All the answers from page 50 to page 5 1 of the sixth grade mathematics method guidance series.

This is the first page of 5 1.

Scheme 1: Assume that the linear distance between the two sides of the inner ring is x meters.

So, according to what is known,

Inner straight line distance: curve distance =x: curve distance =3:2.

So the curve distance =(2/3)x meters.

Because the inner ring ran 300 meters.

So x+(2/3)x=300.

The solution is x= 180.

So both sides of the inner ring know that the distance * * * is 180m.

Solution 2

The total length of the inner runway is 300m.

It is known that the ratio of linear distance to curve distance in the inner ring of runway is 3: 2.

Know that the total inner ring runway is divided into five parts.

The linear distance of the inner ring accounts for 3.

So the linear distance of the inner ring = 300 ÷ 5× 3 = 180m.

We'll continue later. Look at the first two.

The first question: Because the clock turns once from 3 pm to 3 am the next day, we can find the circle-like area. Formula: 3.14× (6× 6) =113.04 (square centimeter);

Second question: Formula: 3. 14×(2×8)=50.24(cm) 3.

Calculate the circumference of the wheel first, then divide the length of the steel wire by the circumference of the wheel, and the result is the number of weeks the wheel rotates. The result is 2.5, which is two and a half weeks. (Pay attention to the conversion between centimeters and meters)

Solution: 50cm = 0.5m.

The circumference of the wheel = 0.5× 3.14 = m.

So to ride a 47. 1 meter-long steel wire, the wheels have to turn.

47. 1÷ =25 (calculate by yourself)

Fourth,

1306-50= 1254

1256/50=25. 12

25. 12/3. 14=8 (cm)

five

(30× 3.14× 50) ÷ (20× 3.14) = 4710 ÷ 62.8 = 75 turns.

six

Diameter of circle: 232.36/3. 14=74.