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Two equal line segments AB and CD coincide in one third. M and N are the midpoint of AB and CD respectively. If Mn = 12cm, find the length of ab
Let AB = CD = 3acm, then BC = ACM,
∵ m, n are the midpoint of AB and CD respectively,
∴BM=1
2AB=3
2acm，CN=1
2CD=3
2acm，
∵MN=12cm，MN=CM+CN=BM-BC+CN，
∴3
2a-a+3
2a=12，
a=6，
3a=18
AB = 18 cm
P is any point of line AB, m, n is the midpoint of PA and Pb, Mn = 10, find ab
MN=PM+PN=1/2(PA+PB)=1/2AB
So AB = 20
It is known that there is a point P, m, N on the line AB with a length of 12 cm, which are the midpoint of PA and Pb respectively, then Mn = ()
Answer: Mn = 6cm
AP+PB=12
Because m and N are the midpoint of PA and Pb respectively, then
MP=1/2AP
PN=1/2PB
therefore
MP+PN=1/2（AP+PB）
That is Mn = 1 / 2Ab
So Mn = 6cm
If the center of the regular hexagon ABCDEF is O, P is any point on the plane ABCDEF that is different from O, and vector OP = m (vector AP + vector BP + vector CP + vector DP + vector EP + vector FP), then real number M =?
So vector AP + vector BP + vector CP + vector DP + vector EP + vector FP = 6op + Ao + Bo + CO + do + EO + fo = 6op (where Ao + Bo + CO + do + EO + fo = 0), so m = 1 / 6
The side length of a regular hexagon ABCDEF is a, P is a point in the hexagon ABCDEF. Find the sum of the distances from point P to each side
Above:
It can be seen from the figure that the distances from P to the six edges are PG, pH, PJ, PK, PM and PN
And PJ + PK = PM + PN = PG + pH = AC
In △ ABX:
AX=√3/2×AB=√3/2×a
∴AC=2AX=√3×a
The sum of distances from point P to each side = 3 √ 3 × a
Given the line Mn and its outer two points AB, and AB two points on both sides to make a point P, so that P on the line Mn, so that the value of PA Pb is the largest Given the line Mn and its outer two points AB, and AB two points are on both sides, find a point P, make P on the line Mn, and make the value of PA Pb maximum .a m__________________________ N .b
Make a point C symmetric about Mn, so that C and B are on the same side of Mn
The intersection point of the straight line connecting C and B with Mn is the point P
prove:
Do a point p1 on Mn, P1 can be any point outside P,
We can get a triangle with C, B, P1 as the vertices
According to the triangle theorem: the difference between the two sides of the triangle is less than that of the third side
｜p1c-p1b｜
Given the line Mn and the two points a and B on both sides of the line Mn, try to find a point P on Mn so that PA = Pb
Point P is a point on the vertical bisector of line ab. how to find this point? Use a compass to draw a circle with point a and point B as the center, and any length greater than AB / 2 as the radius. Two circles will have two intersections, connecting the two intersections. The intersection of this line segment and ab is the point P
Given the line Mn, there are two points AB on the same side of the line Mn RT
As shown in the figure,
① Make a point B about the symmetric point B 'of the line Mn,
② Connect ab ', cross Mn to P
The point P is the point
As shown in the figure, PA and Pb are tangent to point a and point B respectively with ⊙ o, and point m is on Pb, and OM ∥ AP, Mn ⊥ AP, and the perpendicular foot is n (1) Confirmation: om = an; (2) If the radius of ⊙ o is r = 3, PA = 9, find the length of OM
(1) It is proved that: as shown in the figure, if OA is connected, then OA ⊥ AP, ∵ Mn ⊥ AP, ∵ om ∥ AP, ∵ om ∥ AP,
As shown in the figure, two triangular plates with the same shape and size, including 30 degrees and 60 degrees, are placed as shown in the figure. PA, Pb coincide with the line Mn, and the triangle plate pac As shown in the figure, two triangular plates with the same shape and size, including 30 degrees and 60 degrees, are placed as shown in the figure, PA and Pb coincide with the straight line Mn, and the triangle plate PAC and triangle PBD can be counterclockwise around point P
What's the picture? What's the problem?
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