the base of the parallelogram to its top.

enter image description here

Now, a rectangle is a special type of parallelogram all the sides of which make right angles. Hence the width of the rectangle, is, effectively the perpendicular from its base to its top.

So for rectangles, we can use the formula

A = length * width.

Note: I think width, length, base and height are loosely defined in this case, but I hope you get the general idea. You can call a rectangles length "base" and its width "height" if you want. This will probably avoid confusion.

enter image description here


Is your question (or a part of it) why the area doesn't equal $80$ cm${}^2$ (where the $80$ is coming as $8 * 10$)?

If so, imagine what would happen if the parallelogram were tilted more and more; if you like, imagine it as a frame that you are holding up with the bottom resting on the ground, but with the sides tilting (maybe because the screws joining them to the top and bottom aren't tight enough, or something). Eventually the whole thing will collapse, and the top will also be lying on the ground.

Then the whole thing will enclose no area at all!

So it is not really the "width" (the length of the sides) that determines how much area is between the top and bottom of the frame; it is how high the top of the frame is above the bottom that is important.

  • $\begingroup$ Yes this is part of my question but I'm wondering how the actual height of this parallelogram is 6cm instead of 8cm? It's confusing because these terms can be used interchangly throughout math. How do you get 6cm from 8cm? Did the person who made the picture make that number up or is there some sort of calculation that makes 6? $\endgroup$ – W3Geek Dec 16 '13 at 22:08
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    $\begingroup$ @W3Geek: Dear W3Geek, Your question about the height is a good one! Basically, it was made up by the person who drew the diagram. In real life, it would depend on the angle of the "tilt" of the sides. The further the angle is from $90$ degrees, the shorter the height compared to the width. There is a precise formula using trigonometry, but I don't know if you want it. In practice, you could measure the height directly --- say by measuring straight up with a tape measure --- or, if that was not feasible, by measuring the angle directly with a protractor, and then using trig. to pass ... $\endgroup$ – Matt E Dec 16 '13 at 23:37
  • $\begingroup$ ... from the width to the height. (This uses the trig. formula I alluded to before.) I hope this helps. Regards, $\endgroup$ – Matt E Dec 16 '13 at 23:41
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    $\begingroup$ Oh thank you. =) So the person who drew this diagram thought 6cm would be a feasible number to show an example with. I'd like to learn that formula but I'd probably drive myself crazy trying to figure out why it is what it is. If I keep up my studies I'm sure I'll understand. $\endgroup$ – W3Geek Dec 17 '13 at 9:30
  • $\begingroup$ @W3Geek: Dear W3Geek, Exactly! Good luck with your studies. Best wishes, $\endgroup$ – Matt E Dec 17 '13 at 12:11

I've always found terms like height and width to be confusing. Try looking at it this way. You can work with it using the more familiar notions of squares and triangles by doing this. For example, you know the hypotenuse of the right triangle and one side of the right triangle, so you can use the Pythagorean theorem to find the length of the other side. From this, you can find the length of the long side of the rectangle. You already have the length of the other side. From all of this information, you can conclude that the area is equivalent to the area of the rectangle plus two times the area of one of the triangles (since they are the same).

Try this.

  • $\begingroup$ They're conufusing terms indeed. This is one reason why I wasn't able to learn basic geometry in high school. I could really never get past confusing termonology such as this. Unfortunately I don't know much about the Pythagoream Theorm except C^2 = A^2 + B^2. After I ponder on "what's the difference between width and height" for awhile I'll continue into more advanced studies such as this. In the image I posted, how is the width equal to 8cm while the height is only 6cm? $\endgroup$ – W3Geek Dec 16 '13 at 19:21

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