Given that in a circle, the chord AB = 1 / 3 of the root sign of 2 times, and the chord center distance is 1, what is the radius of the circle?

Given that in a circle, the chord AB = 1 / 3 of the root sign of 2 times, and the chord center distance is 1, what is the radius of the circle?

In known circles, chord AB = L = 2 * (1 / 3 ^ 0.5), chord center distance h = 1, and chord center distance h = 1, what is the radius r of the circle? R ^ 2 = (R-H) ^ 2 + (L / 2) ^ 2R ^ 2 = R ^ 2-2 * h * r + H ^ 2 + L ^ 2 / 42 * h * r = H ^ 2 + L ^ 2 / 4R = H / 2 + L ^ 2 / 4 R = H / 2 + L ^ 2 / (8 * h) = 1 / 2 + (2 * (1 / 3 ^ 0.5)) ^ 2 / (8 * 1) = 1 / 2 + (4 * (1 / 3)) / 8 = 8 = 1 / 2 + (4 * (1 / 3)) / 8 = 1 / 2 + (4 * (1 / 3)) / 8 = 1 / 2 + (1(4 / 3) / 8 = 1 / 2 + 1 / 6 = 4 / 6 = 2 / 3

It is known that the radius of ⊙ o is 2cm and the length of chord AB is 2 3 cm, then the distance from the midpoint of the chord to the midpoint of the inferior arc is______ cm．

As shown in the figure,
∵AB=2
3cm，∴AC=
3cm，
In RT △ AOC, OC=
OA2−OC2=
4−3=1cm，
∴CD=2-1=1cm．
So the answer is: 1

Given the radius of circle O 4cm, chord AB = 4 times the root sign 2cm, the distance between the midpoint m of chord AB and the midpoint n of inferior arc of chord AB is () cm

2 times root number 2cm;
Distance from center to chord: √ [4 ^ 2 - (2 √ 2) ^ 2] = 2 √ 2;

(Suining, 2012) if the radius of ⊙ O1 and ⊙ O2 are 4 and 6 respectively, and the distance between centers of circle O1O2 = 8, then the position relationship between ⊙ O1 and ⊙ O2 is () A. Endonuclease B. Intersection C. Exotomy D. Exotropism

The radius of ⊙ O1 and ⊙ O2 are 4 and 6 respectively, and the distance between centers of circle O1O2 = 8,
And ∵ 6-4 = 2, 6 + 4 = 10,
∴6-4＜8＜6+4，
The position relationship between ⊙ O1 and ⊙ O2 is intersection
Therefore, B

(2011 · Tianjin) it is known that the radii of ⊙ O1 and ⊙ O2 are 3cm and 4cm respectively. If O1O2 = 7cm, the position relationship between ⊙ O1 and ⊙ O2 is () A、 Intersect B. Separation C. Endonuclease D. Exotomy

According to the radius of ⊙ O1 and ⊙ O2 are 3cm and 4cm respectively,
R + r = 7,
∵O1O2=7cm，
The relationship between ⊙ O1 and ⊙ O2 is: circumscribed
Therefore, D

As shown in the figure, if ⊙ o is the inscribed circle of the equilateral ⊙ ABC with side length of 2, then the area of ⊙ o is___ .

Let BC cut ⊙ o at point d to connect OC and OD;
∵ Ca and CB are tangent to ⊙ o,
∴∠OCD=∠OCA=30°；
In RT △ OCD, CD = 1
2BC=1，∠OCD=30°；
∴OD=CD•tan30°=
Three
3；
∴S⊙O=π（OD）2=π
3．

As shown in the figure. The high ad of △ ABC intersects with be at h, and BH = AC. it is proved that: ∠ BCH = ∠ ABC

It is proved that the high ad of ABC intersects with be at h,
∴∠ADB=∠AEB=90°，
∠DBH=90°-∠DHB，∠HAE=90°-∠AHE，
∵∠DHB=∠AHE，
∴∠DBH=∠HAE，
∵BH=AC，
∴△ADC≌△BDH，
∴AD=BD，CD=HD，
∴∠BCH=∠ABD=45°．

Circle O1 and circle O2 circumscribed circle O1 radius 3, circle O2 radius 2 and circle O1O2 tangent circle how many can be drawn?

Five

It is known that P and O2 are circles, the two points on ⊙ O1, the circle, ⊙ O1 and ⊙ O2 all pass through two points a and B, and the extension line of PA and Pb are respectively handed over to ⊙ O2 and C and D. It is shown that (1) PO2 bisection ﹤ APB, (2) AC = BD

(1) O2 is the midpoint of the arc ao2b, P is on the circle O1, PO2 bisects ∠ APB
(2) PO2 is ∠ APB bisector, the distance from O2 to PA, Pb is equal, AC = BD

O is a point on the straight line AB, the angle AOC = 1 / 3, the angle BOC, OC is the bisector of the angle AOD. Calculate the degree of angle cod This is mainly about three points and one

AOC+BOC=180°
AOC=1/3 BOC
AOC=45° BOC=135°
AOC=COD COD=45°