A math problem in the second year of junior high school, and a math problem in the second year of ju

No, different triangles have different distances. A is the vertex, do B, C to AD perpendicular, if both lines fall within the triangle, then both are 30 meters, if one falls outside the triangle (assuming the one is C), then the distance from C to the canal AD can not be the perpendicular line, the shortest distance is half of BC, and AD, at this time must be greater than 30 meters.

No, different triangles have different distances.

Or 30mThe snake be perpendicular to Ad and cm perpendicular to Ad, known by the triangle similarity theorem: Bed is similar to Cmd. So be=cm=30m

Since your diagram is not clear, you can't write specific linear equations, so you can only write the principle.

l1：y=x；l2: over (0,1),(1,so y=so the expression of the function between profit and sales is: the difference between the two functions y=x-( x 0

(2) Let the side length of the square be x, then the area of the square is x, which is also the area of the original rectangle.

Using the reverse method, the length of the original rectangle = x+a, width = x-b, then (x+a)(x-b) = x

x²+(a-b)x-ab=x²

a-b)x=ab

x=ab/(a-b)

At this point, we can find that the side length of the resulting square = the product of the reduced length and the increased width divided by the difference between the two, so in the first question, we can directly get the side length of the square = (4 2) (4-2) = 4 cm.

(1) Solution: Let the side length of the square be x.

x^2=(x+4)x(x-2 )

x^2=x^2+4x-2x-8

2x=8x=4(2) x-a=y+b xy=(x-a)(y+b)

(1) Extend DP to G

From the triangle ABC is an equilateral triangle, PE BC, PD AB, PF AC, the quadrilateral pdfc is an isosceles triangle, the quadrilateral PEBG is a parallelogram, and the triangle GPF is an equilateral triangle.

So PE=BG, PG=GF, PD=CF

So PE+PD+PG=BG

There's a data wrong, it's to be handed over to bc at point F. 3 squared plus 4 squared, and then squared: ef = 5

The area is 1

Extend DE to F so that EF=BC, connect FA, and obtain the triangle ABC congruent triangle AEF

So triangle adg area = 1 2 *1*1 = 1 2 and then prove that triangle ACD congruent triangle adg so pentagonal area = 2 * 1 2 = 1

Solution: There are two possibilities for this topic:

1. When the triangle DNM is an isosceles triangle with D as the vertex, there is a triangle with a symmetrical figure, that is, the triangle DCN and the triangle DCE form an axisial symmetry, and its axis of symmetry is DC (proof: since the triangle DNM is an isosceles triangle with D as the vertex, CN=BM can be obtained, and thus CE=CN).

2. When the triangle DNM is not an isosceles triangle, there are no two triangles in a symmetrical shape.