What is the area of the smallest ellipse that can be circumscribed around a 345 triangle?
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What is the area of the largest ellipse that can be inscribed in a 345 triangle?
3.14159
OK, bear with me on this. I fuddled thru.
orientate the triangle with the length 3 at the bottom and 4 on the right.
calculate the internal angles of the bottom corners,
left corner = 53.13 degrees
right corner = 90 degrees
now 1/2 this to give you the angular bisectors.
left corner = 26.565 degrees
right corner = 45 degrees
the point at which these 2 bisectors cross is the "centre" of your triangle and is an equal distance from the 3 sides.
you now need to calculate the height from this point to the base line.
firstly give the left hand angled side a length value of "1"
then calculate the height by using:
height=sin theta(26.565) = 0.4472
this value is also the length from the right hand corner to the
perpindicular line drawn down from the centre point. I will call this
point Z
now calculate the length from the left corner to Z using:
degrees = length theta(26.565) = 0.8944
add these 2 figures together and divide 3 (lenghth of base) by the result
0.8944 + 0.4472 = 1.3416
3 / 1.3416 = 2.2361
multiply this figure by the length from left corner to Z
2.2361 x 0.8944 = 2.000
this means that the length right corner to Z = 1.00
and also that the height of the centre point is 1.00
this means that if the central ellipse were circular, it would have a radius of 1.00, giving it an area of 3.14159
please note that this calculation is based on the ellipse being a perfect circle, if it's not then i haven't got a clue.
Juggler

Posted by Juggler
on 20040518 00:29:35 