Trapezoidal volume calculation formula

V = [S1 + S2 + open radical (S1 * S2)] / 3 * h
Note: V: volume; S1: upper surface area; S2: lower surface area; H: high

Help me explain the trapezoidal volume calculation formula? How can there be volume?

How can there be volume?
There's a length to unfold!
How can we know how to use man hour without digging canal?
Not for 30 years, just think about it
Volume = area * length area = (upper bottom + lower bottom) * height / 2

Volume formula of cube
The edge length of a small cube is 1 cm, so
(1) How many cubic centimeters is its volume? How many cubic meters can it be converted into?
(2) How many such small cubes can fill a cubic meter of space?

(1) Its volume is 1 cubic centimeter, which can be converted into 0.000001 cubic meter
(2) 100000 of these cubes can fill a cubic meter of space

There are three square pools of large, medium and small, with the inner side length of 6m, 3M and 2m respectively. Two piles of gravel are immersed in the small and medium pools, and the water surface of the two pools is smooth
They are raised by 6cm and 4cm respectively. If these two piles of gravel are immersed in the water of the large pool, how many centimeters will the water surface of the large pool rise?

The volume of gravel in the middle pool is 3 * 3 * (6 / 100) = 0.54 m3
Gravel volume of small pond 2 * 2 * (4 / 100) = 0.16 M3
(0.54 + 0.16) / (6 * 6) * 100 = 1.94 cm
So the water level of the pool rises by 1.94 cm

There are three square pools, large, medium and small, whose inner sides are 4m, 3M and 2m respectively. The water surface of the two pools rises by 4cm and 11cm respectively when the two piles of gravel are sunk in the large pool. If the two piles of gravel are sunk in the large pool, how many CM will the water surface of the large pool rise?

4cm = 0.04m, 11cm = 0.11M (3 × 3 × 0.04 + 2 × 2 × 0.11) / (4 × 4), = 0.8 × 16, = 0.05 (m), = 5 (CM); a: the water surface of the pool will rise by 5cm

There are three square pools, large, medium and small. Their inner sides are 6 meters, 3 meters and 2 meters respectively. When two piles of gravel are sunk in the water of the medium and small pools, the water surface of the two pools rises by 6 cm and 4 cm respectively. If these two piles of gravel are sunk in the water of the large pool, how many cm does the water surface of the large pool rise?

6cm = 0.06m, 4cm = 0.04m. 3 × 3 × 0.06 = 0.54 (M3), 2 × 2 × 0.04 = 0.16 (M3), 0.54 + 0.16 = 0.7 (M3). The bottom area of the large pool is 6 × 6 = 36 (M2). The water surface of the large pool has increased by 0.7 △ 36 = 0.736 (m) = 3518 (CM). A: the water surface of the large pool has increased by 3518 cm

To pave a room with floor tiles, 480 pieces are needed if a square with side length of 0.3m is used; how many pieces are needed if a square with side length of 0.4m is used?

If your 480 is whole and there is no loss, it will be 270 yuan for 0.4

There is a 3 m * 4 m rectangular room. If the square tiles with side length of 50 cm are used to fill the floor, at least such tiles are needed____ block

3*4/(0.5*0.5)=48
So you need at least 48 of these tiles

A cuboid warehouse is 8 meters long, 6 meters wide and 3.5 meters high. The warehouse is equipped with a door, which is 1 meter wide and 2 meters high. Now we need to paste tiles on the four walls of the warehouse below 1 meter high from the ground. What is the area of the tile part?

(8 × 1 + 6 × 1) × 2-2 × 1 = 14 × 2-2 = 28-2 = 26 (square meters); answer: the area of tiling is 26 square meters

A cuboid warehouse is 8 meters long, 6 meters wide and 3 meters high. The four walls of the warehouse below 1.5 meters above the ground should be pasted with ceramic tiles. What is the area of the tiles? What is the volume of the warehouse?

Surface area = 2 × 8 × 1.5 + 2 × 6 × 1.5 = 24 + 18 = 42 M2
Volume = 8 × 6 × 3 = 144 M3

Trapezoidal volume calculation formula