Math Cylindrical Cone Elementary School 6th grade 10

1.v cylinder = base area x height, v cone = 1 3 x base area x height, the volume remains the same, so get: x4x4x6 = 1 3x x3x3xh h = 32cm

2. v=πx（20/2)^2x（20-18)=3.Surface area = 2 x bottom area + side area = 2 x r 2 + 2 xrxh minus overlapping surface area, 2 middle size bottom surfaces and 2 minimum bottom surfaces.

So s=2x xr 2+2 xrxh++2 xr in xh++2 xrxh

Substituting it yields s s = 16 + 8 +4 + 2 = 30 = square decimeters.

1.Let the height of the cone be a centimeter, according to the title:

1 then a=32

2.Cubic centimetre.

3.square decimeters.

Question 1: Solve the height of the cone as x centimeters and solve x=32

Question 2: 20 2=10 cubic centimeters.

The third question is not clear.

The volume of the cylinder = the base area is high.

The volume of the cone = 1 3 The base area is high.

The formula for the area of the base area of the hall of the cylinder: s r

The formula for the side area of the cylindrical imitation is: s 2 rh

Formula for the surface area of a cylinder: S2 Rh 2 r The formula for the volume of a cylinder: v r h

The formula for the area of the cone is: v=1 3sh

The one-meter-long cylinder was sawn into two sections, and the standard area was increased by four square decimeters, what was the original volume?

Answer: 4 2 = 2dm2 2 1 = 2dm3

If you dig out the largest cylinder in a cube with an edge length of two cubic meters, what is the volume and surface of the remaining part?

Answer: 2 2 2=8m3 [2 2]2 Volume: Surface area: give points first.

There are two steel bars, one of which has a cross-section of a square with a side length of 4cm, long; The other is circular in cross-section, 4 cm in diameter and long. After the cross-section of these two steel bars is welded together, what is the volume?

Answer: Give points first. Give points.

Question 1: After sawing off the cylinder, there is an upper and lower bottom surface and a lower bottom surface, and the two bottom surfaces are circles as large as the original bottom surface, so the bottom area of the original cylinder is 2 square decimeters, so the original volume is 2 high = 2 10 (decimeters) = 20 square decimeters.

Question 2: The diameter of the bottom surface of the excavated cylinder is equal to the side length of the cube, so the radius of the cylinder is 1, the volume = 1 1 2 = 2, and the surface area of the remaining cube = the surface area of the cube -2 The bottom surface of the cylinder + the side area of the cylinder = ......

Question 3: It is enough to add the volume of two steel bars together.

Analysis: Cutting a cylindrical wooden block into the largest cone means that the cylinder is equal to the bottom of the cone, then the volume of the cut off part is 2 3 of the volume of the cylinder

The volume of this cylinder is:

A: The volume of this cylinder is:

The volume of the cone:

The volume of the cylinder:

Cylindrical volume: Conic volume 1:1

Cylindrical height: 1:1 conical height

Cylindrical base area: The bottom area of the cone is 1 1:1 3 1 1:3, so the base area of the cone is 100 3 300 (square centimeters).

Hello, 107894561230:

Bottom radius of a conical grain pile:

m) The bottom area of the conical grain pile:

5 sqm).

Volume of a conical grain pile:

cubic meters) the bottom area of the cylindrical grain hoard:

5 2) sq.m.).

The volume of the conical grain pile = the volume of the cylindrical grain hoard.

Height: (314 3) m).

A: It can be loaded up to 16 3 meters high.

It's simple:

Solution: Let the plumb bottom area be s

It is obtained by the volume equality: 1 3 xsx8 = 15x2

The solution is s=90 8=45 4=squared breakup).

Handmade I took a look at the answers they have right but the unit is wrong I wish you a happy day and learn to the next level!

The cylindrical glass base is known to have an area of 15 square decimeters. When the conical lead cone is removed from the water, the water drops by 2 decimeters.

So the area of the conical lead cone is:

15*2=30 (square decimeter).

Because the height of the conical lead cone is 8 decimeters.

So the bottom area of the conical lead cone is:

30 8 (1 3) = 90 8 (square decimeter).

The volume of the descending section of the water column in the cylindrical tank is equal to the conical volume, and the bottom area of the plumb hammer is s, and the volume is equal.

15x2=1/3 x8xs

The solution yields s=90 8 (decimeter).

The volume of the column that descends in the cylinder is equal to the volume of the cone, so the bottom area of the cone multiplied by 8 divided by 3 equals 15 times 2, so the bottom area is equal to 90/eighth