triangle area without trig.



#32

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


#33

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.

#34

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


#35

Congratulations! At least you have the right books ;)

#36

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?


#37

"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


#38

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 ;-)


#39

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


#40

Karl,

Great investigatative work!

tm


#41

Quote:
Great investigatative work!

Trent, haven't you watched the movie in the last couple of days???? ;-)


#42

We watch it every year!
*
*
*
You'll shoot your eye out, kid! :^) (I know, different movie, but another Christmas requirement.)

#43

I never understood that "Buffalo Gal" song, George. What is a buffalo gal, and is it supposed to be a compliment?


#44

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


#45

Well now I can die happy. %^)

#46

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)


#47

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 :-)

#48

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 ;-)

#49

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

#50

Thank you everyone for your responses.

Best regards, Hal


#51

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.


#52

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


#53

Quote:
Except if all you know is one side (...) and two angles, you need trigonometry to determine the coordinates of the third vertex.

Except??

#54

"Except... " is short for: "This would work, except for the fact that ...".

A casual-English abbreviation.


#55

Yes, Sir ;) Just IMHO your statement would have been unambiguous without this word.

#56

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?

#57

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.


#58

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.


#59

Karl:

Thank you for the clarification. I was in a hurry and assumed that both lenghts ("a" and "b") were given.

#60

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.

#61

You may find the following interesting:

http://web.maths.unsw.edu.au/~norman/book.htm

Gerson.


#62

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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