The math problems in the second year of junior high school are urgent, and the reward should be high

The cube volume is 10 3

Water volume 40 * 25a = 10 3a

Container volume 40 * 25 * 50 = 5 * 10 4

1. 10 3 + 10 3a> = 5 * 10 4, then the water depth is a 2, 10 3 + 10 3a< 5 * 10 4, then the water depth is.

10^3+10^3a)/(40*25)

Figure 2 should be the outer corner, right?

The answer is. In Figure 1, the intersection is the heart of the triangle. In Figure 2, the intersection point is the side.

The nature of the heart, the side mind, is that the distance to the three sides is equal.

So there are 4 points in question three. One heart, three side hearts.

It should not be required to master the specific proof of intersecting junior high school, but it can be proved in a variety of ways.

The proof upstairs is too sketchy.

Give a more formal and common proof method:

It is known that in the triangle ABC, the two angular bisectors BM and CN intersect at P

Verification: The point p is on the bisector of the angle bac, and the distance from the point p to the three sides is equal.

Proof: PE is perpendicular to E, PD is perpendicular to Ab to D, PF is perpendicular to Ac to F

BM bisects the angle ABC, then PE=PD; CN bisects the angle ACB, then PE=PF(properties of angular bisector).

Therefore PE=PD=PF.The point p is on the bisector of the corner bac.

1. Yes, because the three angular bisectors intersect at one point, the intersection of the two angular bisectors must be on the angular bisector of c, and the proof method is: pass the point o as a perpendicular line to cross bc ac at m, n. Because om=on, the point o is on the angular bisector of c.

2. Yes, ditto.

3. There is only one, and the distance from the point on the bisector of the angle to both sides of the angle is equal.

1) Solution: o is on the perpendicular bisector of c.

Because: o is on the angular bisector of a and b.

So: o on the angular bisector of c (the angular bisector of the triangle intersects at one point).

1.Equal are twice the area of the triangle.

2.Only one rectangle can be drawn, which is a rectangle drawn on the side of the side opposite by the obtuse angle3You can draw 3 of them, one on each side.

The circumference of the side is the smallest, because the area of each rectangle is twice the area of the triangle, so the area is the same, and the perimeter of the rectangle is twice the sum of the height of the triangle side and this side, and the closer the height and the side are to the circumference, about smaller, so the circumference of the rectangle of the AC side is the smallest.

The combination of numbers and shapes is an important method of thought in mathematics, and if you carefully mark the conditions on the diagram according to the problem, you may be able to see the solution.

Solution: (1) =

4) The circumference of the rectangle with ab as the side length is the smallest, and the perimeters of the rectangle bced, achq and abgf are l1, l2, l3, bc=a, ac=b, ab=c and the areas of the three rectangles are equal, and the areas of the three rectangles are equal, and the areas of the three rectangles are equal, and the rectangles are set as s, l1 = +2a; l2= +2b；l3 = +2c l1-l2=2(a-b) and a-b 0, ab-s 0, ab 0 l1-l2 0, l1 l2, the same gives l2 l3

The circumference of the rectangle with ab as the side length is the smallest

x+4x²-800=0

x²+35x-200=0

x+40）（x-5）=0

x1 = -40 is not in line with the topic and is discarded.

x2=52, we know that there is no real root for the equation x, mx -2(m+2)x+m+5=0, then 0

4（m+2）²-4m（m+5）＜0

4m²+16m+16-4m²-20m＜0

16-4m＜0

m 4 equation (m-5) x -2 (m+2) x + m = 0 = 4 (m + 2) -4 m (m-5).

4m²+16m+16-4m²+20m

36m+16

Because m 4 so 36m + 16 0

When m=5, this is a one-dimensional equation with a root.

When m≠5, the equation has two unequal real roots.

Question 1: 2 [40x+(30+2x)x].

That is: 3x(40x+30x+2x 2)=12002x 2+70x=400

x^2+35x=200

Question 2: Because there is no real root for the x equation mx -2(m+2)x+m+5=0, [2(m+2)] 2-4m(m+5)<0m<1 4

In Eq. (m-5) x -2 (m+2) x + m = 0 [2 (m + 2)] 2-4 m (m-5) = 16 m + 36 because m < 1 4

So when -9 40, the equation has two roots.

When m=-9 4, 16m+36=0, then the equation has two equal rootsWhen m<-9 4, 16m+36<0, the equation has no roots.

Question 1:

Solution: If the greening bandwidth is x meters, then according to the topic: (40+x)*(30+x)=40*30

The solution is x=10 or x=-80 (rounded).

Note: The area of the fish pond is a multiple of the area of the green belt, and the area of the fish pond is 40*30, so the unit 1 is the area of the green = 40*30, so the sum of the fish pond and the green area is (1 + times the area of the fish pond, so the above equation is established).

Question 2: Solution: (1) When m=0, it is a one-time equation, and it is found that the equation has a solution, which contradicts the meaning of the topic, so there is no such situation, that is, m must not be equal to 0

2) When m>0, the discriminant formula = (2(m+2)) 2-4m(m+5)=16-4m, when the discriminant formula "0, it is pushed out by 16-4m>0, m<4, so when solving 05 below, the equation (m-5) x -2(m+2)x+m=0 is a quadratic equation, but it does not satisfy 04, so when solving m>4 below, the root of equation (m-5) x -2(m+2)x+m=0:

a2: When m=5, the equation (m-5) x -2(m+2)x+m=0 is a one-time equation, and the equation has a root of x=5 14;

B2: When 40, i.e., m>-4 9, equation (m-5) x -2(m+2)x+m=0 has two mutually different real roots, and m>-4 9 and 45, equation (m-5) x -2(m+2)x+m=0 is a quadratic equation with two mutually different real roots.

3) When m<0, the equation mx -2(m+2)x+m+5=0, that is: -mx +2(m+2)x-m-5=0, its discriminant formula = (2(m+2)) 2-4m(m+5)=16-4m, when the discriminant formula "0, from 16-4m>0, m<4, so when solving m<0 below, the equation (m-5)x -2(m+2)x+m=0 has two mutually different real roots.

In summary: when m<0, the equation (m-5) x -2 (m+2) x + m=0 has two mutually different real roots;

When m>=4 and not equal to 5, the equation (m-5) x -2 (m+2) x + m=0 has two different real roots;

When m=5, the equation (m-5) x -2(m+2)x+m=0 has two equal real roots;

For other values of m, the equation (m-5) x -2 (m+2) x + m = 0 has no real root.