Making a right-angled triangle out of AB, the line going down from B and the line going right from A, we have one angle and one length. The 'opposite' side is 0.8m, the angle is 60° and we want to find the 'adjacent' side. tanθ=o/a, so tan 60=0.8/a, so a=0.8/tan 60=0.461... There are 21 of these, so 9.6995... of the 10m is being taken up by them. That leaves 0.3005... for CD. There are 2 of CD, so they will get 0.1503... each.

Making another right-angled triangle out of CD and the corner, we now know the side length and we want to know the hypotenuse. cosθ=a/h, so cos 60=0.1503.../h, so h=0.1503.../cos 60=0.30m to the nearest cm, which is length CD.

The outer frame has two lengths and two widths on it, so in total it is 10+10+0.8+0.8=21.6m long and one length AB is 0.8/sin 60=0.9238...m, so 21 of them are 19.40m to the nearest cm.