Angle DAE= angle EFC, angle AED and angle CEF are diagonal lines, so they are equal, that is, △ADE and △CEF are equal.
Because DE=EC, the two triangles are congruent.
So the area of △ADE is equal to the area of △CEF, that is, the area of trapezoidal ABCD is equal to the area of triangular ABF.
The second question: Make a straight line CD that passes through point P (AO at point C and BO at point D respectively), so that P is the midpoint of CD, that is, CP=DP.
When MP> is NP, the parallel lines of BO pass through point C and intersect with MN at point E, according to the conclusion of the first question, the area of trapezoidal cone = δ = δ COD, and the area of MON >; The area of a trapezoidal cone.
Similarly, when MP
The third question:
According to the conclusion of the second question, if and only if MP=NP, the area of △MON is the smallest (as long as Q point is circled, it is isolated).
According to the given data, the vertical line of OB passes through point P, and intersects A0 and B0 at points C and D respectively. Obviously, CP
That is, the area of △MON is equal to the area of trapezoidal CENO.
Because OP=4, the angle POB is equal to 30 degrees, OD=2 times the root number 3, and OD=CD=2 times the root number 3. DP=2, so CP=EP=2 times the root number 3-2, so ED=2-(2 times the root number 3 -2)=4-2 times the root number 3, so the △MON area is equal to the △COD area -△DEN area, that is, the square of 1/2(2 times the root number 3)-/kloc-.
Extension problem: According to the problem, AB is perpendicular to the X axis, and OC is a segment on the 45 bisector. Let's assume that the straight line and OC and AB intersect with points M and N respectively, and the coordinate of N is set to (6, s), where the range of S is (0, 3) exclusive, because the straight line is divided into two quadrangles, so point N cannot coincide with point A or B. According to the solution of the straight line, the linear equation Y = (s-) can be obtained by using P (4, 2) and N (6, s). Then the area of the quadrilateral OMPN is the sum of the areas of △OMA and △MAN.
△OAM area is the y-axis coordinate of point 1/2*6*M, and△ △MAN area is1/2 * s * (the x-axis coordinate of point 6-m).
If I have calculated the formula correctly, it should be 14-(4-s)+4/(4-s), and there is a minimum value in brackets if and only if 4-s=4/(4-s), that is, 4-s=2 (I don't need to say why).
Page (abbreviation of page) s: the solution of the minimum value in brackets ... is a+b >; =2 times the root number of ab, according to this principle. . . If you can understand this formula, if you can't deduce the formula for calculating the area, I'll write it down for you. You can calculate it yourself: S =(S+6s-36 squared) /(s-4), because s must be less than 4, so there will be the above formula 14-. It is factorization that is pieced together slowly according to this format. Just be careful and patient. Ask again if you don't understand.