數學問題 40points

hihi

2009-12-27 3:04 am

1.The area of △ABC is 2400cm^2 and the length of the base is 80cm. D,E,F are points on BC,AC,AB respectively. The quadrilateral BDEF is a parallelogram. Let BD=x cm, find the area of BDEF in terms of x.

2.In a golf pratice, a player hits the ball at a certain position and the flying route of the ball satisfies the quadratic function y=-0.2 x^2+10x, where y m is the height of the ball, x m is the horizontal distance of the ball. It is known that the horizontal distance between the ball and the hole is 2 m when the ball reaches the ground.

a.Write down the coordinates of the vertex and the equation of axis of symmetry of the graph of the quadratic function.

b.Find the horizontal distance between the ball and the hitting position when the ball reaches the ground.

c.If the player wants to hit the ball into the hole at the same hitting position with the maximum height of the ball same as previous hitting, then the flying route of the ball should satisfy which quadratic function.

回答 (1)

用戶9112

2009-12-27 5:11 pm

✔ 最佳答案

(1) Since BDEF is a parallelogram, BD // FE and BF // DE, and EF = BD = x
Triangles ABC, AFE and EDC are all similar (AAA)
Let h and H be the heights of the triangles AFE and EDC
The height of triangle ABC = 2400*2/80 = 60
Consider triangles AFE and ABC, h/60 = FE/80
h = 60x/80 = 3x/4
H = 60 - h = 60 - 3x/4
Area of BDEF = BD*H = (60 - 3x/4)(x) = 60x - (3/4)x^2

http://img51.imageshack.us/img51/3376/triangle1.png

圖片參考：http://img51.imageshack.us/img51/3376/triangle1.png

(2)(a) y = -0.2x^2 + 10x
y = -0.2(x^2 - 50x + 625) + 125
y = -0.2(x - 25)^2 + 125
Vertex is (25, 125)
Axis of symmetry is x - 25 = 0
(b) When the ball hits the ground, y = 0
=> -0.2(x - 25)^2 + 125 = 0
0.2(x - 25)^2 = 125
(x - 25)^2 = 625
x - 25 = +/- 25
x = 0(initial condition) or 50
The ball hits the ground at a horizontal distance of 50 m
(c) Let the path be y = -a(x - b)^2 + 125 (same max height of 125)
The horizontal distance when the ball hits the ground should be 0 (initial condition) and 52
0 = -a(0 - b)^2 + 125 ... (1)
0 = -a(52 - b)^2 + 125 ... (2)
(1) - (2) => a[(52 - b)^2 - b^2] = 0
(52 - b + b)(52 - b - b) = 0
b = 26
Sub into (1), 0 = -a(26^2) + 125
a = 125/676
Therefore the equation is y = -(125/676)(x - 26)^2 + 125

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