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Anonymous users2024-02-07First of all, I don't know if I'm answering the right question, I haven't done this for a long time. Hopefully there will be a better answer. First, make a vertical line with fo through the k point.
The area of the shadow part should be the area of the triangle FEC minus the area of the two small triangles FMB and NEC.
That is, the triangle fkc-fmb-nec, and the area formula for the shadow part is obtained: s=1 2fc*ko-1 2fb*mb-1 2ec*ne
Since the right triangle abc is congruent with def, the triangle fmb is equal to the nec area.
s=1 2fc*ko-fb*mb, because the right triangle angle c=angle f=30 degrees.
fc=2*√3ko，fb=√3mb
s=1/2*2√3ko*ko-√3mb*mb=√3ko2-√3mb2=√3(ko2-mb2)
When mb=0, the area is maximum.
So when mb=0, ko gives =1 2a, s = 3ko2 = 3*(1 2a)2= 3a2 4
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Anonymous users2024-02-06Divide the shadow part into 2 parts, square and triangle, let mb be b then get the length of fb, and then get the length of be, calculate the area of the square, the middle of mn is 0, then the length of k0 is half of am, and then get the area of the triangle, the sum of the 2 areas will become a 2 yuan 2 equation, in fact, a is a constant directly 1, you don't have to ignore it, as a unary 2 equation, according to the equation, you can calculate the value of b when the value of heat is a multiple of a, so that you can calculate the value of b, You can also calculate the area. As a student, the answer is also directly untold, just tell you the method, and calculate it yourself.
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Anonymous users2024-02-05a（2，3） d（-2.-3）
b（1，2） e（-1，-2）
c(3,1) f(-3,-1) is characterized by symmetry of the corresponding point with respect to the origin, so a+2a+3=0
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Anonymous users2024-02-04,3）d（－2，－3）b（1,2）e（－1，－2）c（3,1）f（－3，－1）
With respect to (0,0) symmetry, the abscissa and ordinate are opposite to each other, respectively: 4 b 2b 3 0
a＝－1，b＝－1
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Anonymous users2024-02-03（1）bb~⊥om.Find the slope of om (2) find oa amSlope to find the intersection.
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Anonymous users2024-02-02The process is shown in the figure below
If you have any questions, please feel free to ask @
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