Math problems, non 6 years level about circles to join in the fun

The radius of the flower bed is.

m) The outer radius of the path is.

5 1 6 (m).

The area of the trail is:

6 6 sq.m.).

The radius of the flower bed is.

m) The outer radius of the path is.

5 1 6 (m).

The area of the trail is:

6 6 5 5) sq.m.).

Classic Ring Problem!!

Good luck!

5+1=6(m)

of the square - 5 of the square).

square meters) A: The area of the path is square meters.

Note: Annular area formula.

s yin = (square of r - square of r).

It is already known that the formula for the circumference of a circle is 2 r

So the radius of the flower bed is (r is the radius of the circle).

As in Equation 2 r =

r = 5 plus a one-meter-wide circular path.

5 1 6 (m).

Therefore, the area of the Great Circle is (i.e. 6 6 square meters plus the area of the flowerbed on the circular path), and then the area of the flowerbed is subtracted from the area of the Great Circle.

square meters) A: The area of the path is square meters.

To learn math well, you need to listen well in class.

Do your homework carefully, I hope you learn math well.

Divide by equals 10, which is the diameter, 10 divided by 2 plus 1 equals 6, use 6*6* this is the area of the added trail, and finally subtract the area of the circle.

Find the radius of the flower bed first, and then multiply the square of (flower bed radius + 1) minus the difference between the square of the flower bed radius.

First find the radius of the flower bed, then find the large radius and the small radius, and finally use the area of the ring to derive the formula!!

The correct result for this problem should be the tangent connection of the relatively small circle and the large circle.

The conclusion of the circle is not valid, and the Spanish defender's sketch of Hierro, although incorrect, is clear enough.

If you still feel that the evidence is insufficient, you can use the conclusions proven here to make an exact graph to verify it.

The circle is inscribed with hexagons and connects the three lines with the vertices when and only if ab·cd·ef = bc·de·fa

In addition, it requires some computation (in fact, the personal geometry is degraded, and a pure geometric method has not been developed), but fortunately, thanks to the symmetry, the computational effort is quite small.

Proof : Let o have a radius of r, and the radius of each minor circle is a, b, c, d, e, f (corresponding to the tangent letter).

Let's calculate the radius relationship between two adjacent circles, taking p and q as examples.

Let the circumferential angle of ab in o be (the one that is not obtuse), and there is the central angle aob = 2 , ab = 2rsin( ) sinusoidal theorem).

o, p radius r, a, and they are cut into a, op = r-a, and oq = r-b., respectively

and p, q radii are a, b, and they are inscribed, pq = a+b, (a+b) = pq = op +oq -2op·oqcos( poq) = (r-a) +r-b) -2(r-a)(r-b)cos(2 ) cosine theorem), 2ab = (r-a)(r-b)(1-cos(2)) = 2(r-a)(r-b)sin (

sin(θ)= √(ab/((r-a)(r-b)))ab = 2rsin(θ)= 2r√(ab/((r-a)(r-b)))

In the same way, we get Cd = 2R (Cd ((R-C)(R-D)))EF = 2R (EF ((R-E)(R-F)))).

Therefore ab·cd·ef = 8r (abcdef ((r-a)(r-b)(r-c)(r-d)(r-e)(r-f))).

It is easy to see that bc·de·fa is also the same result, that is, ab·cd·ef = bc·de·fa

Therefore, the three lines of AD, BE, CF are common points, and the proof is completed.

Refer to the sufficient and necessary condition for the intersection of three diagonal lines at the same point of the circle and the hexagonal abcdef is (ab bc) * (cd de) * (ef gh) = 1

The diagram you drew satisfies this condition.

The center of the large circle, the center of the small circle, and the tangent point are collinear.

Make tangents of six small circle tangents, according to the tangent length theorem, the length of each tangent in the ring is equal and intersect at the same point, and the angular divisions of the two tangents of the same circle must pass through the center of each small circle and the six angular bisectors must intersect at the same point. To put it bluntly, the tangent point between each small circle must be a common circle, and the center of the circle is the common point.

I've been thinking about it for a long time, and I'm waiting for other people's answers, the topic is very interesting, and I think it's also very meaningful to everyone, I hope the two people I replied to don't think I'm aggressive, I'm just ...... the question

I drew a sketch, a more extreme case, the small circle o1 o2 o3 o4 is about the same size (it seems tangent but a little bit close), and o5 o6 is a little smaller. As shown in the figure below. Therefore, I suspect that this conclusion is wrong......Of course, there must be errors in the drawing, and I'm not particularly sure.

I think your topic has few conditions, and you yourself said, "That's probably all there is to it, and I don't remember much."

I seem to know what the conditions for less are.

Spanish defender Hierro has drawn a counter-example to prove that you have fewer conditions.

Personally, I think your question might look like this.

The condition for you to be less is that all 6 small circles should be cut to the two circles that you have drawn, which is the outer big circle.

The other is the circle inside, which you haven't drawn yet.

And the conclusion is that the center of all 6 small circles is also cocircular.

And not some center of the line co-point.

The method of proof is inverse evolution.

plus: It doesn't have to be 6 circles, it's okay to have several. As long as the end is like a "ball bearing", of course the adjacent beads should be tangent.

The figure formed by the rotation of the minute hand of the clock in a circumference is (circle.

When the distance between the feet of the compass is 3 cm, draw the circumference of the circle is (6) cm, the diameter of the circle is 6 cm, its circumference is (6, the radius of the circle is one dm, and its circumference is (2

The diameter of the circle is 6cm, and its circumference is (6) The radius of the circle is one dm, and its circumference is (2

The circumference of the circle is 25.12 dm (decimeter).

Its diameter is ( and its radius is ( 2 ).

The radius of circle A is three times the radius of circle B, the circumference of circle A is (3) times the circumference of circle B, a semicircle, the radius is five decimeters, and its circumference is (10) decimeters

The radius of the circle expands to 2 times its circumference, and the circumference also expands to 2 times its (right) The circumference of the semicircle is the average (false) circumference of the circle

Practical problems (give me the equation).

The children's park has a circular goldfish pond with a diameter of ten meters, and a stainless steel guardrail should be installed at the meter outside the goldfish pond, and the minimum length of this guardrail is how many meters?

l=2π（10+

Draw the largest circle on a rectangular piece of paper eight centimeters long and six centimeters wide, how many centimeters is the circumference of this circle?

The maximum diameter of this circle is 6, then.

l=6π

The area of the blank part connecting A and C = the area of the triangle ABC = 2 2 2 = 2 square decimeters.

The area of the whole part = 2 4= square decimeters.

So shaded area = square decimeter (or written as -2 square decimeter) I wish you a good day.

If you do a guide to connect the AC, then the area of the whole shadow is a quarter of the circle minus the area of the triangle ABC.

Positive: square meters) Round: 10 10 square meters) square meters 314 square meters.

d =, solution d = 4 cm r =, solution r = 5 m3, s = r 0 5= (20) 0 5 = 1256 square centimeters 4, square area = square meter 2 r=, solution r = 10, circle area 10 0 5 = 314 square meters so the circle is large.

1, m = 4 cm2, m 3, 20 20 square cm4, the area of the square is (square = square meter, the area of the circle is (square square meter so the area of the circle is large.

1 cm ) 2 m) 3 20 20 sq cm ) 4 The area of the square is square meters) The area of the circle is (square meters) The area of the circle is large.

When the circumference is equal, the closer to the plane figure of the circle, the larger the area, and it must be!