David -

My gut reaction is that people are very poor at evaluating equivalent areas on paper, but very good at it in real life.

If you had an apple pie (or, my preference, a pepperoni pizza) cut according to the top/left example, nobody older than a three year old would be fooled into thinking that the inner and outer pieces were of equivalent area. (Just try giving the three year-old the smaller slice)

On paper, however, people use different clues when presented with the same issue presented in a diagram.

As presented, I would guess that most people would favor the top/left (I know too much to have a gut reaction here -- I only have a reasoned reaction). My guess is that for the small group of users that I am acquainted with in my company, problem 3/4 would choose the upper left.

However -- if you showed them the same diagram, with the two slices detached, as if they had been sliced and slightly removed from the pie / pizza, thereby removing the visual comparison to the radius of the circle -- hardly anybody would say the two slices were equivalent.

You going to tell us eventually what's behind your question?

Jim



>Crank up VFP and run the code below. Tell me which of the two diagrams represent 50% in each pie slice. I don't want to bias the opinions by discussing the logic yet. I'm looking for first gut reactions.
>
>

>clear
>_screen.ForeColor = 0
>
>nXCenter = 150
>nYCenter = 150
>nRadius = 100
>_screen.Circle( nRadius, nXCenter, nYCenter, 1 )
>
>_screen.Circle( nRadius * 0.5, nXCenter, nYCenter, 1 )
>
>nTheta = pi() * 0.50
>nXRim = nXCenter + nRadius * sin( nTheta )
>nYRim = nYCenter + nRadius * cos( nTheta )
>_screen.Line( nXCenter, nXCenter, nXRim, nYRim )
>
>nTheta = pi() * 1.25
>nXRim = nXCenter + nRadius * sin( nTheta )
>nYRim = nYCenter + nRadius * cos( nTheta )
>_screen.Line( nXCenter, nXCenter, nXRim, nYRim )
>
>
>nXCenter = 350
>nYCenter = 350
>nRadius = 100
>_screen.Circle( nRadius, nXCenter, nYCenter, 1 )
>
>_screen.Circle( nRadius * 0.7071, nXCenter, nYCenter, 1 )
>
>nTheta = pi() * 0.50
>nXRim = nXCenter + nRadius * sin( nTheta )
>nYRim = nYCenter + nRadius * cos( nTheta )
>_screen.Line( nXCenter, nXCenter, nXRim, nYRim )
>
>nTheta = pi() * 1.25
>nXRim = nXCenter + nRadius * sin( nTheta )
>nYRim = nYCenter + nRadius * cos( nTheta )
>_screen.Line( nXCenter, nXCenter, nXRim, nYRim )
>