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Mathematical problems of chimney
The topic is chimney, so both sides are empty. Only the lateral area is calculated, and whether the length and width are enough is considered. The radius is 2dm, the bottom perimeter can be calculated as 6.28dm, the width of the iron sheet is used as the height and the length is used as the bottom perimeter of the chimney, so that a chimney is completed. General process: π×2=6.28dm.

6.28 decimeter < 15.7 decimeter

4dm=4dm

This iron can be made into a chimney.

5. The topic is cylindrical oil drum, so its area is side area +2 bottom area. And there are five questions to do, and finally multiply them by five to get the answer. The process is as follows:

The height of the oil drum is 24÷2= 12cm.

1 oil drum area π×(24÷2)? × 2+(π× 24 )×12 =1808.64 (square centimeter)

The area of five oil drums is 1808.64×5=9043.2 (square decimeter).

A: It needs at least 9043.2 square decimeters of iron sheet.

6. This problem is equivalent to calculating the side area of this cylinder, plus the area of a round bottom (because there is no cover). The process is as follows:

Transverse area 4 x 5 = 20 (square decimeter)

Bottom area ωω(4/2)? = 12.56 (square decimeter)-π is 3. 14 with a radius of 2dm.

The total area is 20+ 12.56=32.56 (square decimeter).

A: At least 32.56 square decimeters of iron sheet is needed.

π in the above question is 3. 14. I'm only a junior two student, so it's not easy to help you answer questions and type. I hope it can be adopted and praised.