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Mathematical problems in primary schools
1、

The length of AB: 4*3= 12(dm) requires the radius and diameter of isosceles right triangle and circle.

Shaded area:

12 * 4-1/2 * 4 * 3.14 * 4 *1/2 total area-triangular area-semicircular area

=48-8-25. 12

= 14.88 (square decimeter)

2.

Connect two semicircles into a circle.

Shaded area:

10 *10-3.14 * 5 * 5 square area-circular area

= 100-78.5

=2 1.5 (square decimeter)

Shadow part perimeter:

10 * 2+3.14 *10 square upper and lower length+circular circumference

=20+3 1.4

=5 1.4 (decimeter)

As can be seen from the figure, the upper bottom of the trapezoid is equal to the diameter of the circle.

Shadow area:1/2 * (10+20) * 5-1/2 * 3.14 * 5 * 5 trapezoidal area-semicircular area.

=75-39.25

=35.75 (square decimeter)

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