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As shown in the figure, in diamond ABCD, ∠ bad = 60 °, M is the midpoint of AB, P is a moving point on the diagonal AC, if the minimum value of PM + Pb is 3, then the length of AB is () A. 3B. 3C. 6D. 23
Connect BD and AC to o, as shown in the figure: ∵ the quadrilateral ABCD is a diamond, ∵ B and D are symmetrical with respect to the straight line AC, ∵ connect DM and AC to P, then the point P is obtained, BP + PM = PD + PM = DM, that is, DM is the minimum value of PM + Pb (according to the shortest line segment between two points), ? DAB = 60 °, ∵ ad = AB = BD, ∵ m is the middle of ab
As shown in the figure, in diamond ABCD, ∠ bad = 60 °, M is the midpoint of AB, P is a point of AC on the diagonal, if AB = 4, then the minimum value of PM + Pb is____ .
Let m be symmetric with respect to ac. M1 must be the midpoint of AD. let Bm1 intersect with AC at a point, which is the point P
At this point, PM + Pb = PM + PM1 = Bm1 is a straight line segment, so it is the smallest
The triangle ABC is an equilateral triangle, and Bm1 is the height of ad side = 4 * √ 3 / 2 = 2 √ 3
As shown in the figure, in diamond ABCD, ab = 2, ∠ bad = 60 °, e is the midpoint of AB, P is a moving point on diagonal AC, then the minimum value of PE + Pb is ()
A. 1B. 3C. 2D. 5
Connect de and BD, and divide them by the diagonals of the diamond vertically, then we can get that B and D are symmetrical with respect to AC, then PD = Pb, ∧ PE + Pb = PE + PD = De, that is, De is the minimum value of PE + Pb, ∧ bad = 60 °, ad = AB, ∧ abd is equilateral triangle, ∧ AE = be, ∧ de ⊥ AB (the property of isosceles triangle three lines in one). In RT △ ade, de = ad2 − AE2 = 22 − 12 = 3
As shown in the figure, in diamond ABCD, ab = 2, ∠ bad = 60 °, e is the midpoint of AB, P is a moving point on diagonal AC, then the minimum value of PE + Pb is ()
A. 1B. 3C. 2D. 5
By connecting de and BD, and dividing them vertically and equally by the diagonals of the diamond, we can get that B and D are symmetrical with respect to AC, then PD = Pb, ∧ PE + Pb = PE + PD = De, that is, De is the minimum value of PE + Pb, ∧ bad = 60 °, ad = AB, ∧ abd is an equilateral triangle, ∧ AE = be, ∧ de ⊥ AB (the property of isosceles triangle three lines in one)
In diamond ABCD, ab = 1, bad = 600, e is the midpoint of AB, P is a moving point on diagonal AC, then the minimum value of PE = Pb is
When PE = Pb, P is at the midpoint of AC
So PE = Pb = AB / 2 = 1 / 2
The minimum vertical distance from point e to AC = AE / 2 = 1 / 4
The perimeter of diamond ABCD is 8cm, the angle bad = 60 degrees, e is the midpoint of AB side, P is a moving point on diagonal AC, and the minimum value of PE + Pb is obtained
Connect dB, then DB ⊥ AC, and D and B are symmetrical about AC, connect De, intersect AC at P point, then P point is calculated. That is to say, PE + Pb is the shortest at this time. It is proved that if Pb is connected, then Pb = PD, ｜ PE + Pb = de (between two points, the shortest line segment), ⊥ DAB = 60 °, ⊥ DAB is equilateral △ e is the midpoint of AB, ⊥ de ⊥ AB, ⊥ ade = 30 °, ⊥ de = √ 3, that is PE + Pb = √ 3
In diamond ABCD, ab = 4, bad = 60 degrees, e is the midpoint of AB, P is a moving point on diagonal AC, then the minimum value of PE + Pb is?
Connect BD, let AC and BD intersect at o
Because ABCD diamond
So AC vertical BD
So Bo = do
So BP = DP
So BP + PE = DP + PE
In triangle dep
DP+PE>DE
So the minimum value of DP + PE is de
De = ad * sin60 degree = 4 * (radical 3) / 2 = 2 √ 3
So the minimum value of BP + EP is 2 √ 3
In diamond ABCD, point E is the midpoint of AB, ab = 4, and point m is on AC, then the minimum value of EM + BM is______
This problem is simple. If you connect em, BM and DM, it is easy to prove that BM = DM. At this time, you will find that EM + BM is the minimum, that is em + DM is the minimum, and E, m and D are the minimum when they are in a straight line
You didn't give the problem of angle, so it's not easy to find out the result. If you give less conditions, you can figure it out by yourself and know how to do it
In diamond ABCD, the angle DAB = 60 degrees, ab = 1, e is the midpoint of CD, P is any one of AC
Write the process clearly
In diamond ABCD, the angle DAB = 60 degrees, ab = 1, e is the midpoint of CD, P is any point of AC, and the minimum value of PE + PD is obtained
Find the midpoint f on BC, connect DF, the shortest line
That is: PE + PD = DF = the square root of 3 / 2,
Proof: connect PE, PF,
AC is the bisector of angle DCF,
Angle DCA = ACB = 30 degrees
CE=CF,CP=CP,
Similar triangle principle,
Triangle DCP = triangle BCP
PE=PF
Namely: PE + PD = DF
As shown in the figure, in diamond ABCD, diagonal lines AC and BD intersect at point O, ∠ abd = 60 ° BD = 6, find the perimeter of diamond ABCD and the length of diagonal line AC
The four sides of the diamond are equal, ad = Ba, ∠ abd = 60 ° so abd is an equilateral triangle
So the side length of the diamond is 6 and the circumference is 24
The triangle AOB is a right triangle AB = 6 ob = 3 Ao = 3 times root 3
RELATED INFORMATIONS
- 1. As shown in the figure, points E and F are points on edge BC and CD of diamond ABCD respectively, and ∠ EAF = ∠ d = 60 ° and ∠ fad = 45 °, then ∠ CFE=______ Degree
- 2. In rectangular ABCD, BC = 4, BG is perpendicular to diagonal AC and intersects AC, ad and ray CD at points e, F, G, ab = x respectively. (1) when point G coincides with point D, calculate the value of X; (2) when point F is the midpoint of AD, calculate the value of X and the sine value of ∠ ECF
- 3. In the diamond ABCD with side length of 2, ∠ bad = 60 °, e is the midpoint of AB, P is the upper moving point of AC, and the minimum value of Pb + PE is obtained
- 4. In a circle with a radius of 12 decimeters, draw a square, draw a small circle in the square, and calculate the area of the small circle
- 5. In trapezoidal ABCD, AB / / CD, AC is vertical to BD, angle BDC = 60 degrees, AC = 3 root sign 3, find the length of BD and the area of trapezoidal ABCD In the isosceles trapezoid ABCD, ab = 2CD, AC bisects DAB, ab = 4, ab ‖ CD, (1) Find the trapezoid angle; (2) find the trapezoid area
- 6. In the parallelogram ABCD, be ⊥ ad, BF ⊥ CD, the perpendicular foot is e, F, the perpendicular center of △ bef is h, if DG ⊥ BC, the perpendicular foot is g It should be verification: BH = GF
- 7. In the parallelogram ABCD, M is the midpoint of BC, n is on AB, BN = 1 / 2An, NM intersects BD with G, and BG: GD is obtained
- 8. In the trapezoidal ABCD, point P starts from point a and moves along the edge of ad to point d at a speed of 1cm / s, and point Q starts from point C and moves along CB to point B at a speed of 2cm / s. if points P and Q start from two points at the same time, when one of them reaches the end point, the other stops moving. (1) what is the value of T when trapezoidal pbqd is a parallelogram? (2) What is the value of T when trapezoid pbqd is isosceles trapezoid?
- 9. The perimeter of parallelogram ABCD is 75cm, the height is 14cm with BC as the bottom (as shown in the figure); the height is 16cm with CD as the bottom. Calculate the area of parallelogram ABCD
- 10. As shown in the figure below, the area of rectangle ABCD is 12 square decimeters, so the area of circle is () square decimeters Well, there is no picture. Let's give a brief description There is a circle, no matter how big, half of it (that is, a semicircle). Draw a rectangle on the semicircle. The length of the rectangle is equal to the diameter of the circle. Then say that the area of the rectangle is 12 square decimeters. Tell the area of the whole circle. Thank you!
- 11. It is known that the four vertices of the quadrilateral ABCD are a (2,3), B (1. - 1), C (- 1. - 2) and D (- 2.2),
- 12. If the length of a diagonal line of diamond ABCD is 6 sides and the length of AB is a root of the square of equation x - 7x + 12 = 0, then the perimeter of diamond ABCD is In the two known equations x ^ 2 + ax + (A-1) = 0, the absolute value of negative root is greater than that of positive root, and the value range of a? It is known that the equation x ^ 2-mx + M-1 = 0 has two real roots x1, X2 and 1 / X1 + 1 / x2 = m, then M =? One diagonal of the diamond ABCD is 3
- 13. In diamond ABCD, if the angle ABC is equal to 120 degrees, then BD is equal to several times AC
- 14. If the central line and the high line on the hypotenuse of a right triangle are 6cm and 5cm respectively, then its area is
- 15. In rectangle ABCD, ab = 4cm, ad = 3cm, fold the rectangle along the straight line AC, and point B falls at point E, connecting de. what is aced in four sides? Please explain the reason and calculate its area! Very urgent, this is today's home work, tomorrow to hand in, online 30 minutes, please hurry up, good answer bonus!
- 16. M. What is the relationship between the area of Mn parallel to AC, △ ABM and △ BCN?
- 17. As shown in the figure, △ ABC has rhombic ampn, if ammb = 12, then BPBC=______ ．
- 18. As shown in the figure, △ ABC, Ag ⊥ BC is at point G, AB and AC are taken as one side to make rectangle ABME and rectangle acnf outside △ ABC respectively, and the ray GA intersects EF at point h. If AB = KAE, AC = Kaf, try to explore the quantitative relationship between he and HF, and explain the reason
- 19. As shown in the figure, ab = ad, BC = De, BA ⊥ AC, Da ⊥ AE, can you prove am = an?
- 20. In the known triangle ABC, ab = AC, the circle with ab as diameter intersects BC at D and AC at e, and BD = DC = De is proved In the known triangle ABC, ab = AC, the circle with ab as diameter intersects BC at D and AC at e, and BD = DC = De is proved