The radius of circle OA = 10cm, chord AB = 12cm, P is the moving point on AB, find the shortest distance from point P to center O of circle o

The radius of circle OA = 10cm, chord AB = 12cm, P is the moving point on AB, find the shortest distance from point P to center O of circle o

OP '⊥ AB in P'
Because the vertical line is the shortest, the distance from P to center O is the shortest when p is at p '
Vertical diameter theorem AP '= AB / 2 = 6cm
Pythagorean theorem OP '= √ (OA ^ 2-AP' ^ 2) = 8cm
The shortest distance is 8 cm

Circle Q1 is a circle with radius OA of circle O as its diameter, and intersects with chord ab of circle O at point C. given AB = 10cm, find the length of AC

AC=(1/2)AB,∴AC=5（cm）
The reasons are as follows.
Let ⊙ o radius OA = ob = R,
And Ao is the diameter of the circle ⊙ 1,
⊥ OC = 90 ° or OC ⊥ ab,
∴△BOC≌△AOC（H,L）
∴AC=BC=5.
The proof is over

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

As shown in the figure, the diameter AB in circle O intersects with chord CD at point E. given AE = 1cm, be = 5cm, angle DEB = 60 °, find out the length of chord CD

Because AE = 1cm, be = 5cm
So OE = 2cm
Oh through O is perpendicular to CD and H
So ch = DH
Because ∠ DEB = 60 °, EHO = 90 °
So eh = 1, oh = radical 3
Linking Co
Because CO is the radius of circle o
So co = 3cm
Because Oh = root 3cm
So ch = radical 6cm (Pythagorean theorem)
So CD = 2CD = 2 root sign 6
Just made the Ang, hit the music

As shown in the figure, in the parallelogram ABCD, ab = 5cm, ad = 3cm, point E is on the extension line of AB, be: AE = 2:7 Join de and intersect BC at point F Find the value of (1) DF / EF and (2) the length of FC

(1) ∵ in the parallelogram ABCD
BC//AD
∴BF//AD
∴EB：EA=EF：ED=2：7
∴DF/EF=（ED-EF）/EF=5/2
（2）∵BC//AD
∵BF：AD=BE：EA=2：7
∵AD=3
∴BF=6/7
∴FC=BC-BF=3-6/7=15/7

As shown in the figure, the edge ab of rectangular ABCD passes through the center of ⊙ o, e and F are the intersection points of AB, CD and ⊙ o respectively. If AE = 3cm, ad = 4cm, DF = 5cm, then the diameter of ⊙ o is equal to______ .

Connect of, as FG ⊥ AB at point G
Then eg = df-ae = 5-3 = 2cm
Let the radius of ⊙ o be r,
Then of = R, og = R-2
In the right angle △ OFG, of2 = fg2 + og2,
R2 = (R-2) 2 + 42,
The solution is: r = 5
The diameter is 10 cm
So the answer is: 10

If the radius of ⊙ o is 5cm, the chord ab ∥ CD, ab = 6cm, CD = 8cm, then the distance between AB and CD is () A. 1 cm B. 7 cm C. 1 cm or 7 cm D. It's impossible to judge

It can be divided into two cases: ① when AB and CD are on the same side of O, as shown in Fig. 1, OE ⊥ AB in E, Cd in F, OA, OC, ∵ ab ∥ CD, ∵ of ⊥ CD, ⊥ from the vertical diameter theorem, AE = 12ab = 3cm, CF = 12CD = 4cm, in RT △ OAE, from the Pythagorean theorem, OE = oa2 − AE2 = 52 − 32 = 4 (CM) is obtained

In ⊙ o, chord AB = 24cm, chord CD = 10cm. If the distance between center O and ab is 5cm, then the distance from point O to string CD is______ cm．

Through O, make the vertical lines of AB and CD respectively, and the vertical feet are e, F, connected with OA and OC, as shown in the figure,
∴AE=BE，CF=DF，
AB = 24cm, CD = 10cm,
∴AE=12cm，CF=5cm，
And OE = 5,
In RT △ AOE, OA=
AE2+OE2=13（cm）；
In RT △ OCF, of=
OC2−CF2=12（cm）；
The distance from point O to CD is 12cm
So the answer is: 12

If the radius of ⊙ o is 5cm, the chord ab ∥ CD, ab = 6cm, CD = 8cm, then the distance between AB and CD is () A. 1 cm B. 7 cm C. 1 cm or 7 cm D. It's impossible to judge

There are two cases: ① when AB and CD are on the same side of O, as shown in Figure 1,
O as OE ⊥ AB in E, Cd in F, OA, OC,
∵AB∥CD，
∴OF⊥CD，
According to the vertical diameter theorem, AE = 1
2AB=3cm，CF=1
2CD=4cm，
In RT △ OAE, from Pythagorean theorem, OE is obtained=
OA2−AE2=
52−32=4（cm）
Similarly, of = 3cm is obtained,
EF=4cm-3cm=1cm；
II.
When AB and CD are on both sides of O, as shown in Fig. 2, OE = 4cm, of = 3cm can be obtained by the same method,
EF = 3cm + 4cm;
The distance between AB and CD is 1cm or 7cm,
Therefore, C

It is known that AB and CD are two parallel chords of circle O. the radius of circle O is 5cm, ab = 8cm, CD = 6cm. Find the distance between AB and CD

7 cm or 1 cm