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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),
The slope k = (y1-y2) / (x1-x2). X1, X2, Y1 and Y2 are the coordinates of the point respectively
1. The four vertices of the quadrilateral ABCD are a (2,3), B (1, - 1), C (- 1, - 2) and D (- 2,2). Find the slope and inclination angle of the straight line where the quadrilateral is located
1. (1) if and only if what is the value of M, the slope of the straight line passing through two points a (- m, 6) and B (1,3m) is 12?
(2) If and only if M is the value, the inclination angle of the straight line passing through two points a (m, 2) and B (- m, 2m-1) is 60 degrees
k=(y2-y1)/(x2-x1)=tanа
A is the inclination angle
It is known that the four vertices of the quadrilateral ABCD are a (0,0), B (2, - 1), C (4,2), D (2,3). (1) try to judge the shape of the quadrilateral ABCD and give the proof; (2) find the solution
It is known that the four vertices of the quadrilateral ABCD are a (0,0), B (2, - 1), C (4,2), D (2,3). (2) find the area of the quadrilateral ABCD
Parallelogram
If the vertex a (- 4,3) B (2,5) C (6,3) d (- 3,0) of quadrilateral ABCD is known, its shape can be judged
emergency
Using vector to determine vector AB = (6,2) vector CD = (- 9, - 3) because vector AB = - 3 / 2 vector CD, then ab ∥ CD vector ad = (1, - 3) vector BC = (4, - 2) it is obvious that AD and BC are not parallel, so we get that ABCD is trapezoid, and the scalar product of vector AD and vector AB = (6,2) (1, - 3) = 6-2 * 3 = 0, so ad ⊥ AB can know ab
Given the vertices a (- 7,0), B (- 2, - 3), C (5,6), D (- 4,9) of the quadrilateral ABCD, we can judge the shape of the quadrilateral
A (- 7,0), B (- 2, - 3), C (5,6), D (- 4,9), quadrilateral ABCD is a general quadrilateral. If the coordinate of B is changed to B (2, - 3), then quadrilateral ABCD is a square. The slopes of AB and CD are equal, which is - 1 / 3. The slopes of AD and BC are equal, which is 3
The coordinates of three vertices A.B.C of a diamond are a (- 2,5), B (4,8), and C (10,5). What is the coordinate of the fourth vertex D?
The sum of coordinates corresponding to diamond ABCD, a and C is equal to the sum of coordinates corresponding to B and D. that is to say, the sum of coordinates corresponding to two diagonal ends of parallelogram is equal. The coordinates of vertex D are (x, y)
X+4=10-2,y+8=5+5,
x=4,y=2
So D (4,2)
Let a (3,4), B (- 2, - 1) and C (4,1) be right triangles
Find out the length of three sides
AB²=5²+5²=50
AC²=1²+3²=10
BC²=6²+2²=40
Accord with Pythagorean theorem
So it's a right triangle
In diamond ABCD, the angle bad is 80 degrees, the vertical bisector of ad intersects the diagonal AC at point F, e is the perpendicular foot, connecting DF, then the angle CDF is equal to ()
A.80 B.70 C.65 D.60
Choose D
Because EF is the vertical bisector of AD
So AF = DF and angle CAD = angle FDA
Because AC is the diagonal of diamond ABCD
So angle CAD = 1 / 2 angle bad = 40 degrees
Angle ADC = 180 degrees - 80 degrees = 100 degrees
So the angle is 40 degrees
So angle CDF = angle ADC - angle FDA = 100 degrees - 40 degrees = 60 degrees
Find video: in diamond ABCD, the angle bad = 80 degrees, the vertical bisector of AB intersects the diagonal AC at point F, e is the perpendicular foot, connecting DF, then the angle CDF is equal to []
If the perimeter of the diamond is 40 cm and the ratio of two adjacent angles is 1:2, the length of the longer diagonal is 40 cm
According to the meaning of the title, the diagonal lines AC and BD of the diamond are divided vertically and equally,
DF=BF
AF = BF
∠ADF=∠DAC=∠BAD/2=80°/2=40°
And ∠ ADC = 180 ° - bad = 100 °
So ∠ CDF = 100 ° - ADF = 100 ° - 40 ° = 60 °
②
If the two adjacent angles of the diamond are complementary, the smaller angle is 180 °× 1 / (1 + 2) = 60 ° and the larger angle is 180 °× 2 / (1 + 2) = 120 °
The side length of diamond is 40 △ 4 = 10cm
Then the length of the longer diagonal is √ (2 * 10 ^ 2-2 * 10 * 10 * cos 120 °) = 10 √ 2 (1 + cos 60 °) = 10 √ 2 (1 + 1 / 2) = 10 √ 3
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 the two points), ? DAB = 60 °, ∵ ad = AB = BD, ∵ m is the midpoint of AB, ? DM X AB, ∵ PM + Pb = 3, ∵ DM = 3, ∵ AB = ad = dmsin60 ° = 332 = 23 ．
RELATED INFORMATIONS
- 1. 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
- 2. 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
- 3. 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
- 4. 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
- 5. 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
- 6. 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
- 7. 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
- 8. 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
- 9. 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?
- 10. 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
- 11. 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
- 12. In diamond ABCD, if the angle ABC is equal to 120 degrees, then BD is equal to several times AC
- 13. If the central line and the high line on the hypotenuse of a right triangle are 6cm and 5cm respectively, then its area is
- 14. 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!
- 15. M. What is the relationship between the area of Mn parallel to AC, △ ABM and △ BCN?
- 16. As shown in the figure, △ ABC has rhombic ampn, if ammb = 12, then BPBC=______ ．
- 17. 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
- 18. As shown in the figure, ab = ad, BC = De, BA ⊥ AC, Da ⊥ AE, can you prove am = an?
- 19. 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
- 20. It is known that: as shown in the figure, in △ ABC, ∠ C = 90 °, cm ⊥ AB in M, at bisection ∠ BAC intersects cm in D, intersects BC in T, de ∥ AB through D intersects BC in E, verification: CT = be