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As shown in the figure, AB is a chord of ⊙ o, OD ⊥ AB, perpendicular foot is C, intersect ⊙ o at point D and point E on ⊙ o (1) If ∠ AOD = 52 °, calculate the degree of ∠ DEB; (2) If OC = 3, OA = 5, find the length of ab
(1) ∵ AB is a string of ⊙ o, OD ⊥ ab,
Qi
AD=
DB，∴∠DEB=1
2∠AOD=1
2×52°=26°；
(2) ∵ AB is a string of ⊙ o, OD ⊥ ab,
/ / AC = BC, i.e. AB = 2Ac,
In RT △ AOC, AC=
OA2−OC2=
52−32=4，
Then AB = 2Ac = 8
AB is the diameter of circle O, BC is the chord, OD ⊥ CB is at point E, and intersection bcfu is at point D If BC = 8, ed = 2, find AC length!
∵ OE ⊥ BC ᙽ e is the midpoint of BC ᙽ be = CE = 4
Let the radius be r, then od = R OE = od-ed = R-2
In triangle OBE
There is ob 2 = be 2 + OE 2
That is, R 2 = 4 2 + (R-2) 2
The solution is r = 5
∴AB=10
∵ AB is the diameter ᙽ ACB = 90 ° C
In the right triangle ABC
AB? 2 = BC? 2 + AC? 10? 2 = 8? 2 + AC? 2, AC = 4
As shown in the figure, if the diameter of ⊙ o is CD = 10cm, AB is the chord of ⊙ o, ab ⊥ CD, the perpendicular foot is m, OM: OC = 3:5, then AB is=______ cm．
∵ circle O diameter CD = 10 cm,
The radius of circle O is 5cm, that is, OC = 5cm,
∵OM：OC=3：5，
∴OM=3
5OC=3cm，
Connect OA,
∵AB⊥CD，
/ / M is the midpoint of AB, i.e. am = BM = 1
2AB，
In RT △ AOM, OA = 5cm, OM = 3cm,
According to Pythagorean theorem: am=
OA2−OM2=4cm，
Then AB = 2am = 8cm
So the answer is: 8
As shown in the figure, if the diameter of ⊙ o is CD = 10cm, AB is the chord of ⊙ o, ab ⊥ CD, the perpendicular foot is m, OM: OC = 3:5, then AB is=______ cm．
According to the Pythagorean theorem, we get that: am = oa2 − om2 = 4cm, then AB = 2am = 8cm
In the circle O, CD is the diameter and ab is the chord AB perpendicular to m CD = 10. If om: OC = 3:5, what is the length of chord AC?
If AB vertical CD is on OC side, AC length is 2 √ 5; if AB vertical CD is on OD side, AC length is 4 √ 5
As shown in the figure, if the diameter of ⊙ o is CD = 10cm, AB is the chord of ⊙ o, ab ⊥ CD, the perpendicular foot is m, OM: OC = 3:5, then AB is=______ cm．
According to the Pythagorean theorem, we get that: am = oa2 − om2 = 4cm, then AB = 2am = 8cm
As shown in the figure, AB is the chord of ⊙ o, and the radius OC and OD intersect AB at points E and f respectively, and AE = BF. Please find out the quantitative relationship between line OE and of and give proof
OE=OF，
Proof: connect OA, ob,
∵OA=OB，
Ψ OAB = ∠ oba, that is ∠ OAE = ∠ oBf
In △ OAE and △ oBf,
OA=OB
∠OAE=∠OBF
AE=BF ，
∴△OAE≌△OBF（SAS）．
∴OE=OF．
As shown in the figure, the diameter ab of the circle O and the chord CD intersect with E. we know that AE = 6cm, EB = 2cm, and CEA = 30 ° to find CD
In RT △ OEB, ∵ CEA = 1cm, OC = OA = 4cm, ∵ CF = within the root (OC square of square) = 15 cm
If the diameter ab of O intersects with the chord CD at point E, given AE = 6cm, EB = 2cm, ∠ CEA = 30 °, then the length of chord CD is () A. 8cm B. 4cm C. 2 Fifteen D. 2 Seventeen
O is used as om ⊥ CD, and OC is connected,
∵AE=6cm，EB=2cm，
∴AB=8cm，
∴OC=OB=4cm，
∴OE=4-2=2（cm），
∵∠CEA=30°，
∴OM=1
2OE=1
2×2=1（cm），
∴CM=
OC2−OM2=
42−12=
15，
∴CD=2CM=2
15．
Therefore, C
If the diameter ab of O intersects with the chord CD at point E, given AE = 6cm, EB = 2cm, ∠ CEA = 30 °, then the length of chord CD is () A. 8cm B. 4cm C. 2 Fifteen D. 2 Seventeen
O is used as om ⊥ CD, and OC is connected,
∵AE=6cm，EB=2cm，
∴AB=8cm，
∴OC=OB=4cm，
∴OE=4-2=2（cm），
∵∠CEA=30°，
∴OM=1
2OE=1
2×2=1（cm），
∴CM=
OC2−OM2=
42−12=
15，
∴CD=2CM=2
15．
Therefore, C
RELATED INFORMATIONS
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