A 55 meter long and 45 meter wide rectangular green space is built with two mutually vertical paths of the same width in the middle, and the rest can be used for greening, with an area of 2000 square meters

A 55 meter long and 45 meter wide rectangular green space is built with two mutually vertical paths of the same width in the middle, and the rest can be used for greening, with an area of 2000 square meters

The width of the path is set as X, the shape of the path is regarded as a "ten", the area of the cross part of the "ten" is 55 * x, the area of the vertical part of the "ten" is 45 * x, the area of the path is rectangular minus the green area of 45 * 55-2000, because the area of the shadow part of the path (x * x) is calculated twice, so there is 55x + 45x-x * x = 45 * 55-2000, it is OK to solve this binary linear equation

There is a piece of green space in the park. Its shape is parallelogram. Several straight paths should be built on the green space, as shown in the figure, ab = 15cm, ad
There is a green space in the park. Its shape is a parallelogram. There is a green space in the park
It is necessary to build several straight paths, as shown in the figure, ab = 15cm, ad = 12cm,
AC⊥BC.
(1) Find the length of BC, CD, OC,
(2) Calculate the area of green space
(3) Calculate the distance between AB and CD
PS. the process of solving the third problem

BC=12,CD=15,AC=9,OC=OA=4.5
Area of ABCD = BC * AC = 12 * 9 = 108 cm ^ 2
Let the distance between AB and CD be H
Area of ABCD = BC * AC = AB * H (bottom * height = another bottom * another height)
15 * H = 108
h=7.2
The distance between AB and CD is 7.2 cm

As shown in the figure: to build a road with the same width on a rectangular green space with a length of 100 m and a width of 90 m, and the total area of six green spaces is 8448 m2, the width of the road is___ ．

Let the width of the road be XM, according to the equation of the meaning of the problem, we can get ｛ (100-2x) (90-x) = 8448, we can get x2-140x + 276 = 0, we can get X1 = 2, X2 = 138 (not practical, omit); so the answer is 2m