Ask two math cylinder questions and one math question about cylinders

1.If you want to be as high as possible, let the length of the rectangle be the height of the cylinder, and the width is the circumference of the bottom surface, and the radius of the bottom surface can be found according to the perimeter, and of course, the bottom area can also be found.

Equation: This is the radius of square centimeters.

2.The sum of the volume of the cuboid plus the cube is the volume of the cylinder, knowing the volume of the cylinder, and then finding the base area according to the diameter of the bottom surface, you can find the height, so that the surface area can be found according to the formula.

Formula: cylindrical volume: 7x3x9 + 5x5x5 = 189 + 125 = 314 cubic centimeters.

Bottom radius: 20 2 = 10cm

Area: square centimeters.

Height: 314 314 = 1cm

Surface area: square centimeters.

Question 1: Divide by 2 and multiply by 4 to get it.

Question 2: 7 times 3 times 9 plus 5 times 5 times 5 times 5 to get 314

Multiply 10 squared to get 314 314 divided by 314 to get 1 centimeter.

314 + 314 + 20 multiplied by 1.

1. Make a cylinder as high as possible.

Just take the length of the rectangle as the height, the width as the circumference of the bottom circle of the cylinder can be obtained that is, the circumference of the cylinder 2 r = , autumn pond r, you can get the area of the ground circle.

2. It can be seen that the volume of the cylinder is the sum of the volume of the cuboid and the volume of the cube, that is, 7*3*9+5 3

The volume of the cylinder is equal to the base area multiplied by the height, so the height of the cylinder can be obtained, and then the surface area is obtained by multiplying the bottom circumference by the height of the surface area formula.

1.There is a rectangle of centimeters long and centimeters wide, roll it into the side with it, and how many square centimeters of cardboard can be used to make a cylinder as high as possible?

x22.There is a cuboid with a length of 7 cm, 3 cm and a height of 9 cm, and a cube with an edge length of 5 cm, which is cast into a cylinder (excluding loss). If the bottom surface of this cylinder is 20 cm in diameter, what is its surface area?

Base radius: 20 2 = 10 (cm).

Area: square centimeters.

Height: 314 314=1 (cm).

Surface area: square centimeters.

1.The cylindrical is the highest; wide as the circumference of the upper and lower bottoms; Circumference 3 r= gives r= ; Then use r = area (' is the radius squared); This is a bottom one, multiply by 2 to get the highest cylinder.

2.Length multiplied by width and then height: 7x3x9 + 5x5x5 = volume;

Then find the height of the cylinder: the height of the cylinder = the volume of the bottom area;

Base area = (20 2).

Volume Base Area = High;

2 r = bottom circumference;

Circumference x height = side area;

Side area + 2 base area = surface area.

Thanks.

1) 2 kinds of height, and, take as high.

Centimeter.

1, to two areas is 50...A circle of 215 square centimeters, if it is a square, it will have two areas of 63,935 square centimeters.

1.Solution: 2 r=, r=4, s=2· r 2=

2 solution: v=7*3*9+5*5*5=314, r=10, bottom area s= r 2=314, h=v s=1, surface area s'=2*s+2 r*h=

The surface area of the cylinder = upper and lower area + lower base area + perimeter * height.

So there's a girth, right?

Joints should be on both ends, right? Cylindrical circumference2 total length (

A, 16/9 decimeter! 2/15 of a meter!

Dude gives you a formula for you to do the math lately! It's not interesting for me to tell you directly!

Cylindrical side. Area formula.

Side area. The circumference of the underside is high.

The formula for the surface area of a cylinder: surface area = 2 r2 + bottom perimeter of the tall cylinder.

Volume formula. Volume = Volume =

The base area is high.

Radius of the bottom surface of a cylindrical sink: 6 2 = 3 dm.

Radius of the bottom surface of a cylindrical iron block: 4 2 = 2 dm.

1. (decimeter.)

2. (decimeter.)

Because the reduced area is the side area, the formula for finding the side area is the bottom circumference * height, and the truncated 4 cm is the height, so the bottom perimeter can be used.

This is a question of a cylindrical side view.

Decimeter = meter, 2 cm = meter.

The length of the rectangle is 2 meters.

Width = meters. Area = square meters.

1.The volume of a cylinder is the base area multiplied by the height.

2.The height of the cylinder is.

3.The base area of the cylinder is.

Simple..

A: The diameter is 8 cm.

1：π*1^2*2*

2: Cone volume: , cylindrical volume: 3*3=9

1.x1 rounded up to 116 days.

2.The volume of the cone is.

Because the volume ratio of the cylinder to the cone of the equal-bottomed ascending is 3

So the volume of cylindrical steel is required to be 3x3 = 9 cubic decimeters.

1 Arithmetic: The volume of the granary is approximately equal to 116

2 arithmetic: cone = 1 3 cylindrical cone x3 = cylindrical weight in kilograms divided by the volume of the cylinder.

The top area of the crown of the hat = square decimeters.

The side area of the top of the hat = square decimeters.

The area of the brim = the area of the ring = the square of the square - the square of 1] = the square decimeter to make this hat at least with a flower cloth = square decimeter = square meter.

The crown and brim area is (

Cap side area is 2**

The total area is:

【Analysis】 The total area should be equal to the circle with a radius of 2 decimeters, plus the side area of the cylindrical shape, followed by the area of the rectangle with the height (1 decimeter) as the width and the circumference of the circle with a radius of 1 decimeter as the length.

So s pix2 2+2pix1x1=6pi square decimeter = square meter.

1。Base area = r=

The circumferential area of the pool = 3*2 r = 6

In total, 7 meters is equal to 30 decimeters.

2 * 2 * 30 * 2 roots = 120

1.The area of the plastering part is *1 +2 *1*3=7 m 22 m

2.Painted area = 2 * 2 * 2 * 30 = 240 754d m

Agree with the two big brothers in front of you.

Set up a straight line. l is: y=k(x+1)+4

k 0).

l and the negative half axis of the x-axis and the positive half axis of the y-axis intersect a(-4 k-1,0), b(k+4, respectively

Two points, then ao=4 k+1, bo=k+4;

The inscribed round round of the AOB', then the radius of the inscribed circle is . r=1.

ao^2bo^2

ab 2 Pythagorean theorem).

ao+bo-ab=2r=2

It can be seen from the above three formulas.

k = double root number 2) (rounded negative).

Then the functional expression of the straight line l is.

kx-y+k+4=0