1. Given that the circumference of the diamond is 16 cm and the length of a diagonal is 4 cm, the four corners of the diamond are () respectively.
a . 30 150 30 150 b . 45 135 45 135
C.60, 120,60, 120 D none of the above are correct.
2. As shown in the figure, in diamond-shaped ABCD, M and N are on AB and CD respectively, AM = CN, and MN and AC intersect at point O to connect BO. If ∠ DAC = 28, the degree of ∠OBC is ().
A.28 B.52 C.62 D.72
3. As shown in the figure, in the diamond-shaped ABCD, point E is a point on AB, point O connects DE and AC, connects BO, and ∠ AED = 50, then ∠ CBO = _ _ _ degrees.
4. As shown in the figure, in the diamond ABCD, ∠ ABC = 120, diagonal AC and BD intersect at point O, AE bisects ∠CAD, OD and CD intersect at points F and E respectively, and the degree of ∠AFO is found.
5. As shown in the figure, in the rhombic ABCD, AB = 13cm, the height of BC = 5cm, and the length of diagonal AC is ____cm.
6. As shown in the figure, the quadrilateral ABCD is a diamond, AC = 8, DB = 6, and DH⊥AB is at point H, so the length of DH is ().
A.245 B. 125 C.5 D.4
7. As shown in the figure, in the diamond-shaped ABCD, diagonal AC = 6 and BD = 10, then the area of the diamond-shaped ABCD is _ _ _ _.
8. As shown in the figure, the quadrilateral ABCD is a diamond, and O is the intersection of two diagonals. The diamond is divided into a shadow part and a blank part through three straight lines at the O point. When the lengths of the two diagonal lines of the diamond are 10 and 4 respectively, the area of the shadow part is _ _ _ _.
9. As shown in the figure, O is the intersection of AC and BD, CD=5 cm, OD=3 cm, intersection C is CE∨DB, intersection B is BE∨AC, and CE and BE intersect at point E. 。
(1) Find the length of oc;
(2) Find the area of quadrilateral OBEC.
10. As shown in the figure, in the diamond ABCD, ∠ bad = 44, the diagonal AC of the median vertical line of AB is at point F, the vertical foot is E, and the connecting line DF is equal to ∠CDF ().
a . 1 12 b . 1 14 c . 1 16d . 1 18
1 1. In the rhombic ABCD, ∠ A = 30, and in the same plane, for an isosceles triangle BDE with diagonal BD as the base and vertex angle of120, the degree of ∠EBC is.
12. As shown in the figure, the quadrilateral ABCD is a diamond, the extension line from CE⊥AB to AB is at point E, and the extension line from CF⊥AD to AD is at point F, which proves that DF = BE.
13. As shown in the figure, in rhombic ABCD, AB = 4, E is the midpoint of BC, AE⊥BC, AF⊥CD is at point F, cg∨AE, CG and AF intersect at point H, and AD intersects at point G. 。
(1) Find the area of rhombic ABCD;
(2) Find the degree of ∠CHA.
14. As shown in the figure, in the rhombic ABCD, f is any point on BC, connecting the diagonal BD of AF to point E and connecting EC.
(1) verification: AE = EC
(2) When ∠ ABC = 60 and ∠ CEF = 60, where is the point F on the BC line? Please explain the reason.
15. As shown in the figure, rotate two rectangular paper strips with a length of 4 and a width of 1 crosswise, so that the overlapping part becomes a diamond. In the process of rotation, when two pieces of paper are vertical, the minimum value of diamond perimeter is 4, so the value of diamond perimeter is _ _ _ _.
16. As shown in figure 1, in the rhombic ABCD, point E and point F are the midpoint of AB and AD respectively, connecting CE and CF. 。
(1) verification: CE = CF
(2) As shown in Figure 2, if H is a point above AB, connect CH to make ∠ CHB = 2 ∠ ECB, and verify: CH = AH+AB.
Reference answer
1.C
2.C
3.50
4.∫ In the diamond ABCD, ∠ ABC = 120, ∴∠ Bad = 60, ∫ diagonal AC, BD intersect at point O, ∴∠ BAC = ∠CAD = 30.
5.26
6.A
7.30
8. 10
9.( 1)∵ quadrilateral ABCD is a diamond, and ∴AC⊥BD, ∴,
OC=CD2-OD2=52-32=4 (cm)
(2)∫ce∨db, be∨AC, ∴ quadrilateral OBEC is a parallelogram.
∵AC⊥BD, that is ∠ COB = 90, ∴ parallelogram OBEC is rectangular,
Ob = od, ∴S quadrilateral OBEC = OB? OC=4×3= 12(cm2)
10.B
1 1.45 or 105.
12. Connect AC, ∫ quadrilateral ABCD is a diamond, ∴AC bisects ∠DAB, CD = BC, ∵CE⊥AB, CF⊥AD, ∴ CE = CF, ∠CFD =∞.
13.( 1) Connect AC and BD, AC and BD intersect at point O, ∵AE⊥BC.
And AE divides BC equally, ∴ AB = AC = BC, ∴ BE = 12bc = 2,
∴AE=42-22=23,S=BC？ AE=4×23=83，
The area of diamond ABCD is 83.
(2) ∵ AC = AB = AD = CD, △ADC is an equilateral triangle, ∵AF⊥CD.
∴∠ DAF = 30, while ∵CG∨AE, AE⊥BC,
∴ Quadrilateral AECG is a rectangle, ∴∠ AGH = 90,
∴∠AHC=∠DAF+∠AGH= 120
14.( 1) connects AC, ∫BD is also the diagonal of diamond ABCD, ∴BD divides AC vertically.
∴AE=EC
(2) Point F is the midpoint of line BC. Reason: in diamond ABCD, AB = BC,
∫∠ABC = 60, ∴△ABC is an equilateral triangle, ∴∠ BAC = 60,
∵AE=EC,∴∠EAC=∠ACE,∵∠CEF=60，
∴∠eac= 12∠cef=30 ,∴∠eac= 12∠bac，
∴AF is the bisector of △ABC, and ∵AF and BC intersect at point F,
∴AF is the center line on the BC side of △ABC, and ∴ point F is the midpoint of line segment BC.
15. 172
16.( 1) Yi Zheng△ BCE△ DCF (SAS), ∴ CE = CF
(2) Extension BA and CF intersect at point G, ∫ quadrilateral ag∨CD is a diamond, ∴∠ B = ∠ D, AB = BC = CD = AD, af∨BC, ab∨CD, ∴∠. ∵∠chb=2∠ecb,∴∠chb=2∠g,∵∠chb=∠g+∠hcg,∴∠g=∠hcg,∴gh=ch,∴ch=ah+ag=ah+ab