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Given that the two diagonals of the diamond are 6 and 8 respectively, what is the perimeter of the diamond?
The characteristics of the diagonals of the diamond are: the diagonals are equally divided and perpendicular to each other! One is 6, one is 8, half is 3 and 4, then the side length is 5 (because the side length is the hypotenuse of the right triangle), so the perimeter is 4 × 5 = 20
Thank you
If the length of the two diagonals of the diamond is 6 and 8 respectively, what is its perimeter
20
Point O is the middle point of AC. translate the diamond ABCD with the perimeter of 4cm along the diagonal AC direction to get the diamond ob'c'd ', then the perimeter of the quadrilateral OECF is?
It should be translational Ao length
It's not the length
Quadrilateral OECF doesn't know which quadrilateral it is
The intersection of trapezoid ABCD, AD / / BC, ab = AC, BC = BD, AB, vertical AC, diagonal AC, BD and point O is proved; OC = DC
It is proved that AE = DF because AB = AC, ab ⊥ AC so ∠ ABC = ACB = 45 ° and ⊥ BAE = CAE = 45 ° so AE = be = CE = BC / 2 and because BC = BD so DF = AE = BC / 2 = BD / 2 so ⊥ DBF = 30 ° and ⊥ BCD = BDC
In trapezoidal ABCD, ad is parallel to CD, diagonal AC and BD intersect at O, OB = OC, try to explain AB = DC
Wrong. Is ad parallel to BC
First of all, there is something wrong with your question. AD and CD can only intersect, because both have D. it should be that ad is parallel to CB
It is proved that ∵ ob = OC
∴∠OBC=∠OCB
∵AD∥CB
∴∠OAD=∠OCB,∠ODA=∠OBC
∴∠OAD=∠ODA
∴△OAB≌△ODC
∴AB=DC
As shown in the figure, in rectangular ABCD, AC and BD intersect at o AE bisector angle bad intersects at e angle CAE = 15 degrees
The triangle AOB is an equilateral triangle
Finding the degree of angle aoe
Certification:
The quadrilateral ABCD is a rectangle
∴OA=OB,∠BAD=90°
∵ AE bisection ∠ bad
∴∠BAE=45°
∵∠CAE=15°
∴∠BAO=45+15=60°
The AOB is an equilateral triangle
∵∠OBE=30°,BE=BA=BO
∴∠BOE=75°
∴∠AOE=60+75=135°
A mathematical problem: in ladder ABCD, points E and F are on AB and CD respectively, EF is parallel to AD. suppose EF moves up and down in parallel
In trapezoidal ABCD, points E and F are on AB and CD respectively, and EF is parallel to AD
(1) If AE / EB = 1 / 2, prove 3ef = BC + 2ad
(2) Please explore the general conclusion, that is, if AE / EB = m / N, what conclusion can be drawn?
1. Make auxiliary line from point a, parallel CD, intersect EF with G, intersect BC with H
Because triangles are similar
AE/EB=1/2
3EG=BH
3EF=3EG+3GF=BH+3GF=BC+2AD
2. AE / EB = m / n
（m+n）EF=mBC+nAD
As shown in the figure, in trapezoidal ABCD, ad ‖ BC, ∠ C = 90 °, e is the midpoint of CD, EF ‖ AB intersects BC at point F (1) to prove: BF = AD + CF; (2) when ad = 1, BC = 7, and be bisects ∠ ABC, the length of EF is calculated
(1) Proof: proof 1: as shown in figure (1), extend the extension line of ad-fe to n ∫ ad ∥ BC, ∫ C = 90 ° At the point, d make DN 8741; AB as DN ∥ AB, BC to n \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\theresults show that EF = 1 + 7-bf, 2BF = 8, BF = 4, EF = 4 The length is 4
As shown in the figure, in quadrilateral ABCD, ad = BC, AC = BD, ab ≠ DC, it is proved that (1) od = OC, (2) quadrilateral ABCD is isosceles ladder
It is proved that: (1) if
RELATED INFORMATIONS
- 1. As shown in the figure, in the triangle ABC, ab = BC, angle ABC = 90, there is a straight line PQ passing through point B, ad is made through a and C respectively, PQ is vertical to D, CE is vertical to E (1) (2) if the line PQ intersects with the line AC, other conditions remain unchanged, whether the conclusion in (1) is still valid, if not, write the corresponding conclusion and prove it
- 2. As shown in the figure, in the isosceles trapezoid ABCD, ad ‖ BC, ad = 3, BC = 5. AC and BD intersect at point O, and ∠ BOC = 60 °. If AB = CD = x, then the value of X is______ ．
- 3. As shown in the figure, the two diagonals of rectangle ABCD intersect at point O, through O for of ⊥ ad at point F, of = 2, through a for AE ⊥ BD at point E, and be: BD = 1:4, find the length of AC
- 4. As shown in the figure, it is known that in the trapezoidal ABCD, ab ∥ DC, ∠ d = 60 °, C = 45 °, ab = 2, ad = 4, the area of trapezoidal ABCD can be calculated Who answers fast and well? I'll give another 50 points
- 5. In the parallelogram ABCD, AE is perpendicular to BC, AF is perpendicular to CD, and the angle EAF is 75. Calculate the degree of each internal angle of the parallelogram ABCD
- 6. As shown in the figure, in the known trapezoid ABCD, ad is parallel to BC, ab = CD = 5, AD / BC = 2 / 5, CoSb = 3 / 5, P is the moving point on the edge BC, ∠ Apq = ∠ B The distance from point P to point B is X 1) Find the length of BC 2) if △ Apq is an isosceles triangle, find the value of x 3) can △ CPQ and △ ABP be similar? If so, please write the value of X directly. If not, please explain the reason
- 7. In rectangular ABCD, ab = 3, ad = 4, P moves on AD and DC, let ∠ ABP = θ, fold △ ABP along BP, make dihedral angle a-bp-c become straight dihedral angle, when θ is (), AC length is minimum A. 30°B. 45°C. 60°D. 75°
- 8. The plane of square ABCD and square abef intersects AB, and there are two points P and Q on AE and BD respectively, and AP = DQ
- 9. As shown in Figure 1, in the square ABCD, there is a point P satisfying AP = AB, Pb = PC, connecting AC and PD (1) Verification: △ APB ≌ △ DPC; (2) verification: ∠ PAC = 12 ∠ BAP; (3) if the square ABCD in the original question is changed into isosceles trapezoid ABCD (as shown in Figure 2), ad ‖ BC, and Ba = ad = DC, a point P in the shape still satisfies AP = AB, Pb = PC, is the conclusion in question (2) still valid? If yes, please give proof; if not, please give reasons
- 10. As shown in the figure, the known point P is a point on the square ABCD, PE ⊥ BC, PF ⊥ CD, and the perpendicular feet are e and F, respectively There is no picture,
- 11. In the isosceles trapezoid ABCD, the diagonal AC ⊥ BD of ad ∥ BC is perpendicular to o, AE and DF, and the perpendicular feet are e and F. if ad = a, BC = B, the perimeter of quadrilateral aefd can be calculated
- 12. In rectangular ABCD, point m is the midpoint of CD, ab = 8cm, BC = 10cm. Fold the rectangular ABCD so that a and M coincide. Crease EF intersects ad at point E, BC at point F, Find the length of AE
- 13. As shown in the figure, in square ABCD, e is on CD, and AE = EC + BC, M is the midpoint of CD, and the proof is: ∠ BAE = 2 ∠ dam The map address is D: / / my documents / doc1.mht
- 14. In the parallelogram ABCD, e is the midpoint of diagonal AC, EF is perpendicular to F, angle B = 60 degrees, ab = 4, angle ACB = 45 degrees, find the length of DF
- 15. In the quadrilateral ABCD, ab = CD, e and F are the midpoint of BC and ad. the extension lines of Ba and EF intersect at m, the extension lines of CD and EF intersect at n, proving that angle ame is equal to angle dne
- 16. Known: as shown in the figure, in rectangular ABCD, points E and F are on the edge of AD, CE and CF intersect with BD at points m, N, AE = EF = FD = 4cm, ab = 16cm, respectively, to find the length of Mn
- 17. Given that B is a point on line AC, M is the midpoint of line AB, n is the midpoint of line AC, P is the midpoint of Na, q is the midpoint of Ma, find Mn: PQ Because PQ = ap-aq = an ﹣ 2-AM ﹣ 2 = (an-am) ﹣ 2 = Mn ﹣ 2 So Mn △ 2 = PQ Mn = 2pq So Mn: PQ = 2:1
- 18. As shown in the figure, there are free moving metal rods PQ and Mn on the two horizontal smooth guide rails. When PQ moves under the action of external force, Mn moves to the right under the action of magnetic force, the movement of PQ may be () A. Uniform acceleration to the right B. uniform acceleration to the left C. uniform deceleration to the right D. uniform deceleration to the left
- 19. As shown in the figure, CD ‖ EF, ∠ C ＋ f = ∠ ABC, verification: ab ‖ GF
- 20. As shown in the figure, it is known that the straight line AB and CD are cut by the straight line EF. If ∠ BMN = ∠ DNF, ∠ 1 = ∠ 2, then MQ ‖ NP. Why?