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As shown in the figure, the diagonals AC and BD of rectangular ABCD are called o, and E F are the midpoint of OA and ob respectively If ad = 4cm, ab = 8cm, find the length of DF of EF
prove:
∵ the diagonals of the rectangle are equally divided and equal, the four corners are 90 ∵ and the opposite sides are equal
∴OA=OB,
∴∠OAB=∠OBA
The remaining angles of equal angles are equal
∵ e, f are the midpoint of OA, ob
∴AE=BF
And ∵ ad = BC
∴⊿ADE≌⊿BCF（SAS）
∵AB=8cm,AD =4cm,∠DAB=90º
∴BD=√(AB²+AD²)=4√5
∵ OB=½BD ,
∴OF =½OB=¼BD=√5
DF=BD-BF=¾BD=3√5
∵ E F is the midpoint of OA and ob respectively
Ψ EF is the median of ⊿ OAB
∴EF=½AB=4
If the point O is the center of gravity of the diamond ABCD, and OA = 1cm, OB = 2cm, then the perimeter of the diamond is______
Because o is the center of gravity of the diamond, the angle AOB is a right angle. So AB is equal to the root 5 and the circumference is four times the root 5
Let DP = xcm (x ＞ 0) and the area of bcdp be ycm2. (1) find the functional relation of Y with respect to X; (2) when x is the value, the perimeter of △ PBC is the smallest, and the value of Y is obtained
(1) In RT △ ABC, ac2 + BC2 = AB2, AC = AB2 − BC2 = 152 − 92 = 12. In △ ACD, ∵ Da = DC, DF ⊥ AC, CF = AF = 6, y = 12 (x + 9) × 6 = 3x + 27 (x > 0); (2) ∵ BC
1. If the perimeter of a parallelogram ABCD is 20 and the heights of adjacent sides AB and BC are 4 and 6 respectively, its area is equal to_____ .
2. If the perimeter of the diamond ABCD is 48 and the degree ratio of its two adjacent inner angles is 1:5, the area of the diamond is 2_____ .
3. In rectangle ABCD, the horizontal shadow part is rectangle, and the other shadow part is parallelogram. According to the data marked in the drawing, the area of the blank part in the drawing is calculated_____ .
Picture of question 3
A____________ B
│___ \＊\ __ │￣↑
d¯ │_____ ＊__ │b
¯│ \＊\ │↓
C ¯¯¯¯¯¯¯¯¯¯¯D￣
What is * is the shadow
Copy and paste the picture to see clearly
1. The area is 24
two point seven two
3. I think the method is to remove the shadow and make it a rectangle
RELATED INFORMATIONS
- 1. It is known that the perimeter of the parallelogram ABCD is 32, ab = 4, then BC is equal to
- 2. The parallelogram ABCD, e, f are the trisection points of BC, connecting BD, AE, AF to intersect BD with m, N, finding: Mn: Nd It is known that: the parallelogram ABCD, f is the trisection point of BC, connecting BD, AE, AF with BD in M, and finding: Mn: Nd
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- 4. As shown in the figure, it is known that in the parallelogram ABCD, e and F are respectively the midpoint of AD, G and H are two points on the diagonal BD, and BG = DH, what conclusion can be obtained?
- 5. As shown in the figure, in the parallelogram ABCD, points E and F are on CD and ab respectively, and DF / / BF is on the point
- 6. As shown in the figure, fold a square piece of paper ABCD along EF, AD / / BC. If the angle EFG = 52 °, calculate the degree of angle deg
- 7. As shown in the figure, after a rectangular piece of paper ABCD is folded along EF, points D and C respectively fall at d 'C', the intersection of ED 'and BC is g, if ∠ EFG = 65 degrees Find the degree of ∠ 1 and ∠ 2
- 8. When a rectangular piece of paper ABCD is folded along EF, the intersection points of ED and BC are g, D and C at M and N, respectively. If ∠ EFG = 55 °, then ∠ 1=______ °，∠2=______ °．
- 9. (Mianyang, 2011) as shown in the figure, fold a rectangular piece of paper ABCD with a length of 8cm and a width of 4cm so that point a and point C coincide, and the length of crease EF is equal to 2525cm
- 10. The edge length of rectangular paper ABCD is ab = 4, ad = 2. Fold the rectangular paper along EF so that point a and point C coincide. After folding, color it on one side (as shown in the figure), the area of the colored part is () A. 8B. 112C. 4D. 52
- 11. As shown in the figure, the diagonals AC and BD of rectangle ABCD intersect at O, e and F are the midpoint of OA and ob respectively. (1) prove: △ ade ≌ △ BCF; (2) if ad = 4cm, ab = 8cm, find the length of CF
- 12. If vector AB = vector DC, then ABCD is a parallelogram? Is that right? Maybe it's not collinear
- 13. It is known that in the quadrilateral ABCD, ab = ad = 4, BC = 6, CD = 2,3 vector AB * vector AD + 4 vector CB * vector CD = 0, we can find the circumcircle radius r of triangle ABC
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