The circle O of the equilateral triangle ABC with circumference of 6 π and radius of 1 starts from the position where AB is tangent to point D Then: on the outside of triangle ABC, roll clockwise along the triangle and return to the position of point d tangent to ab. how many cycles does circle O rotate? What do you mean by turning the outer angles of three triangles? Why can you turn one angle by turning an outer angle? Why not turn 240? You'd better take the picture to solve it! How to turn it?

Turning along three sides, it's three turns
6π/(2π)=3
At the vertex of the triangle, the circle rotates around the vertex, and the rotation angle of the circle center is 360 ° - 60 ° - 90 ° = 120 °
Three angles are 360 degrees. This is another circle
So it's a total of four laps

In △ ABC, B = 10, area of △ ABC is 10, radius of circumscribed circle r = 13

2R = B / SINB --- > SINB = B / 2R = 5 / 13; B < 60 --- > CoSb = 12 / 13 s = (1 / 2) acsinb -------- > AC = 2S / SINB = 52 B ^ = a ^ + C ^ - 2accosb = (a + C) ^ - 2Ac (1 + CoSb) - > A + C = 10 √ 3 --- > triangle perimeter = a + B + C = 10 + 10 √ 3

It is known that △ ABC ∽ △ A'B'C', and the ratio of the three sides of △ ABC is 3:5:7, and the maximum side length of △ A'B'C' is 15cm. Find the circumference of △ A'B'C'

∵ △ ABC ∵ a'b'c ', and the ratio of the three sides of ∵ ABC is 3:5:7, and the maximum side length of  a'b'c' is 15cm
The ratio of three sides of ABC is 3:5:7
Let the three sides of △ ABC be 3xcm, 5xcm and 7xcm respectively
7x=15
x=15/7
3x=45/7 cm 5x=75/7cm

We know that the side length of △ ABC is a, B, C, the perimeter is 60 cm, and a: B: C = 7:5:3, find the length of a, B, C There should be a detailed reason, but also a reason, so

Let a, B, C be 7x, 5x, 3x respectively
So a + B + C = 15x = 60, x = 4
So a = 4 * 7 = 28, B = 4 * 5 = 20, C = 4 * 3 = 12

The ratio of the lengths of the three sides of the triangle ABC is 3:5:7, and the maximum side length of the similar triangle a'b'c 'is 12 RT

The trilateral ratio of a'b'c 'is also 3:5:7
The largest proportion of the circumference 7 × (3 + 5 + 7) = 7 / 15
Max = 12
So perimeter = 12 ÷ 7 / 15 = 180 / 7

Given that △ ABC ～ △ a'b'c ', the three side lengths of △ ABC are 3, 4 and 5 respectively, and the minimum side length of △ a'b'c' is 7, calculate the circumference of △ a'b'c '

Let the other two sides of the triangle △ a'b'c 'become a, B. then 3 ∶ 7 = 4 ∶ a = 5 ∶ B. A = 28 / 3, B = 35 / 3. Therefore, the circumference of △ a'b'c' is 7 + A + B = 7 + 28 / 3 + 35 / 3 = 28

It is known that the circumference of △ ABC is even, where the lengths of two sides are 3 and 7, and the third side of △ ABC is even, then the circumference of this triangle is______ .

Let the third side of the triangle be X,
According to the meaning of the title, 7-3 < x < 7 + 3 is obtained,
That is 4 < x < 10,
∵ the third side x is even,
ν x = 6 or 8
The circumference is 7 + 3 + 6 = 16 or 7 + 3 + 8 = 18,
So the answer is: 16 or 18

If we know that the equal ratio sequence an satisfies a1 + A2 + a3 + A4 + A5 = 3, then the value of a1-a2 + a3-a4 + A5 is Known also includes: A1 square plus A2 square plus A3 square plus A4 square plus A5 square = 12

Let the common ratio be q, Q ≠ 1 ∵ a1 + A2 + a3 + A4 + A5 = 3A1  A2 + A2

If a1 + A2 = 1, A3 + A4 = 9, A4 + A5 =? Is there a simple way?

Let the common ratio be Q. a1 + A2 = 1A1 (1 + Q) = 1 (1) A3 + A4 = 9a1 (q? 2 + Q) = 9a1q? 2 (1 + Q) = 9 (2) (2) / (1) q? = 9q = 3 or q = - 3A4 + A5 = A1 (q? + Q ⁴) = a1q? (1 + Q) =

If a1 + A2 + a3 = 2, A3 + A4 + A5 = 8, then A4 + A5 + A6 is known=______ .

Let the common ratio of the equal ratio sequence is Q, from a1 + A2 + a3 = 2, then A3 + A4 + A5 = Q2 (a1 + A2 + a3) = 2q2 = 8, that is, Q2 = 4, q = ± 2;
So A4 + A5 + A6 = Q3 (a1 + A2 + a3) = ± 8 × 2 = ± 16
So the answer is: ± 16