A cylindrical wooden stick with a base diameter of 4cm, a height of10cm and a surface area of () cm2. If it is sawed into two equal pieces along the bottom diameter, the surface area of one piece is () square centimeters.
① A cylindrical wooden stick with a bottom radius of 2 cm and a height of 3 cm. If cut longitudinally along the bottom diameter, the sum of surface areas will increase by () square centimeters.
a、6 b、 12 c、24 d、48
(2) Cut a cylindrical wooden stick with a diameter of 2 cm and a height of 4 cm into two small cylinders, and the surface area is increased by () square cm.
a、 16 b、3. 14 c、8 d、6.28
(3) Cut a cylindrical steel into three sections along the direction parallel to the bottom surface, and the sum of the surface areas will increase by 12 cm2, and the bottom area of the steel should be () cm2.
a、6 b、4 c、3 d、2
1, the height of cylinder and cone is equal, the sum of their volumes is 36 cubic decimeter, and the volume of cone is () cubic decimeter.
① 12 ②9 ③27 ④24
2. The volume of a cone is n cubic centimeters, and the volume of a cylinder with the same height as its bottom surface is () cubic centimeters.
① n ②2n ③3n ④ 13 n
3. Cut a section of round steel into the largest cone. The cutting part weighs 8kg, and this section of round steel weighs () kg.
①24 ② 16 ③ 12 ④8
4. The volume of a cylinder is larger than that of a cone with the same height as its bottom ().
①23 ② 1 time ③2 times ④3 times
5. The volume difference between a cylinder with equal bottom and equal height and a cone is 16 cubic meter. The volume of this cylinder is () cubic meters, and the volume of this cone is () cubic meters.
1. If the top and bottom of a trapezoid remain unchanged, the height will increase by 2 cm and the area will increase by 32 cm. If the bottom and height remain the same, the upper bottom will increase by 4 cm and the area will increase by 20 cm. The area of the original reminder is () cm?
① From "If the top and bottom of a trapezoid remain unchanged, the height will increase by 2 cm, and the area will increase by 32 cm." The sum of the upper and lower bottoms of this trapezoid is 32 cm.
Find the center line with the formula: (a+b)÷2=32÷2= 16.
I gave three examples through this sentence and found that the increase in area is the sum of the upper and lower bottoms of this trapezoid.
② The bottom and height are unchanged, and the upper bottom is increased by 4 cm, and the area is increased by 20 cm.
(a+b)h \u 2 →( a+4+b)h \u 2
=(ah+bh)÷2 =(ah+bh+4h)÷2
Therefore, the area increase number is 4(h÷2) and the height is 20÷4×2= 10cm.
③ Find the trapezoidal area (taking the center line as m) with the formula: MH =16×10 =160cm2.
◆ Math graphics problems in primary schools
Reward score: 0- analysis time: February 9, 2008 15: 18.
△-□ = 2.8, □+○ = 9.2, △+○ = △△△ So, they are △, △, □?
Questioner: the wings of a kind angel-the best answer during the probation period
Delta+○ = Delta Delta Delta, subtract Delta from both sides at the same time, and you can get ○ = 2 Delta.
Convert □+○=9.2 to □+2△=9.2.
Using formula 1+ formula 2,
δ-□+□+2δ= 2.8+9.2
3△= 12
△=4
According to △=4 and ○=2×4=8.
4-□=2.8,
□= 1.2
A rectangle, if its length increases by 2 cm and its width increases by 5 cm, then its area increases by 60 cm 2, which is exactly a square. How many square centimeters was the original rectangle?
Let the length be a and the width be b.
a+2=b+5
(a+2)(b+5)=ab+60
The solution is a=8 b=5.
ab=40
The original rectangular area is 40 square centimeters.
1 problem: plant corn in a rectangular field with a width of 34 meters and a length of less than 8 meters. The plant spacing of corn is 0.2 meters and the row spacing is 0.3 meters. It is estimated that each corn will get 0.225 kg. How many kilograms of corn can this field harvest?
1, I think everyone's consideration is too biased. It is not necessary to add "1" to both rows and columns. Is it possible to grow corn as a sideline? Therefore, the mode of planting corn should be "Tian"-shaped planting in the grid, not on the line, just like chess in the grid, not China chess walking on the line.
Solution: 34/0.2× 60/0.3× 0.225 = 34000× 0.225 = 7650kg.
Question 2: The side lengths of the two squares on the right are 10 cm and 14 cm respectively. One vertex of a big square is in the center of another square. What are the coverage areas of these two squares?
Question 2: It is a kind of graphic thinking. Think of it this way: 1. After connecting the diagonals of the small squares, put a diagonal of the small square and an edge of the big square, and the covered part is an isosceles right triangle, which is a quarter of the small square with a side length of 10cm, that is, 25 square centimeters.
So what if it doesn't look like the one above?
Think of it this way. After connecting the diagonals of the small squares, we will twist the situation a little. You will find a small triangle moved out of the big square, but you will find an identical small triangle moved into the big square, so the coverage area is still 25 square centimeters.
You can cut two squares by yourself and look around. You will find that after the auxiliary lines are drawn, the ones that have moved out of the big box will always be the same as those that have moved in, and the coverage area will remain unchanged.
The side length of square ABCD is 20CM, e and f are the midpoint of AB and BC respectively, and CE and d f intersect at G. Find the area of quadrilateral BEGF (? )CM2 .
Answer: 5 * (4 2) = 80 (cm2)
If the length and width of a rectangle increase by 8 meters, the area will increase by 208 meters. What is the length and width of the original rectangle?
Add two rectangles with a width of 8m and a square with a side length of 8m.
Area of rectangle: 208-8×8= 144 (square meter)
The sum of the length and width of the original rectangle: 144÷8= 18 (m).
The unit is square centimeters.
49,13,35 is the area of each graph, and the largest graph is a rectangle. Find the area of the shaded part?
S 1+S2+S3=S7+S5
49+ 13+S2=35+S5
S5=S2+27
2*(S 1+S2+S3)=S2+X+S5
2*(49+S2+ 13)=S2+X+S2+27
X=2*49+2* 13-27=97
The area of the triangle ABC is 12cm2, and there is a point e and BE=2EC on the side of BC, which connects AE, and a point d on the side of AB connects CD, which intersects with the point f on the side of AF, and f is the midpoint of CD. What's the area of BEFD?
This problem is solved by the idea of differential application problem, which is generally not easy to think of. The first time I did it, I didn't make it.
The answer is 5 square centimeters.
According to the meaning of the question, it can be concluded that:
The area of 1 and triangle ABE is twice that of triangle AEC.
2. The area of triangular FBE is twice that of triangular FEC.
3. The area of triangular ADF is equal to that of triangular AFC.
4. The area of triangle DBF is equal to the area of triangle BFC.
Let the area of triangle FEC be a, then the area of triangle FBE should be 2a, and the area of triangle BDF should be 3a;
Triangle area ABE = 2a+3a+ triangle area ADF (2 copies)
Area of triangle AEC = A+ area of triangle AFC (1 copy) (note: the area of triangle ABE is twice that of triangle AEC).
Triangle ABE- triangle AEC = 4a (note: the area of triangle ADF is equal to the area of triangle AFC).
4a ÷1= area of 4a triangle AEC.
4a-a=3a Area of triangular AFC and triangular ADF
The area of the whole graph = a+2a+3a+3a =12a =12.
a= 1
The area of quadrilateral BEFD = 2a+3a = 5a = 5 square.
This is all I found. Sorry, not too much. I hope it can be adopted.