Here's a simple one, so speed counts?

area exposed in the center = area of square - 4 X (a quarter area of disk)

= area of square - area of disk

= (2r)^2 - piXr^2

= 4r^2 - pi(r^2)

= (4-pi)(r^2)

= (4 - 3.14)(7.5^2)

= 0.86(56.25)

= 48.375 cm^2

Well, i'm assuming the circles touch the ends of a square table... you didn't really specify how much area around the circles there are... but supposing the touch the edges, then the table has an area of 16R^2, then subtract the area of the disks which is 4PiR^2 and we get 4R^2(4-Pi), assuming pi=3.14 and R=7.5cm, 4(56.25)(.86)=193.5Sq.Cm, which further more means that 193.5/852=22.71% exposed, remainder, covered by the disks. This is also assuming that the disks are whole, and not rings with empty centers.

Radius of disc = R

Area of disc = pi x R^2

Side of square = 2R

Area of square = (2R)^2 = 4R^2

Area of square covered by 1 disc = 1/4 x (area of 1 disc)

Area of square covered by 4 discs = 4 x 1/4 x (area of 1 disc) = area of 1 disc

Area of the table exposed = Area of square - Area of square covered by 4 discs = 4R^2 - pi x R^2 = (4 - pi)R^2 = (4 - 3.14159)R^2 = .858407346 R^2

When R = 7.5cm,

Area of the table exposed = .858407346 x R^2 = .858407346 x7.5^2 = 48.28541324 cm^2

The area of the square less the area of the incircle, or (4 - pi) times the radius. Consider a tiling -- makes it easier to see.

It is area of square - 4 * (1/4 * area of disk)

so, area of square - area of disk

ie. (7.5*2)^2 - pi*7.5^2

ie 48.2854 sq.cm

one side would be 7.5 x 2 = 15cm

the area would be 15 * 15

= 225 sq cm

225cm squared