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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
It can't be vertical
As shown in the figure, in the quadrilateral ABCD, ab = ad, AC bisects ∠ BCD, AE ⊥ BC, AF ⊥ CD. If there are triangles congruent with △ Abe in the figure, please explain the reason
In the graph, △ ADF and △ Abe are congruent. ∵ AC bisects ∠ BCD, AF ⊥ CD, AE ⊥ CE; ∵ AF = AE, ∠ AFD = ∠ AEB = 90 ° in RT △ ADF and RT △ Abe, ∵ AB = ADAF = AE, ≌ RT △ ADF ≌ RT △ Abe
As shown in the figure, in the trapezoidal ABCD, AD / / BC, angle ABC = 90 °, ab = 20cm, CD = 25cm
In trapezoidal ABCD, AD / / BC, angle ABC is 90 degrees, ab = 20cm, CD = 25cm, moving points P and Q start from point a at the same time, point P moves along the route of a-d-c at the speed of 3m / s, point Q moves along the route of a-b-c at the speed of 4m / s, and PQ two points arrive at point C at the same time
(1) Finding the area of trapezoid ABCD
(2) Let the movement time of PQ two points be t (s) and the area of quadrilateral apcq be s (cm2). Try to find the functional relationship between S and T, and find the value range of independent variable t
(1) Through point D, make de ⊥ BC at point e. from the known ad = be, de = AB = 20cm
In RT △ Dec, according to Pythagorean theorem, EC = 15cm,
(AD + 25) / 3 = (20 + AD + 15) / 4
The area of ladder shaped ABCD = [(AD + BC) * ad] / 2 = [(5 + 20) * 20] / 2 = 250 (cm2)
(2) When the time of P and Q is t (seconds), the distance of P and Q is 3T (CM) and 4T (CM)
① When 0 ＜ t ≤ 5 / 3, P moves on AD and Q moves on ab
In this case, the area of quadrilateral apcq is s, and trapezoid abcd-s △ bcq-s △ CDP is 70t
② When 5 / 3 ＜ t ≤ 5, P moves on DC and Q moves on ab
At this time, the area of quadrilateral apcq is s = s, trapezoid abcd-s △ bcq-s △ ADP = 34T + 60
③ When 5 ＜ t ＜ 10, P moves on DC and Q moves on BC
At this time, the area of quadrilateral apcq is s = s, trapezoid abcd-s △ abq-s △ ADP = - 46t + 460
As shown in the figure, in the parallelogram ABCD, ad is perpendicular to DB, AC and BD intersect with point O, OD = 1, and CAD = 30 degrees. Calculate the length of AC and DC
In RT triangle AOD, from od = 1 angle, CAD = 30 degree, Ao = 2, ad = root 3
Then AC = 4 and DB = 2, then in RT triangle abd, hypotenuse AB = CD = radical 7
As shown in the figure, in the parallelogram ABCD, e and F are two points on AD and BC respectively, and BF = De, connecting be
DF, verification: be = DF,
A quadrilateral ABCD is a parallelogram
∴AD‖BC
∵ E and F are two points on AD and BC respectively
∴BF‖DE
And ∵ BF = De
The bfde is a parallelogram
∴BE=DF,BE//DF
As shown in the figure, in the parallelogram ABCD, be is perpendicular to point E, BF is perpendicular to point F, be = 2, BF = 3, the perimeter of the parallelogram is 20, then the area of the parallelogram is
Let CD = x, BC = y
S parallelogram = BC * be = 2Y
S parallelogram = CD * BF = 3x
So 3x = 2Y
C parallelogram = 2 * (x + y) = 20
So x + y = 10
solve equations
3x=2y
x+y=10
X=4
y=6
As shown in the figure, it is known that in rectangular ABCD, e is the midpoint of AD, CE ⊥ BD, the foot is f, passing through point F, FG is parallel to BC, and intersection be is at point g. verify: BG & # 178; = BF * FD
It is proved that rectangle ABCD
RELATED INFORMATIONS
- 1. 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
- 2. 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?
- 3. 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
- 4. 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!
- 5. In the quadrilateral ABCD, BD bisects ∠ ABC, AC ⊥ BD at R, PQ intersects BC and ad at points Q and P respectively, and ∠ bad = ∠ BQP
- 6. As shown in the figure, ad ∥ BC in ladder ABCD, point E is the midpoint of edge ad, connecting be with AC at point F, the extension line of be with CD at point G, verify Ge / GB = AE / BC
- 7. As shown in the figure, in ladder ABCD, ab ‖ CD, e is the midpoint of DC, the line be intersects AC at F, and the extension line of ad intersects g; verification: EF · BG = BF · eg
- 8. Take a point G on the extension line of one side of the square ABCD and connect it with a, intersect BC at point F, connect diagonal BD and intersect AF at point E. AE = 5, AF = 3 are known, and calculate the length of FG chart
- 9. In quadrilateral ABCD, ad = BC, AE, CF are perpendicular to BD, be = DF. It is proved that quadrilateral ABCD is parallelogram I can't find the last condition in the verification. Please give me some advice. Thank you
- 10. In square ABCD, point G starts from point C and moves along CB; point h starts from point a and moves along Ba, and two points move at the same speed and stop at the same time
- 11. 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
- 12. 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
- 13. 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
- 14. 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
- 15. 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
- 16. 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
- 17. 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),
- 18. 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
- 19. In diamond ABCD, if the angle ABC is equal to 120 degrees, then BD is equal to several times AC
- 20. If the central line and the high line on the hypotenuse of a right triangle are 6cm and 5cm respectively, then its area is