What is the area of the smallest ellipse that can be circumscribed around a 3-4-5 triangle?
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What is the area of the largest ellipse that can be inscribed in a 3-4-5 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 2004-05-18 00:29:35