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Math problems in the sixth grade last semester
The following calculations all take π as 3.14;

1, as shown in the figure, the binding part is equivalent to four 1/4 circumferences and four line segments, the length of which is the diameter; The total length is π * 5 * 2+4 * 5 * 2+20 = 60+25 π ≈ 91.42 cm;

2. The minute hand walks 24 times a day and night, and the hour hand walks 2 times a day and night (according to the ordinary 12 hour clock);

Walking distance of splitting needle tip = π * 60&; sup2* 24 = 86400π≈27 1296cm;

Distance moved by needle tip = π * 50&; sup2*2=5000π≈ 15700 cm

3. As shown in the figure: the sum of the diameters of all small circles is the diameter of the big circle;

Let the diameter of the big circle be r and the diameter of the small circle be R 1, R2...rn;; There are:

The circumference of the great circle =πR, and the length of the small circle = π r1+π R2+... π rn; Because r = r1+R2+... rn; So the circumference of the big circle is the sum of the circumference of the small circle;

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