Heron's Formula, Cartesian Coordinate PlaneDate: 11/01/2001 at 07:36:16 From: Jacob Kohl Subject: Geometry If a triangle has sides 5, 6, and the square root of thirteen, what is the area of the triangle? Thank you for your help. Date: 11/01/2001 at 12:18:50 From: Doctor Greenie Subject: Re: Geometry Hi, Jacob - Cool problem! Thanks for sending it to us. When I am given the lengths of the three sides of a triangle and am asked to find the area, I have the following thoughts: (1) Maybe the lengths of the sides satisfy the Pythagorean theorem, indicating the triangle is a right triangle; then finding its area is easy. (2) If the triangle is not a right triangle, then maybe I can draw it on a Cartesian coordinate plane in such a way that its area is easy to determine. (3) If (1) and (2) don't get me anywhere, there is always Heron's formula: Area = sqrt[(s)(s-a)(s-b)(s-c)] where a, b, and c are the lengths of the sides and s is the semi- perimeter, s = (a+b+c)/2 For the particular problem you give.... (1) The given lengths do not satisfy the Pythagorean theorem, so the triangle is not a right triangle. (2) I looked at this for a while and didn't see anything right away. So I went on to Heron's formula, with the idea of coming back to look at this approach some more. (3) The use of Heron's formula looks formidable, with one side having length sqrt(13) - but the arithmetic - thanks to "(a+b)(a-b) = a^2-b^2" - turns out not to be at all bad. Here it is: With a = 5, b = 6, and c = sqrt(13), we have s = (11+sqrt(13))/2. Then we have A = sqrt[(s)(s-a)(s-b)(s-c)] (11+sqrt(13)) (11-sqrt(13)) sqrt(13)+1 sqrt(13)-1 A = sqrt( ------------- * ------------- * ---------- * ---------- ) 2 2 2 2 121-13 13-1 A = sqrt( ------ * ---- ) 4 4 108*12 A = sqrt( ------ ) 16 A = sqrt( 27*3 ) A = sqrt( 81) A = 9 Now let's go back and look at approach (2). I had noticed that the side with length sqrt(13) could be represented by the hypotenuse of a right triangle with sides of length 2 and 3; so I drew a right triangle on a Cartesian coordinate plane with vertices A(0,0), B(2,0), and C(0,3): | C | * * * * * * * * * * * * * --------------------------*-*-*-*-*-*-*-*------------- A | B | Then I thought about how I could use BC, with length sqrt(13), as one of the sides of the triangle in the problem. With a little playing around, I discovered that, if I chose point D(-4,0), I would have AD = 4, which with AC = 3 means CD = 5; then DB = DA+AB = 4+2 = 6. So then triangle BCD is a triangle with sides 5, 6, and sqrt(13), as required: | C | * * * * * * * * * * * * * * * * * * * ----*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*------------- D A | B | The area of this triangle is easy to calculate, because it has a base of 6 and a height of 3: A = one-half base times height = (6*3)/2 = 9. Thanks again for a nice problem. It is always fun for me to find an unusual problem that has a very neat solution. Write back if you have any questions about any of this. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/ |
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