Re: triangle area without trig (Approximate solution) Message #31 Posted by Chris Dean on 10 Jan 2008, 3:29 a.m., in response to message #30 by Gerson W. Barbosa
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.
