# How to decide: Sergey Petrovich decided to build a greenhouse of 6 m on the site?

1) the length of the greenhouse 6 m or 600 cm. Distance between arcs 70 cm. Need to share:

600/70 = 8.6, round the larger direction – 9 arc + 1 in front. A total of 10 arcs.

2) There will be three beds in the greenhouse, between them – two paths. Each has a width of 60 cm, the length of the paths is equal to the length of the greenhouse, that is, 600 cm. Find the area:

60*600 = 36000 cm² – area of one track.

36000*2 = 72000 cm² – the area of two tracks.

20*20 = 400 cm² – the area of one paving slabs.

72000/400 = 180 (pcs.) – tiles will be needed.

180/8 = 22.5, round to the whole – 23 packs of tiles you need to buy.

3) the width of the greenhouse can be found based on the semicircle known to us, which is 5 m. We will carry out calculations according to the formula for the location of the circumference L = 2πR, the semicircle is 5 m, which means that the circumference length is 5*2 = 10 m.

10 = 2πr

5 = 3.14R

R = 1.59 m, that is, AD = 1.59*2 = 3.18 m

Since points B, O and C are divided by AD into 4 parts, then in = OS = 0.795, and BC = 1.59 m.

Thus, the width of the entrance to the greenhouse is 1.59 m.

4) The width of the greenhouse is AD = 3.18 m = 318 cm, the width of the two paths 60*2 = 120 cm.

318-120 = 198 cm – the width of all three beds.

Let the narrow bed X cm, then wide – 2x cm. Then:

x+x+2x = 198

4x = 198

x = 49.5 cm – width of a narrow bed.

49.5*2 = 99 cm – width of a wide, central bed.

5) Width of beds without paths 198 cm, length 600 cm. The area for the beds is:

198*600 = 118800 cm²

The width of the site under the greenhouse is 318 cm, length 600 cm. The area of the site under the greenhouse:

318*600 = 190800 cm²

118800/190800 *100%= 62.3%