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As shown in the figure, AB is the diameter of semicircle o, and point C is a point on the semicircle. Through point C, make CD perpendicular to D, AC = 2 times, root sign 3cm, ad: DB = 3:1, and find the length of AD As shown in the figure, AB is the diameter of semicircle o, and point C is a point on the semicircle. Through point C, make CD perpendicular to D, AC = 2 times, root sign 3cm, ad: DB = 3:1, and find the length of AD
Let DB = a, then ad = 3A, then AB = 4A
Because the triangle ACD is similar to the triangle BCD, and CD / ad = BD / CD, that is, CD? 2 = ad * BD = 3A
CD? 2 + ad? 2 = AC? I.e. 3A? 2 + 9A? 2 = 12
Then a = 1, that is, ad = 3A = 3
AB is the diameter of semicircle o, C is any point on the semicircle. The crossing point C is CD, perpendicular to AB, and the perpendicular foot is D, ad = a, DB = B. according to the graph, a + B = 2 root sign AB is verified, The condition of equal sign
Let the diameter be d
d=a+b
b=d-a
A + B = 2 roots ab
D = 2 root sign a (D-A)
Square on both sides
d*d=4a(d-a)
4a*a-4ad+d*d=0
(2a-d) square = 0
So it holds when a = half the diameter
It is known that, as shown in the figure, AB is the diameter of semicircle o, and point C is a point on the semicircle. Through the point C, CD is perpendicular to point D, AC = 2, radical 13, ad: DB = 9:4, and find the length of AD
AB is the diameter, then ∠ ACB = 90 degrees
CD is perpendicular to AB, so ∠ BDA = 90 degrees
In the right triangle ABC
CD²=AD×BD
Because ad: DB = 9:4
So let ad = 9a, DB = 4A
that
CD²=36a²
CD=6a
In right triangle ACD
AC²=AD²+CD²
4×13=81a²+36a²
117a²=52
a²=4/9
a=2/3
AD=9a=2/3×9=6
AB is the diameter of circle O, CD is perpendicular to AB, ad = 9 cm, DB = 4 cm. Find the length of CD and AC
Connecting AC, BC
∵ AB is the diameter
∴∠ACB=90°
∵CD⊥AB
∴CD²=AD*BD=9*4=36
∴CD=6
In △ ACD, ad = 9, CD = 6
According to Pythagorean theorem, AC = 3, radical 13 can be obtained
As shown in the figure, OA and ob are the two radii of circle O. points D and C are on OA and ob respectively, AC and BD intersect with point E and ad = BC Proof angle a = angle B
Is this what the original picture looks like?
Because ad = BC
Because OA = ob
So OA - ad = ob - BC
So od = OC
Because od = OC, < DOB = < COA, OB = OA
So triangle AOC is all equal to triangle BOD
So < a = < B
(< this symbol for the angle)
Known: as shown in the figure, OA, OB are the radius of ⊙ o, C and D are the midpoint of OA and ob respectively. Verification: ad = BC
It is proved that: ∵ OA, OB are the radius of ⊙ o, C and D are the midpoint of OA and ob respectively,
∴OA=OB，OC=OD．
In △ AOD and △ BOC,
A kind of
OA＝OB
∠O＝∠O
OD＝OC ，
∴△AOD≌△BOC（SAS）．
∴AD=BC．
As shown in the figure, AC＝ CB, D and E are the midpoint of radius OA and ob respectively. What is the relationship between Cd and CE? Why?
CD=CE．
The reason is: connect OC,
∵ D and E are the midpoint of OA and ob respectively,
∴OD=OE，
And ∵
AC＝
CB，∴∠DOC=∠EOC，
OC=OC，∴△CDO≌△CEO，
∴CD=CE．
It is known that ab crosses ⊙ o in C and D, and AC = BD. please prove that OA = ob
prove:
O is used as OE ⊥ AB to E,
∵ OE is over the center o,
∴CE=DE，
∵AC=BD，
∴AE=BE，
∵OE⊥AB，
∴OA=OB．
As shown in the figure, the radius of ⊙ o is 3cm, point B is a point outside ⊙ o, OB intersects ⊙ o at point a, and ab = OA, starting from point a, the moving point P moves anticlockwise on ⊙ o at the speed of π cm / s, returns to point a and stops immediately A. 1 B. 5 C. 0.5 or 5.5 D. 1 or 5
Connect OP,
∵ the line BP is tangent to ⊙ o,
∴OPB=90°，
∵AB=OA=OP，
∴OB=2OP，
∴∠PBO=30°，
∴POB=60°，
The length of the arc AP is 60 π· 3
180=π，
That is, the time is π △ π = 1 (second);
When point P ', the line BP is tangent to ⊙ o,
The length of the arc app 'is (360 − 60) π· 3
180=5π，
That is, the time is 5 π - π = 5 (seconds);
Therefore, D
As shown in the figure, AB is the tangent line of ⊙ o, the radius OA = 2, OB intersects ⊙ o at C, ∠ B = 30 °, then the inferior arc What is the length of AC______ (results retain π)
∵ AB is the tangent of ⊙ o,
∴∠OAB=90°，
∵ radius OA = 2, OB intersection ⊙ o at C, ∵ B = 30 °,
∴∠AOB=60°，
The inferior arc
It's 2 π 60 long
180=2
3π，
So the answer is: 2
3π．
RELATED INFORMATIONS
- 1. As shown in the figure, the radius of ⊙ o is 3cm, point B is a point outside ⊙ o, OB intersects ⊙ o at point a, and ab = OA, starting from point a, the moving point P moves anticlockwise on ⊙ o at the speed of π cm / s, returns to point a and stops immediately A. 1 B. 5 C. 0.5 or 5.5 D. 1 or 5
- 2. It is known that: as shown in the figure, take the oblique edge ab of RT △ ABC as the diameter to make ⊙ o, and D is the point on ⊙ o, and there is AC = CD. Pass through point C to make the tangent of ⊙ o, and connect CD with the extension line of BD at point E (1) Try to judge whether be and CE are perpendicular to each other, please explain the reason; (2) If CD = 2 5，tan∠DCE=1 2. Find the radius length of ⊙ o
- 3. Given that AB and CD are two diameters of circle O, chord CE / / AB, angle COE = 50 degrees, then the acute angle DOB =?
- 4. It is known that, as shown in the following four graphs, circle O1 and circle O2 intersect at two points a and B, the line CD passing through point a intersects with circle O1 at C, intersects with circle O2 at D, and straight line ef? 2 passing through point B As shown in the following figure, circle O1 and circle O2 intersect at two points a and B, the line CD passing through point a intersects with circle O1 at C, and with circle O2 at D. the line EF passing through point B intersects with circle O1 at e and circle 02 at F, so as to verify CE ∥ DF Picture in space photo album
- 5. As shown in the figure, the diameter of ⊙ o is ab = 10, C and D are two points on the circle, and AC＝ CD＝ Let the tangent ed of point d cross the extension line of AC at point F. connect OC to ad at point G (1) Confirmation: DF ⊥ AF (2) Seek the length of OG
- 6. As shown in the figure, it is known that AB is the chord of ⊙ o, the radius OA = 20cm, ∠ AOB = 120 ° and the area of ⊙ AOB is obtained
- 7. In the RT triangle ABC, the angle ABC = 90 degrees, the circle O with the diameter of AB intersects AC with D, and the tangent passing through D intersects BC with e. it is proved that de = 1 / 2 BC, If Tan angle c = root 5 divided by 2, de = 2, find ad
- 8. As shown in the figure, the straight line DF intersects with AB and AC on both sides of △ ABC at two points D and e respectively, and intersects with the extension line of BC at point F, ∠ B = 50 °, 1 = 76 °, f = 3
- 9. As shown in the figure, in △ ABC, D is a point on AC, e is a point on CB extension line, and AC BC＝EF DF， Confirmation: ad = EB
- 10. As shown in the figure, it is known that AB is the diameter of ⊙ o, AC is the chord, and bisection ⊥ bad, ad ⊥ CD, and the perpendicular foot is d Proof: CD is tangent of ⊙ o
- 11. As shown in the figure, ad is the diameter of circumscribed circle of △ ABC, CE ⊥ ad intersects ad with F and ab with E. try to explain that AC square = ab × AE
- 12. As shown in the figure of triangle ABC, ad is the center line on BC side, f is a point on ad, and AF: FD = 1:3, extend BF, cross AC to e, and find AE: EC
- 13. It is known that AB is the diameter of circle O, PD tangent circle O to C, and the extension line of Ba intersects PC at P Angle P = 26 degrees, calculate angle BCD
- 14. As shown in the figure, it is known that ⊙ o is the circumscribed circle of ⊙ ABC, CE is the diameter of ⊥ o, CD ⊥ AB and D are perpendicular feet. It is proved that ⊙ ACD = ∠ BCE
- 15. As shown in the figure, the triangle ABC is inscribed with the circle O, ad bisects ∠ BAC, intersects ⊙ O and D, passes through D as de ∥ BC, and intersects the extension line of AC with E The triangle ABC is inscribed with the circle O, ad bisects ⊙ BAC, intersects ⊙ O and D, passes D as de ∥ BC, and intersects the extension line of AC with E Try to judge the position relationship between de and circle O and prove your conclusion 2. If ∠ e = 60 degrees, the radius of circle O is 4, and find the length of ab
- 16. O is the diameter of a perpendicular point on each other (1) Verification: AC bisection ∠ DAB; (2) If CD = 4, ad = 8, try to find the radius of ⊙ o
- 17. As shown in the figure, the vertical bisector of chord AB intersects at point C and intersects chord AB at point D. It is known that ab = 24cm, CD = 8cm (1) Find the circle where the fragment is located (keep the drawing trace and write the method); (2) Find the radius of the circle in (1)
- 18. As shown in Figure AC is the diameter of circle O, PA is perpendicular to AC, connecting OP, chord CB / / op, diameter BC intersects straight line AC at D, BD = 2PA, prove that BP is tangent of circle O, OP is related to BC Verify the value of sin angle OPA I 1 and can not insert the picture The diameter BC in the problem is changed to the straight line Pb
- 19. As shown in the figure, the radius of ⊙ o is 1cm, and the lengths of strings AB and CD are respectively 2 cm, 1 cm, then the acute angle α between the strings AC and BD=______ Degree
- 20. It is known that the circle centered on point P passes through points a (- 1,0) and B (3,4), and the vertical bisector of line AB intersects the circle P at points c and D, and | CD | = 4 10． (1) Find the equation of straight line CD; (2) Find the equation of circle P; (3) Set the point Q on the circle P. how many points Q have the area of △ qAB equal to 8? Prove your conclusion