1. Classical examples of primary school students' Olympic mathematics geometry problems
There are two rectangles. The length of rectangle A is 98769 cm and the width is 98765 cm. The length of the rectangle is 98768 cm and the width is 98766 cm. Which of these two rectangles has the largest area?
Analyze and solve? It is feasible to calculate the area directly by using the rectangular area formula and then compare it, but the calculation is too complicated.
You can use the law of multiplication and distribution to deform the formula, and then compare the areas of two rectangles, which is much simpler.
The area of the rectangle is:
98769×98765
=98768×98765+98765
B the area of the rectangle is
98768×98766
=98768×98765+98768
Comparing the dimensions of 98768×98765+98765 and 98768×98765+98768, it can be seen at a glance that the area of rectangle A is small and that of rectangle B is large.
2. Classical examples of primary school students' Olympic math problems
1. Draw a circle with a perimeter of 12.56 cm, mark the center and radius with letters, and then calculate the area of the circle. There is a circular lawn in the school. Its diameter is 30 meters. What is the area of this lawn? If you put a pot of chrysanthemums every 1.57 meters around the lawn, how many pots of chrysanthemums should you prepare?
3. The radius of a circle and a sector is equal, the area of the circle is 30 square centimeters, and the central angle of the sector is 36 degrees. Find the area of the sector.
In the distance of 720 meters, the front wheel turns 40 times more than the rear wheel. The circumference of the rear wheel is 2m, so find the circumference of the front wheel.
5. The diameter of the circular flower bed is 10 cm, and there are 2-meter-wide paths around it. What is the area of this path?
6. The school has a circular open space with a diameter of 40 meters. It is planned to build an annular flower bed in the center, and the rest will be paved with 6m wide cement pavement. What is the area of cement pavement?
3. Classical examples of primary school students' Olympic math problems
1, there are two rectangles, the length of rectangle A is 98769 cm and the width is 98765cm；; The length of the rectangle is 98768 cm and the width is 98766 cm. Which of these two rectangles has the largest area? 2. There are 50 cubes coated with red paint, and their side lengths are 1 cm, 3 cm, 5 cm, 7 cm, 9 cm, ... and 99 cm respectively. Saw these cubes into cubes with a side length of 1 cm. At least one cube is red. How many cubes are there?
3. There are 102 sides 1, 2, 3, ..., 99, 100,10, 102 cm. Paint their surfaces with red paint and dry them.
4. There is a cuboid block with a length of 125cm, a width of 40cm and a height of 25cm. Saw it into several small cubes of equal volume, and then put these small cubes together to make a big cube. What is the surface area of this large positive body in square centimeters?
4. Classical examples of primary school students' mathematical Olympic geometry problems
There is a cuboid block with a length of 125cm, a width of 40cm and a height of 25cm. Saw it into several small cubes of equal volume, and then put these small cubes together to make a big cube. What is the surface area of this large positive body in square centimeters? Analysis and solution Generally speaking, the surface area of a cube is required, and the side length of the cube must be known. It is known that the length, width and height of a cuboid are not directly related to the side length of a cube, which brings difficulties to the solution. We should consider the problem as a whole.
According to the meaning of the question, the cuboid block was sawed into several small cubes with equal volume, and then put together to form a big cube. The volume of this big cube is equal to the volume of the original cuboid. Knowing the length, width and height of a cuboid, we can find out the volume of a cuboid, that is, the volume of a big cube. Then the side length of the cube can be calculated, and the surface area of the cube can be calculated.
The volume of a cuboid is
125×40×25= 125000 (cubic centimeter)
Decompose 125000 into prime factors:
125000=2×2×2×5×5×5×5×5×5
=(2×5×5)×(2×5×5)×(2×5×5)
As you can see, the side length of the big cube is
2×5×5=50 cm
The surface area of a large cube is
50×50×6= 15000 (square centimeter)
A: The surface area of this big cube is 15000 square centimeters.
5. Classical examples of primary school students' mathematical Olympic geometry problems
The sum of the front and upper areas of a cuboid is 209 square centimeters. The length, width and height of this cuboid are prime numbers in centimeters. What is the volume and surface area of this cuboid? Mental navigation
The sum of the areas of the front and top of a cuboid is length × width+width × height = length × (height+width). Since the numbers in centimeters of a cuboid are prime numbers, there are 209 =11×19 =11. Knowing the length, width and height, it is easy to calculate the volume and surface area.
209= 1 1× 19= 1 1×( 17+2)
1/kloc-0 /×17× 2 = 374 (cubic centimeter)
(11×17+1× 2+17× 2 )× 2 = 486 (square centimeter)
Practice (1) a cuboid. The sum of its front and upper areas is 1 10 square centimeter, and its length, width and height are prime numbers. What is the volume of this cuboid?
Exercise (2) The length, width and height of a cuboid are three consecutive even numbers, and its volume is 960 cubic centimeters. Find its surface area.
Exercise (3) The edges of a cuboid and a cube are equal. Given that the length, width and height of a cuboid are 6 decimeters, 4 decimeters and 2 decimeters respectively, find the volume of the cube.