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The cylinder problem in the second volume of sixth grade mathematics
"If a 3-meter-long cylindrical steel is cut into three sections, the surface area will increase by 12.56 cubic decimeter", that is:

Cut twice and increase the area of four round bottoms, so the area of one round bottom is:

12.56/4=3. 14 square decimeter

So, the bottom circumference is:

2 * 3.14 * √ (3.14/3.14) = 6.28 decimeter.

The transverse area is:

6.28*30= 188.4 square decimeter

Surface area of cylinder = side area+two bottom areas.

So, the surface area of this steel is:

188.4+2 * 3.14 =194.68 square decimeter.