It is known that in circle O, the length of chord AB is three times the root of radius OA, and point C is the midpoint of arc ab. what figure is the quadrilateral oacb and why?

It's a quadrilateral, OA = ob, AC = BC, neither parallelogram nor rectangle
It is proved that if OC and line AB intersect with point E, OE is perpendicular to AE, and AE = be = AB / 2 = 3 times OA of 2 / 2
So the angle AOE = 30 degrees, so the angle AOB = 2 * 30 = 60 degrees
Because OA = OC, the angle OAC = (180-30) / 2 = 75 degrees
Because arc AC = arc BC, the line AC = BC

It is known that in circle O, the length of chord AB is three times that of the root of radius OA, and the point C goes to the midpoint of arc ab. what figure is the quadrilateral oacb and why?

Let AB and co intersect with D, because C bisects the arc AB, CO is perpendicular to AB, Da = AB / 2 = radical 3 / 2 * OA, do = 1 / 2 * OA, DC = OA = od = 1 / 2 * OA, that is, DC = do, so acbo is a diamond (the quadrilateral whose diagonals are perpendicular to each other is a diamond)

In center O, the diameter is 30cm, the chord AB / / CD, and ab = 18cm, CD = 24cm

We can draw a graph on the basis of AB / / CD, draw a diameter parallel to AB and CD, then connect AO and Co, and then make a vertical line of AB through point O (obviously this vertical line is also perpendicular to CD), intersect AB and e respectively, CD at F. the vertical line thus made also bisects AB and CD (why I don't prove it here) after three

A. B and C are three points on the sphere. Given the chords (the line connecting the two points on the sphere) AB = 18cm, BC = 24cm, AC = 30cm, the distance between plane ABC and the center of the ball is exactly half of the radius of the ball. The surface area and volume of the ball are calculated

There are three points a, B and C on the sphere, plane ABC and the sphere intersect in a circle, and three points a, B and C are on this circle
∵AB=18，BC=24，AC=30，
 ac2 = AB2 + BC2,  AC is the diameter of the circle, AC is the center of O ′ circle
The distance from the center of the sphere o to the plane ABC is OO '= half of the radius of the ball = 1
2R
In △ OO ′ a, ∠ OO ′ a = 90 ° and OO ′ = 1
2R，AO′=1
2AC=30×1
2=15，OA=R
According to Pythagorean theorem (1
2R）2+152=R2，3
4R2=225
R = 10
3．
The surface area of the sphere s = 4 π R2 = 1200 π (cm2);
And volume v = 4
3πR3＝4
3×π× (10
3)3=4000
3π（cm3）．

A. B and C are three points on the sphere. Given the chords (the line connecting the two points on the sphere) AB = 18cm, BC = 24cm, AC = 30cm, the distance between plane ABC and the center of the ball is exactly half of the radius of the ball. The surface area and volume of the ball are calculated

Three points a, B, C on the sphere, plane ABC and the sphere intersect in a circle, three points a, B, C on the circle

The radius of the circle is 12cm and the chord AB is 16cm. If the two ends of chord AB slide on the circumference, what kind of figure does the midpoint of chord AB form?

Because the vertical diameter theorem makes OD ⊥ AB through the center of a circle O, it is called AB and E. in △ AEO, Ao is radius 12, AE is 16 / 2 = 8. In Pythagorean theorem, OE is 4 times the root sign five. Then you take a compass and slide it over the length of AB to form a regular pentagon~

If the diameter of O is 20cm, the chord ab ∥ C, ab = 12cm and CD = 16cm, then the distance between chord AB and CD is

Using Pythagorean theorem to calculate the height of the base edge in the isosceles triangle AOB: √ (10? - (12 / 2) 2) = 8
Using Pythagorean theorem to calculate the height of the base edge in the isosceles triangle cod: √ (10? - (16 / 2) 2) = 6
In both cases, the distance between AB and CD is 8-6 = 2 when AB and CD are on the same side of o point, and 8 + 6 = 14 when AB and CD are on both sides of o point

If the radius of ⊙ o is 10cm, the chord ab ∥ CD, ab = 12cm, CD = 16cm, then the distance between AB and CD is______ .

(1) In RT △ OAE, OA = 10cm, AE = 6cm;
According to Pythagorean theorem, OE = 8cm;
In the same way, of = 6cm;
Therefore, EF = oe-of = 2cm;
(2) As shown in Fig. 2; the same as (1), OE = 8cm, of = 6cm;
Therefore, EF = OE + of = 14cm;
So the distance between AB and CD is 14 cm or 2 cm

Given that AB is the diameter of circle O, the chord CD is perpendicular to AB, CD = 4, root sign 3. If AP: Pb = 1:3, find the brightness of chord CD Degree of arc

AP: Pb = 1:3, (AO PO): (AO + P) = 1:3 to get AP = Po, so obviously Po = AP = 1 / 2ao = 1 / 2CO
According to the trigonometric sine function, the angle cod is only half when it is 120 degrees, which is the degree of the arc corresponding to CD,
Basic mathematical problems, more summary methods, basically not too much change, come on

As shown in the figure, the radius of circle O is OA = 5cm, chord AB = 8cm, and point P is the moving point on chord AB, then the shortest distance between point P and center O is______ cm．

When op ⊥ AB, OP is the shortest,
∴AP=1
2AB=1
2×8=4（cm），
∴OP=
OA2−AP2=
52−42=3（cm）．
The shortest distance from point P to center O is 3cm
So the answer is: 3

It is known that in circle O, the length of chord AB is three times the root of radius OA, and point C is the midpoint of arc ab. what figure is the quadrilateral oacb and why?