A square iron sheet, four corners, cut off four small squares of the same size, and fold them into a cuboid box with a bottom side length of 50 cm and a volume of 45000 cubic cm. Find out the side length of the original square iron sheet. (use equation solution)

Suppose: the side length of the cut small square is X
X=180
The side length of the original square is 180 + 180 + 50 = 410cm

In a 50 cm, 40 cm wide rectangular cardboard four corners of each cut out a side length of 5 cm cube, it will be

V＝(50－5×2)×(40－5×2)×5＝6000㎝³
S＝[(50－5×2)×(40－5×2)＋(50－5×2)×5＋(40－5×2)×5]×2＝3100㎝²

The side length of the big and small squares is 8 cm and 6 cm respectively. Calculate the area of the shadow part

8 × 8 - 6 × 6 = 64 - 36 = 28 (square centimeter). A: the area of shadow is 28 square centimeter

As shown in the figure, the side of the square is 8 cm long. What is the area of the shadow?
There are two semicircles in the square. One is in the south of the square, and the other is in the east of the square. There is a shadow where the two semicircles overlap. Except that the area of the two semicircles that do not overlap is white, the others are shadows

2 [a ^ 2 / 2 - (3.14 * a ^ 2) / 8] is actually a triangle minus the area of a semicircle and multiplying by 2

The sides of the two squares in the figure below are 8 cm and 5 cm respectively

As shown in the figure, the area of trapezoid abdc is: (5 + 8) × (5 + 8) △ 2 = 13 × 13 △ 2, = 169 △ 2, = 84.5 (square centimeter), the area of triangle Abe is: 8 × 8 △ 2 = 64 △ 2, = 32 (square centimeter), the area of shadow part is: 84.5-32 = 52.5 (square centimeter), a: the area product of shadow part is 52.5 square centimeter

As shown in the figure, what is the relationship between the areas of the three semicircles whose diameters are the three sides of RT △ ABC? Please give reasons

The area of semicircle D is equal to the sum of the area of semicircle E and the area of semicircle F. it is proved that in the right angle △ ABC, ac2 = BC2 + AB2, ∵ the area of semicircle D is 12 π· (ac2) 2, the area of semicircle e is 12 π· (AB2) 2, the area of semicircle f is 12 π· (BC2) 2, and the sum of the areas of semicircle E and semicircle f is 12 π· (AB2) 2 + 12 π· (BC2) 2 = 12 π· (ac2) 2 = the area of semicircle D, so the area of semicircle D is equal to the area of semicircle E The sum of the area and the area of the semicircle F

As shown in the figure, two identical right triangles are stacked together to find the area of the shadow part. (unit: cm)

Because the area of Figure 1 + the area of Figure 2 - the area of Figure 2 + the area of Figure 3, so: the area of Figure 3 = the area of Figure 1, figure 1 is a trapezoid, the upper bottom is 12 cm, the lower bottom is 12-3 = 9 (CM), the height of the trapezoid is 6 cm, so the shadow area of Figure 1 is: (12 + 9) × 6 △ 2 = 21 × 6 △ 2 = 126 △ 2 = 63 (square cm). Answer: the area of the shadow part is 63 square cm ．

Two same right triangles are overlapped together, as shown in the figure below. Calculate the area of the shadow part

OC = 12-4 = 8 (CM) area of trapezoidal oefc: (8 + 12) × 2 / 2 = 20 × 2 / 2 = 20 (square cm) because the area of shadow part + area of triangle ODC = area of triangle def (triangle ODC + trapezoidal oefc), so the area of shadow part = area of trapezoidal oefc, so the area of shadow part = 20 square cm. A: the area of shadow part is 20 square cm

As shown in the figure, if two identical right triangles are overlapped, the area of the shadow part in the figure is () square centimeter
A. .24B. .30C. 60

(12-4 + 12) × 3 / 2, = 20 × 3 / 2, = 30 (square centimeter); answer: the area of shadow in the picture is 30 square centimeter

As shown in the figure, if two identical right triangles are overlapped, the area of the shadow part in the figure is () square centimeter
A. .24B. .30C. 60

(12-4 + 12) × 3 / 2, = 20 × 3 / 2, = 30 (square centimeter); answer: the area of shadow in the picture is 30 square centimeter

A square iron sheet, four corners, cut off four small squares of the same size, and fold them into a cuboid box with a bottom side length of 50 cm and a volume of 45000 cubic cm. Find out the side length of the original square iron sheet. (use equation solution)