Rotate the right triangle ABC along the right angle sides AB and BC for 1 cycle respectively to get two different cones. Which one is larger and how much larger? I want to explain it carefully. I can't copy it on the Internet

Rotate the right triangle ABC along the right angle sides AB and BC for 1 cycle respectively to get two different cones. Which one is larger and how much larger? I want to explain it carefully. I can't copy it on the Internet

The right triangle ABC is rotated 1 cycle along the right sides AB and BC respectively to obtain two different circular vertebrae
When the axis is rotated with BC side, ab = 6 becomes the radius of rotation, and there will be a new geometric cone,
At this time, the volume of its bottom circular vertebrae is 6x6x3.14x1/3x3 = 113.04 cubic centimeter
When the axis is rotated along the AB side, BC = 3 becomes the radius of rotation, and there will be a new geometric cone,
At this time, the volume of its bottom circular vertebrae is 3x3x3.14x1/3x6 = 56.52cm3
Therefore, the volume of the cone formed by using the long right angle side as the radius is larger than that formed by the short right angle side. The volume difference between the two cones is 113.04-56.52 = 56.52 cubic centimeters

A right triangle has two right angles, the right sides are 3cm and 7cm, one or two right angles and one axis. What kind of figure will you get when you rotate them around quickly? How many centimeters are the diameter and height of its bottom surface?

The diameter of the bottom of the cone is 6 cm and the height is 7 cm when rotating along the edge of 3 cm
When rotating around the edge of 7 cm, the diameter of the bottom of the cone is 14 cm and the height is 3 cm

ABC. Find the shadow area of the triangle with the diameter of 20 cm

[3.14×（20÷2）2÷2-20×（20÷2）÷2]÷2，
=[3.14×50-100]÷2，
=[157-100]÷2，
=57÷2，
=28.5 (square centimeter);
A: the shadow area is 28.5 square centimeters

In RT △ ABC, ∠ C = 90 ° and take AB, AC, BC as the sides to make semicircles outwards. It is proved that the area of semicircles with oblique side as diameter is equal to the area of the other two semicircles

It is proved that: let BC = a, AC = B, ab = C
The conclusion is that
1/2∏（c/2）²=1/2∏（a/2）²+1/2∏（b/2）²
Left formula = 1 / 8 Π C
Right formula = 1 / 2 Π (a? 2 / 4 + B? 2 / 4) = 1 / 8 Π (a? + B?)
In a right triangle
c²=a²+b²
So the conclusion is proved

In the right triangle ABC, ∠ ABC = 90 ° and ab = 4, respectively AC.AB Make two semicircles for the diameter, then the area of the two semicircles

Let AC be x and BC be y, then x ^ 2 + y ^ 2 = 4 ^ 2 = 16
The area of AC circle is [(x / 2) ^ 2] π
The area of BC garden is [(Y / 2) ^ 2]
AC circle + BC circle = [(x / 2) ^ 2] π + [(Y / 2) ^ 2] π, extract x ^ 2 + y ^ 2, and get 4 π
The area of the two gardens is 4 π

As shown in the figure, in △ ABC, ∠ C = 90 °, BC = 3, AC = 4___ .

∵ in ᙽ ABC, ∵ C = 90 °, BC = 3, AC = 4,
According to Pythagorean theorem, ab = 5,
That is, the radius of the semicircle is 5
2，
So the area of the semicircle is 1
2×π×（5
2）2=25
8π，
So the answer is: 25
8π．

It is known that △ ABC is a right triangle with ∠ ACB as the right triangle and AB, BC and Ca as the diameters respectively. If the sum of the areas of the three semicircles is 64

1/2π(AB^2+BC^2+CA^2)=64π
AB^2+BC^2+CA^2=128
AB ^ 2 = BC ^ 2 + Ca ^ 2
SO 2 * AB ^ 2 = 128
AB^2=64
AB=8

In the RT triangle ABC, the angle c is equal to 90 degrees, AC = 4, BC = 3. Take the diameter of the hypotenuse AB as the semicircle, and calculate the area of the semicircle

The triangle is a right triangle AC = 4, BC = 3 according to the Pythagorean theorem AB = 5
Because the diameter of semicircle is ab = 5
So the semicircle area s = (1 / 2) π R ^ 2 = (1 / 2) π × (5 / 2) ^ 2 = 25 π / 8

In the right triangle ABC, the angle a = 90 ° BC = 12cm, s triangle = 30cm, ab =?

(ab)^2+(ac)^2=144
(ab)*(ac)=30*2
After solving the equation, it will be OK to solve ab

In the right triangle ABC, ab = 20cm, BC = 60cm. Make the largest square in the triangle. The area of the square is () square centimeter

225 square centimeter