High reward points for help!! Who has the Olympiad math problem for the 5th grade of primary school

2.In the known quadrilateral ABCD, B= D=90°, C=135°, AD=12 cm, BC=4 cm, what is the area of the quadrilateral ABCD in square centimeters?

4.Draw an inscribed equilateral triangle within the circle, another inscribed circle in the equilateral triangle, a second inscribed equilateral triangle within the second circle, and so on (as shown in the figure below). If the area of the first triangle is 512 square centimeters, then what is the area of the fifth triangle in square centimeters?

5.As shown in the figure, the right-angled trapezoidal ABCD, AD is 15 cm long, the height DC is 30 cm long, and the area of the triangle BOC is 150 square centimeters larger than the area of the triangle AOD.

6.If you spray it with red paint and disassemble it after drying, how many pieces are there on 1, 2, 3, 4, and 5 sides dyed with red paint? How many pieces are there that are not dyed with red paint?

7.As shown in the figure, ABCD is a right-angled trapezoid, AEFC is rectangular, BC-AD=6 cm, CD=8 cm, and the trapezoidal area is 80 square centimeters. Find the area of the shaded part.

8.Find the area of the shaded part in the image below.

Sweat! Don't you have to be half dead?

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Well, it's convenient in the bookstore and there are answers.

Carried away = 5 7 Total carried away 5 12 The first time it was carried 3 8 Then the second time it was 5 12-3 8 = 1 24 The second time it was 50 50 (1 24) = 1200.

I don't know if you have learned the equation x, ySo I don't use it, the first time I take away 3 8, then there is 1 3 8 5 8;The second time 50 is transported, and the ratio of what is taken away and what is left is 5:7, then the proportion of what is taken away and the rest are 5 12 and 7 12 of the total tonnage, respectively, because the remaining 7 12 is equal to the remaining 3 8 after the first shipment and subtract 50

Then the total proportion of 50 transported is 3 8 7 12 1 24, so the total tonnage is 50 (1 24) 1200

To sum up: the total tonnage is 1200 tons.

1200 tons.

Set up the original goods in the warehouse x tons, 3x 8 + 50) :(5x 8-50) = 5:7 to solve x = 1200

The rest accounted for seven-twelve, and the first time it was transported away, it was three-eighths, so the second time it was transported away, it was one-twenty-four, and it turned out to be 50*24=1200... Give it the best...

Suppose there is x tons of the original equation from the proposition:

3/8×x+50）/(x-3/8×x-50)=5/7

Solve the equation: x=1200.

Starting with 9, the remaining 3 digits are any three of the remaining 4 numbers, and there are 4*3*2*1=24 kinds of arrangement.

Similarly, starting with 7, there are 24 types of arrangement.

The two largest numbers starting with 5 are 5973, 5970, and the 50th number is 5970