triangle area without trig. « Next Oldest | Next Newest »

 ▼ Hal Bitton in Boise Unregistered Posts: 291 Threads: 43 Joined: Jun 2007 01-07-2008, 01:06 AM Hi folks I have an isoceles (non-equilateral) triangle, with givens of the angles and the base. Is there any way to find the area without using a trig function. It's my contention that one cannot...but then what do I know. Best regards, Hal ▼ Walter B Unregistered Posts: 4,587 Threads: 105 Joined: Jul 2005 01-07-2008, 01:48 AM For the area of an arbitrary triangle you need its base and height. On first glance, I do not see a simple way to get the height even of your isosceles triangle without trig. Karl Schneider Unregistered Posts: 1,792 Threads: 62 Joined: Jan 2005 01-07-2008, 01:50 AM Hi, Hal -- I remember from Geometry 30 years ago that congruence of two triangles can be proved by equivalence of three sides, so I figured that it could be done -- i.e., a triangle of three given side lengths is unique. However, the expression for area isn't quite as simple as I thought it might be: From Schaum's Mathematical Handbook, Equations 4.5 and 4.6: ```Area = sqrt [s*(s-a)*(s-b)*(s-c)] where s = (1/2)*(a + b + c) and {a, b, c} are the lengths of the three sides of the triangle ``` Both the HP-41 Math Pac and the HP-11C Owner's Handbook and Problem Solving Guide include triangle-solutions programs, but neither uses the formula provided above to calculate area. -- KS Edited: 7 Jan 2008, 2:13 a.m. after one or more responses were posted ▼ Walter B Unregistered Posts: 4,587 Threads: 105 Joined: Jul 2005 01-07-2008, 02:00 AM Congratulations! At least you have the right books ;) George Bailey (Bedford Falls) Unregistered Posts: 335 Threads: 12 Joined: Dec 2007 01-07-2008, 02:31 AM Quote: and {a, b, c} are the lengths of the three sides of the triangle [/pre] How to compute the side lenghts if base and angles are given - without trigonometry, as the OP needs to do? ▼ Karl Schneider Unregistered Posts: 1,792 Threads: 62 Joined: Jan 2005 01-07-2008, 04:52 AM "George" (most Americans will recognize that this is a pseudonym) posted: Quote: How to compute the side lengths if base and angles are given - without trigonometry, as the OP needs to do? Oh! I read what I expected to see, i.e., calculate the area of a triangle without trig functions, given only the sides. Why would anyone not want to use -- or expect to avoid -- trigonometry when one or more angles, but not all sides, are given? For an isosceles traingle, knowing one angle is enough to deduce the other two. Every calculation of area, height, or side length, will require a trig calculation, though. -- KS ▼ George Bailey (Bedford Falls) Unregistered Posts: 335 Threads: 12 Joined: Dec 2007 01-07-2008, 08:21 AM Quote: "George" (most Americans will recognize that this is a pseudonym) If only you were right...;-) My parents lived near Bedford Falls, Madison, IN, by the time I was conceived. As my father's given name is George he wanted his first son to be named after him but being only 18 didn't like the idea of going by George senior ... they came up with the idea of this unique middle name ;-) ▼ Karl Schneider Unregistered Posts: 1,792 Threads: 62 Joined: Jan 2005 01-07-2008, 10:30 PM George -- Ah! That's rather clever; my apologies. I should have at least included "probably" or "likely" in the 'pseudonym' statement. I suppose that, for example, a Jesus Cordero of Nazareth, Pennsylvania could use "Jesus of Nazareth" as a technically-correct moniker, but people would likely doubt its authenticity. ;-) For those who are unfamiliar with the American cultural reference, here's an excerpt from http://en.wikipedia.org/wiki/It's_a_Wonderful_Life "It's a Wonderful Life is a 1946 American film... (which) takes place in the fictional town of Bedford Falls shortly after World War II and stars James Stewart as George Bailey... Best regards, -- KS ▼ Trent Moseley Unregistered Posts: 406 Threads: 47 Joined: Jul 2005 01-07-2008, 11:16 PM Karl, Great investigatative work! tm ▼ George Bailey (Bedford Falls) Unregistered Posts: 335 Threads: 12 Joined: Dec 2007 01-08-2008, 04:43 AM Quote: Great investigatative work! Trent, haven't you watched the movie in the last couple of days???? ;-) ▼ Ron G. Unregistered Posts: 70 Threads: 9 Joined: Jul 2007 01-08-2008, 09:49 AM We watch it every year! * * * You'll shoot your eye out, kid! :^) (I know, different movie, but another Christmas requirement.) Ron G. Unregistered Posts: 70 Threads: 9 Joined: Jul 2007 01-08-2008, 01:10 PM I never understood that "Buffalo Gal" song, George. What is a buffalo gal, and is it supposed to be a compliment? ▼ George Bailey (Bedford Falls) Unregistered Posts: 335 Threads: 12 Joined: Dec 2007 01-08-2008, 01:32 PM Quote: I never understood that "Buffalo Gal" song, George. What is a buffalo gal, and is it supposed to be a compliment? ...my guess has allways been that it refers to girls from Buffalo, NY ▼ Ron G. Unregistered Posts: 70 Threads: 9 Joined: Jul 2007 01-09-2008, 12:43 AM Well now I can die happy. %^) Howard Boardman Unregistered Posts: 30 Threads: 10 Joined: Jul 2007 01-07-2008, 03:31 AM Using equation list in a HP 33s or HP 35s A=SQRT(((X+Y+Z)/2)×(((X+Y+Z)/2)-X)×(((X+Y+Z)/2)-Y)× (((X+Y+Z)/2)-Z) ▼ Steve (Australia) Unregistered Posts: 241 Threads: 29 Joined: Jan 2008 01-12-2008, 05:37 AM I think that is more clearly stated at http://www.clarku.edu/~djoyce/trig/area.html: "Another is Heron's formula which gives the area in terms of the three sides of the triangle, specifically, as the square root of the product s(s – a)(s – b)(s – c) where s is the semiperimeter of the triangle, that is, s = (a + b + c)/2" Same answer, but provides a little more insight :-) George Bailey (Bedford Falls) Unregistered Posts: 335 Threads: 12 Joined: Dec 2007 01-07-2008, 04:41 AM A very easy way to get the area of said triangle without trig functions would be to plot it and measure... You might also consider using a Taylor series for sin and cos, so you would technically use only powers and factorials ;-) John Keith Unregistered Posts: 46 Threads: 2 Joined: Feb 2007 01-07-2008, 09:27 AM If you can specify the X,Y coordinates of the verteces of the triangle, you can use the polygon area formula, which uses only addition, subtraction, and multiplication. Go to this site: http://www.faqs.org/faqs/graphics/algorithms-faq/ and navigate to section 2.01 for an explanation. The link given in that section has further formulas which may be useful. HTH, John Hal Bitton in Boise Unregistered Posts: 291 Threads: 43 Joined: Jun 2007 01-07-2008, 10:42 AM Thank you everyone for your responses. Best regards, Hal ▼ Juan J Unregistered Posts: 195 Threads: 25 Joined: Jul 2005 01-07-2008, 12:06 PM Hello, I remember a formula that involved a 3x3 determinant, given the coordinates of each point. If (x1, y1), (x2, y2) and (x3, y3) are the coordinates, then the area is given by Coordinates must be designated clockwise. And no trigonometry is involved. Hope this helps. ``` |x1 y1 1| ``` ```A = |x2 y2 1| = x1*(y2-y3)-y1*(x2-x3)+1*(x2*y3-x3*y2) ``` ``` |x3 y3 1| ``` Edited: 7 Jan 2008, 12:11 p.m. ▼ Stefan Vorkoetter Unregistered Posts: 217 Threads: 21 Joined: Aug 2007 01-07-2008, 01:12 PM Except if all you know is one side (without loss of generality, one can assert that the ends of the side are at 0,0 and x,0) and two angles, you need trigonometry to determine the coordinates of the third vertex. Stefan ▼ Walter B Unregistered Posts: 4,587 Threads: 105 Joined: Jul 2005 01-07-2008, 05:08 PM Quote: Except if all you know is one side (...) and two angles, you need trigonometry to determine the coordinates of the third vertex. Except?? ▼ Stefan Vorkoetter Unregistered Posts: 217 Threads: 21 Joined: Aug 2007 01-07-2008, 08:15 PM "Except... " is short for: "This would work, except for the fact that ...". A casual-English abbreviation. ▼ Walter B Unregistered Posts: 4,587 Threads: 105 Joined: Jul 2005 01-08-2008, 12:27 AM Yes, Sir ;) Just IMHO your statement would have been unambiguous without this word. Egan Ford Unregistered Posts: 1,619 Threads: 147 Joined: May 2006 01-07-2008, 05:51 PM Quote: I have an isoceles (non-equilateral) triangle, with givens of the angles and the base. Is there any way to find the area without using a trig function. If a is the base the B and C are the two angles, then: ```Area = a*a*sin(B)*sin(C)/(2*sin(B+C)) ``` You can replace sin() with: ```sin x = (exp(i*z) - exp(-i*z)) / (2*i) ``` or ```sin x = sum( (((-1)^(n-1)) / (2*n - 1)!) * x^(2n-1) ) ``` Why would you want to do this? Andrés C. Rodríguez (Argentina) Unregistered Posts: 1,193 Threads: 43 Joined: Jul 2005 01-08-2008, 09:18 PM I apologize in advance if I'm missing something already discussed in the thread. If we have an isosceles triangle, and divide the uneven side at its half, we can plot the height as a segment from that midpoint to the opposite vertex. Then, taking the uneven side as the base, which we will call "b" and each of the other sides as "a", the height "h" is h = sqrt (a^2 -(b/2)^2) , so the area is A = (b/2) * sqrt (a^2 - (b/2)^2) ... which is obtained without trigs. ▼ Karl Schneider Unregistered Posts: 1,792 Threads: 62 Joined: Jan 2005 01-08-2008, 10:03 PM Hi, Andres -- Indisputable mathematics, but not enough data: Only one side length (the original base) and the angles were given. If both unique side lengths had been given, then we wouldn't even need the angles -- the general formula that I (and John Howard) provided, or the determinant formula provided by John Keith and "JuanJ", could also be utilized to calculate area. Yours is a bit simpler, though, for the special case of an isosceles traingle. -- KS Edited: 9 Jan 2008, 1:45 a.m. ▼ Andrés C. Rodríguez (Argentina) Unregistered Posts: 1,193 Threads: 43 Joined: Jul 2005 01-09-2008, 08:32 PM Karl: Thank you for the clarification. I was in a hurry and assumed that both lenghts ("a" and "b") were given. Dave Boyd Unregistered Posts: 33 Threads: 1 Joined: Nov 2006 01-09-2008, 05:23 PM Well, it doesn't say you have to calculate the area, only "find" it. So you could draw it on a large sheet of foil, cut it out and weigh it, and the divide the weight by the weight of a square of unit dimension... You could probably get three figures? Otherwise, I don't think so. Gerson W. Barbosa Unregistered Posts: 2,761 Threads: 100 Joined: Jul 2005 01-09-2008, 06:43 PM You may find the following interesting: Gerson. ▼ Chris Dean Unregistered Posts: 120 Threads: 9 Joined: Aug 2005 01-10-2008, 03:29 AM To turn the problem on its head consider the base of the triangle, length b, to be a chord in a circle and the two equal sides of the triangle representing radii. The angle at the centre of the circle, theta (used in radians), is given as part of the problem. Then an approximation to the required area can be given by Area = b^2 / (4 * theta) * (sqrt(4 – theta^2) / 2 + 1). This actually represents the average between the triangular area with a base length of b (calculate the radius or side using s=r*theta) and the sector area of arc length b (Area= r^2 * theta / 2). This formula can be used for theta values up to approximately 114 degrees (< 2 radians) and gives a maximum error less than 5% to the true area value. The only point trig has been used is to determine the true value for analysis purposes only! Below is a listing for an HP17bII+ AREA=L(T:THETA X PI / 180) X 0 + L(B:BASE) X 0 + SQ(G(B)) / (4 X G(T)) X SQRT(4 – SQ(G(T))) / 2 + 1) where / means divide and X multiply. Enter THETA in degrees and then BASE value. Click on AREA to Solve. Chris Dean Edited: 10 Jan 2008, 11:03 a.m.

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