Many college students can't calculate the shadow area with the Olympic series diagram of primary school, so they don't even have a diagram. Not to mention college students, nor can the immortal master.
Primary schools urgently need to practice the shadow area and perimeter of the Olympic circle and other graphics (2)
The triangle, rectangle, square, parallelogram, trapezoid, diamond, circle and sector we have learned are generally called basic graphics or regular graphics. Their area and perimeter are directly calculated by the corresponding formula. For the calculation of the area and perimeter of irregular graphics, most of us are transformed from regular graphics!
This lecture will mainly introduce the main methods and common skills of linear graphics problems:
The example shows a rectangular grassland, with a length of 12 and a width of 8, and a road with a width of 2 in the middle. Find out the area of the grass (shaded part).
Analysis: divide the road to a certain extent, as shown in the left figure below, and then push 1, 3,5 to the left end of the rectangle, and push 2,4,6 to the upper end of the rectangle, and you can get the right figure below. The area of the shaded part is: (12-2) × (8-2) =10.
The second lecture on circles and sectors
In order to calculate the area of irregular figure composed of circle, sector, arch, triangle, rectangle, parallelogram, trapezoid, etc. Sometimes, it is often necessary to change the position of the graph or divide, rotate and repair the graph to make it a regular graph whose area can be calculated.
There are seven plastic pipes with a diameter of 5 cm. Tie them into a bundle with a rubber band (as shown on the right). What is the length of the rubber band at this time? (π takes 3)
Analysis: According to the picture on the right, the length of the rope is equal to the sum of the arc lengths of six line segments AB and six BC.
Translating and splicing the central angles of six arcs similar to BC arc in the figure. If the sum of the six angles is 360, the central angle of BC arc is 60, and the six BC arcs are equal to the circumference of a circle with a diameter of 5 cm.
The line segment AB is equal to the diameter of the plastic pipe, so we know that the length of the rope is 5x6+5x3 = 45 (cm).
You don't have a picture of the shadow area of the circle in elementary school mathematics. It's hard to tell.
But I can tell you that finding the shadow area of a circle in primary school mathematics is a method.
It is to subtract the area of a figure that can be found inside from the area of a circle, and then get the area of the shadow, all of which are obtained by this subtraction (⊙o⊙).
How to find the vertical line of mathematical shadow area!
In primary school mathematics, which figures have the same shadow area? If you transfer the bow shadow above to the blank below, the shadow part will become half of the big right triangle. Shadow area S=4x4÷2÷2=4cm?
A math problem, find the shadow area difference in the picture:
4×4×3. 14 ÷4 - 4×2
= 12.56 -8
= 4.56 square centimeters
Just do it.
How to learn math shadow area well. Remember the area formula and skillfully use it to cut and paste and fill it into graphics. Just add one more when it's all done.
What is the thinking method of finding shadow area or perimeter in primary school mathematics?
It is mainly divided into the sum of the areas of several regular figures or the difference of the areas supplemented by several regular figures.
Circumference:
Split a line into a straight line, or cut a three-dimensional figure flat.
How to calculate the shadow area of 1 in the third grade? Total area-outer area = shadow area.
2. Total area * number of copies = shadow area
3. Length * width (large)+length * width (small) = shadow area
I only think so much, I'm really sorry!