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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
Let DG ∥ AC intersect the BC extension line at point G, then BDG is an isosceles right triangle, BG = a + B, DF ⊥ BG
So: DF = 1 / 2 * (a + b),
The perimeter of quadrilateral aefd is 2A + A + B = 3A + B
In rectangle ABCD, ab = 6, ad = 8, P is the point on ad. if the distance from point P to AC is 2, what is the distance from point P to BD
AC=BD=10
According to the conditions, PF is perpendicular to AC and F, PE is perpendicular to BD and E
Because pf: PA = 6:10, PF = 2, PA = 10 / 3, PD = 8 - 10 / 3
Similarly, PE: PD = 6:10
PE = 2.8
ABCD is a rectangle, P is the last moving point of AD, ad = 15, ab = 8, find the distance sum of point P to AC, BD. sorry, no illustration
The sum of these distances is equal to the distance between point D and ac. in a right triangle ADC, the distance between D and AC using triangle area is 120 / 17
rectangle ABCD.AB=a , ad = B fold along the diagonal AC, let ABC and ADC into 60 degrees, calculate DB distance
Explain the process clearly
As shown in the figure, in known quadrilateral ABCD, ab = ad, ∠ ABC = ∠ ADC
It's a diamond with a cross in the middle like a kite
Certification:
∵AB=AD
∴∠ABD=∠ADB
∵∠ABC=∠ADC
℅ ∠ CBD = ∠ CDB [equal quantity minus equal quantity]
∴CB=CD
And ∵ AC = AC
∴⊿ABC≌⊿ADC（SSS）
∴∠BAC=∠DAC
That is, AC is the bisector of ∠ bad
Abd is an isosceles triangle
⊥ AC ⊥ BC [three lines in one]
As shown in the figure ad = AB, CD = CB, verify: ∠ B = ∠ D
It is proved that: connecting AC, in △ ADC and △ ABC, ∵ ad = abdc = BCAC = AC, ≌ △ ADC ≌ △ ABC (SSS), ∩ B = D
Known: as shown in the figure, rectangular ABCD, AB is 8cm long, and the diagonal is 4cm longer than ad side. Calculate the ad side length and the distance AE from point a to BD
Let ad be x, then BD be (x + 4). Pythagorean theorem: (x + 4) square - x square = 8, square = 64, the solution is x = 6, x + 4 = 10, that is, ad = 6cm, BD = 10cm. Then s triangle abd = AB * ad / 2 = BD * AE / 2, that is, 24 = 5ae, AE = 4.8cm
How did 5ae get it
The area of the same right triangle is calculated with the right side and the hypotenuse as the bottom
AE is the high edge of BD
10AE=6X8
AE=4.8
It is known that, as shown in the figure, the length of rectangle ABCD is 8cm, and the diagonal is 4cm longer than the side of AD. find the length of AD and the distance AE from point a to BD
The solution is: x = 6, ad = 6cm, BD = 10cm, AE = ab · addd = 4.8 (CM)
If the length of the two diagonals of a diamond is 6 and 8 respectively, the perimeter of the diamond is ()
A. 20B. 16C. 12D. 10
As shown in the figure, in diamond ABCD, AC = 8, BD = 6. ∵ ABCD is diamond, ∵ AC ⊥ BD, Bo = 3, Ao = 4. ∵ AB = 5. ∵ perimeter = 4 × 5 = 20
If the length of the two diagonal lines of the diamond is 6 and 8, the perimeter of the diamond is? And the area is?
How many?
The perimeter is 20. The area is 24
RELATED INFORMATIONS
- 1. Given that the two diagonals of the diamond are 6 and 8 respectively, what is the perimeter of the diamond?
- 2. 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
- 3. 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______ ．
- 4. 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
- 5. 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
- 6. 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
- 7. 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
- 8. 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°
- 9. 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
- 10. 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
- 11. 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
- 12. 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
- 13. 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
- 14. 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
- 15. 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
- 16. 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
- 17. 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
- 18. As shown in the figure, CD ‖ EF, ∠ C ＋ f = ∠ ABC, verification: ab ‖ GF
- 19. 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?
- 20. As shown in the figure, the straight lines AB and CD are cut by the straight line EF. If the apposition angle ∠ 1 = ∠ 3, is the internal stagger angle ∠ 2 equal to ∠ 3? Is the inner corner ∠ 3 and ∠ 5 complementary? Please give reasons