It is known that the expanded side view of a cone is a semicircle, and the ratio of its side area to its bottom area is calculated Try to explain it to me. I can't understand the answer like this The length of conical generatrix L and the radius of bottom R Conical generatrix length = developed circle radius = L Base circumference = 2 * pi * r = arc length of expanded circle = pi * l r=L/2 Side area s' = pi * L * L / 2 Bottom area s = pi * r * r S':S=1:2 Or such an answer Side area = expanded area = C (bottom perimeter) XL / 2 = 2 x R (bottom radius) XL / 2 And C = pie XL = 2 pie XR, i.e. L = 2R So the side area is 2 x R ^ 2 Bottom area = x R ^ 2 So the final result is 2:1

1. If the side view of a cone is a semicircle, then the arc length of the semicircle is the bottom circumference of the cone. Let the length of the arc length and the circumference be L
2. The arc length L of the side of the cone, because it is a semicircle, let the radius of the semicircle be r (this R is actually the length of the generatrix of the cone), then the area of the semicircle, that is, the side area S1, is π R ^ 2 / 2
3. We know that the bottom circumference of a cone is also L. for example, the formula of the bottom circumference of a cone is L = 2 π R (let the bottom radius of the cone be r), which is equal to the arc length of the side expansion L = π D / 2 = π R, so 2 π r = π R, and R = R / 2
4, bottom area S2 = π R ^ 2 = π (R / 2) ^ 2 = (π R ^ 2) / 4
5. Conclusion
The ratio of side area to bottom area is (π R ^ 2 / 2) / [(π R ^ 2) / 4] = 1 / 2
Brother, (I'm afraid I can't see clearly that "π" is the one of "3.1415926")

In division without remainder, the divisor x quotient of the dividend is 164, and the divisor is ()

The divisor = quotient multiplied by the divisor 164 / 2, the divisor multiplied by the quotient, that is 84, decomposes the prime factor, 84 = 2 * 2 * 3 * 7
The results are as follows: 2,3,7,4,6,21,14,12,42,28,84

The perimeter of a regular shape is equal to that of a circle. Given that the side length of a square is 3.14 cm, what is the area of the circle?
There is also a question. In a square with a circumference of 20 cm, draw the largest circle. What are the circumference and area of the circle?

(1) Circumference of circle = 3.14 * 4 = 3.14 * r & # 178;
R=2
S = 3.14 * r & # 178; = 3.14 * 4 = 12.56 square centimeter
(2) The circumcircle of a square
R=10
Perimeter = 2 * 3.14 * 10 = 62.8 cm
Area = 3.14 * 10 * 10 = 314 square centimeters

As shown in the figure, it is known that s is a point out of the plane of the parallelogram ABCD, m and N are points on SA and BD respectively, and SMMA = bnnd. Then the line Mn______ Plane SBC

It is proved that bnnd = bgag can be obtained by making ng ∥ ad through N, intersecting AB with G and connecting mg. According to the known condition bnnd = SMMA, SMMA = bgag, ∥ mg ∥ sb. ∩ mg ⊄ plane SBC, sb ⊂ plane SBC, ⊂ mg ∥ plane SBC. Ad ∥ BC, ∥ ng ∥ BC, ng ⊄ plane SBC, BC ⊂ plane SBC ∩ plane SBC ∩ plane MNG, ⊂ plane MNG, ∥ plane SBC The answer is ‖

It is known that the perimeter of the sector is 20cm. When we ask the value of the central angle a of the sector, the sector area s is the largest, and the maximum value of S is obtained

Let R be the radius
Arc length is ar, perimeter = Ar + R + r = 20, r = 20 / (a + 2)
Area = 1 / 2R (AR)
Area = 200A / (a + 2) ^ 2
If the derivative of a is 0, we can get
a=2
Where does the landlord not understand

Let the sum of the first n terms of the arithmetic sequence an be SN. We know that A5 = - 3 and S10 = - 40. (1) find the general formula of the arithmetic sequence an. (2) if the arithmetic sequence an is an arithmetic sequence,
And B1 = 5, B2 = 8, find the first n terms of BN and TN (the second question is not,

Common ratio q = 8 / 5, TN = 5 * (1 - (8 / 5) ^ n) / (1-8 / 5) = - 25 / 3 (1 - (8 / 5) ^ n)

I'm looking for an English speech about 100 words

Wild animalsAnimals are ou friends,but many people kill "friends" for their fur and meat.The people have killed a lot of wild animals.So many kinds of animals are in danger.Meanwhile ,the green house e...

It is known that the coordinates of the three vertices of the triangle ABC are a (- 2,3), B (1,2), C (5,4)
It's very urgent to have a specific process!

Vector 0A = (- 2,3) ob = (1,2) 0C = (5,4)
Vector Ba = 0a-0b = (- 2-1,3-2) = (- 3,1)
Vector BC = 0c-0b = (5-1,4-2) = (4,2)

The bottom of a cuboid is a square with a circumference of 16 cm and a height of 3 cm. What is the volume of the cuboid?

16 △ 4 = 4 (CM), 4 × 4 × 3 = 48 (CC). A: the volume of this cuboid is 48 CC

If cos (a - π / 6) = 12 / 13 and π / 6 ＜ a ＜ π / 2, Tana is obtained

π/6＜a＜π/2
0＜a-π/6＜π/3
sin(a-π/6)=5/13
sina=sin(a-π/6+π/6)=sin(a-π/6)cos(π/6)+cos(a-π/6)sin(π/6)
=(12+5*3^(1/2))/26
cosa=cos(a-π/6+π/6)=cos(a-π/6)cos(π/6)-sin(a-π/6)sin(π/6)
=(12*3^(1/2)-5)/26
tana=sina/cosa=(12+5*3^(1/2))/(12*3^(1/2)-5)