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Trigonometry: Heron's Formula


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About this Lesson

  • Type: Video Tutorial
  • Length: 7:24
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 79 MB
  • Posted: 07/01/2009

This lesson is part of the following series:

Trigonometry: Full Course (152 lessons, $148.50)
Trigonometry: Applications of Trigonometry (14 lessons, $26.73)
Trigonometry: The Law of Cosines (4 lessons, $5.94)

Taught by Professor Edward Burger, this lesson comes from a comprehensive course, Trigonometry. This course and others are available from Thinkwell, Inc. The full course can be found at The full course covers trigonometric functions, trigonometric identities, application of trig, complex numbers, polar coordinates, exponential functions, logarithmic functions, conic sections, and more.

Edward Burger, Professor of Mathematics at Williams College, earned his Ph.D. at the University of Texas at Austin, having graduated summa cum laude with distinction in mathematics from Connecticut College.

He has also taught at UT-Austin and the University of Colorado at Boulder, and he served as a fellow at the University of Waterloo in Canada and at Macquarie University in Australia. Prof. Burger has won many awards, including the 2001 Haimo Award for Distinguished Teaching of Mathematics, the 2004 Chauvenet Prize, and the 2006 Lester R. Ford Award, all from the Mathematical Association of America. In 2006, Reader's Digest named him in the "100 Best of America".

Prof. Burger is the author of over 50 articles, videos, and books, including the trade book, Coincidences, Chaos, and All That Math Jazz: Making Light of Weighty Ideas and of the textbook The Heart of Mathematics: An Invitation to Effective Thinking. He also speaks frequently to professional and public audiences, referees professional journals, and publishes articles in leading math journals, including The Journal of Number Theory and American Mathematical Monthly. His areas of specialty include number theory, Diophantine approximation, p-adic analysis, the geometry of numbers, and the theory of continued fractions.

Prof. Burger's unique sense of humor and his teaching expertise combine to make him the ideal presenter of Thinkwell's entertaining and informative video lectures.

About this Author

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Heron's Formula
I thought that since this is the last little thing about the Law of Cosines, I would just tell you a really neat fact. I'm not going to actually prove this to you. I want to tell you a really amazing consequence of the Law of Cosines. What I mean is that if you want to prove what I'm about to tell you, all that it requires is a little bit of ingenuity, some careful manipulation of some algebra and trig, and the Law of Cosines, but here's what you can get out.
You know, a great question to ask is, how do you find the area of a triangle if you just know the length of the sides? So if a triangle looks like this, which is a right triangle--suppose that this is a 3,4,5 right triangle--well, finding the area of a right triangle is not a big deal because I know the base and I know the height or the altitude. So actually the area is just going to be the base times the height. One way of thinking about that, by the way, is just thinking about the fact that it really is a rectangle cut in half. So you take the area of the rectangle, base times height, and cut it in half. So in this case it would be , which would be 6. So this has area of 6. Okay, fine.
It turns out that if you have a triangle that looks like this--in fact, this is one that we saw a little bit earlier--10,8,3--it's harder to find the area of that. I know the base, but finding the height requires me to do a little bit of trig or something. I've got to figure out the height. I just don't take the 8 there because that's not actually going to work. Well, it turns out there's this great formula, and it's called Heron's Formula. Heron was a mathematician, Heron of Alexandria in fact. He was in olden days around, I think, 75 A.D., and he came up with this really, really cool formula. And the formula, to verify it, just requires some ingenuity and the Law of Cosines. Now, I'm not going to actually verify the formula for you but what I want to do is just tell you what the formula says.
So if you have any triangle at all, whether it's right or wrong--heh, heh, I made a joke again--and suppose the sides are a, b and c, and just knowing the sides I want to figure out a formula for the area without doing any work, it turns out that here's what you do. First of all, you compute the perimeter. The perimeter is easy. You just add this, add this, and add this. And I want half of that. So let's just let s be half of the perimeter. So I'll just take and divide it by 2. Add up those numbers and divide by 2; that's half the perimeter. Just from that number, here's the area. This is Heron's Formula. Area of a triangle equals . That is the formula for the area of a triangle. Absolutely amazing and very, very simple.
Let's try it in this example where we already know the answer. If you wanted to find the area of that triangle, let's use that formula. First of all, let's find the semi-perimeter. So if you add up the perimeter here, you have , so what we see is , which equals 6, so that's half the perimeter. Now let's plug it into this formula. The area should equal the square root of 6 times (6-a), so that's (6-3). (6-3) is 3. Then (6-b), so (6-4) which is 2. And then this last thing is (6-c), which is (6-5) which is 1. Well, what's 6 x 3 x 2? That's 36. The square root of 36 is 6. Isn't that cool? This is so cool! You can actually get the area for this triangle just using a simple formula. Know base times height.
Or this one. This one, of course, we couldn't even do before. Now let's see if we can do it. Let's see, we'd have 10 and 8 is 18--19, 20, 21. So the perimeter is 21. And so s would half the perimeter. So in this case what we see is s being half the perimeter, so s would be . And so what's the area? Well, the area would equal the square root of that number, , times , so , so what is that? Okay, now I've got to do a little arithmetic. Well, 3 is actually , and is what? Well, you can do this in your head. , over 2. Then you take and you subtract b, let's say that's 8. So 8 is , so you have , which is . And the last thing you multiply by is s, which is , minus c, which is 10. 10 is , so is just . And that whole thing is under a square root.
And so what do I see here? Well, I see on the bottom 16, so I'm taking the square root of that 16, which is 4. And what do I have on the top? On the top I have 21 x 15 x 5, and that equals . And so what we see is--which, by the way, you can work out a little bit if you want to--and you can see it's around 39.6 on top, and you divide that by 4, and what you get is roughly 9.9 something. But anyway, that's the rough estimate. The exact value is . That is the exact area of this thing, which is roughly 9.9 something. And notice all I did was use this really simple formula.
Well, I haven't proved this for you but it turns out that this formula is a consequence of the powerful--and now I hope you believe wonderful--Law of Cosines. Anyway, enjoy this formula and have some fun.

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