It is known that the radii of circle O1 and circle O2 are 3cm and 5cm respectively, and if they are inscribed, the position relationship between the center of circle and O1O2? What is the positional relation? Which is the extrinsic intersection and tangent? It is known that the radius of circle O1 and circle O2 are 3cm and 5cm respectively, and if they are inscribed, the center distance of circle is 5cm!

Two circles intersect when the center distance is less than radius and greater than radius difference
The distance between the centers of a circle is greater than the radius and the distance between two circles
The distance between centers is less than the difference of radius
The distance between the centers of a circle is equal to the radius and the two circumscribed circles
The distance between centers is equal to the difference of radius

As shown in the figure, if the chord AB = AC = 5cm in circle O, then the radius of circle O is equal to?

There is no picture
But it should be connected Ao angle circle O to D, then called abd and angle ACD are right angles, and then use the correlation of right angles to find out the answer
If it is a little more complicated, then we have to use ad perpendicular to BC. At this time, it will be more convenient to set ad to BC and e to assist calculation
complete

In the circle O, the radius is 5cm, the chord AB is parallel to CD, ab = 6cm, CD = 8cm

Make ef through o point, AB perpendicular to e, cross CD to F, connect OA and OC
Because AB is parallel to CD
So EF vertical CD
In RT △ AOE
OA = 5 AE = 1 / 2 ab = 3
So OE = root (OA square - AE Square) = 4
In RT △ OCF, OC = 5, CF = 1 / 2, CD = 4
So of = root (OC square - CF Square) = 3
So EF = OE + of = 4 + 3 = 7

Given that ⊙ O1 and ⊙ O2 are circumscribed with radii of 1cm and 3cm, then circles with radius of 5cm and tangent to ⊙ O1 and ⊙ O2 can be made______ One

The radius of ⊙ O1 and ⊙ O2 are 1cm and 3cm respectively. The distance between the centers of two circles is 4cm. There are two circles with radius of 5cm;
And a circle circumscribed and a circle inscribed have two;
Two are inscribed with two circles;
If two circles are tangent to each other, we can see that there are six

(2002, Guangzhou) if the radii of ⊙ O1 and ⊙ O2 are 1 and 3 respectively, and ⊙ O1 and ⊙ O2 are circumscribed, then the circle with radius 4 on the plane and tangent to ⊙ O1 and ⊙ O2 has () A. 2 B. Three C. Four D. Five

The radius of ⊙ O 2 is 1 and ⊙ 2 respectively,
1+3=4，
There are 5 circles tangent to ⊙ O1 and ⊙ O2;
There are two circles which are tangent to the two circles, one of which is circumscribed and the other is inscribed; the other is that the two circles are tangent to the two circles, each of which is circumscribed
Therefore, D

As shown in the figure, ⊙ o is the inscribed circle of the equilateral triangle ABC with side length of 2, then the area of the shadow part in the figure is______ .

Connect OA, OD (internal tangent point on AB)
Since the inner part of an equilateral triangle is its outer center, we can get ad = 1
2AB=1，∠OAB=1
2∠CAB=30°；
In RT △ oad, Tan 30 ° is OD
Ad, i.e
Three
3=OD
1, get 0d=
Three
3．
The area of the shaded part in the graph is equal to s △ abc-s ⊙ o=
Three
4×22-π（
Three
3）2=
3−1
3π．

As shown in the figure, find the area ratio of inscribed circle and circumscribed circle of regular triangle ABC

Area ratio of inscribed circle to circumscribed circle of regular triangle ABC = square of radius ratio
Two radii in the same right triangle, and one angle is 30 degrees, 1 / 2
So the area ratio of inscribed circle and circumscribed circle of regular triangle ABC is 1 / 4

If circle 0 is the inscribed circle of the equilateral triangle ABC with side length 2, the area of the shadow part in the figure is The shadow part is the area of the triangle divided by the inscribed circle

Triangle area circle area = (1 / 2) × 2 × 2 ×√ 3 / 2-3.14 × 2 √ 3 / 3 × 2 √ 3 / 3 = √ 3 / 10

In triangle ABC, inscribed circle I and edges BC, CA, AB are respectively tangent to points D, e, F. find out the relationship between angle FDE and angle a, and explain the reason!

The inscribed circle and the edges BC, CA, AB are tangent to points D, e, F, connecting OE and of, (o is the center of the circle)
Then ∠ AFO = ∠ AEO = 90 °
Because ∠ foe + ∠ a + ∠ AFO + ∠ AEO = 360 °
Because the center angle is twice the circumference angle, we can know that ∠ foe = 2 ∠ FDE
Therefore, 2 ∠ FDE + ∠ a + ∠ AFO + ∠ AEO = 360 °
And ∠ AFO = ∠ AEO = 90 °
Therefore, 2 ∠ FDE + ∠ a = 180 °, that is to say ∠ FDE and ∠ a are complementary

As shown in the figure, ⊙ I is the inscribed circle of ⊙ ABC, which is tangent to points D, e and F with AB, BC and Ca respectively, and ﹤ def = 50 °, then the degree of ∠ A is______ .

Connect Di, fi,
∵∠DEF=50°，
∴∠DIF=2∠DEF=100°，
⊙ I is the inscribed circle of ⊙ ABC,
∴∠ADI=∠AFI=90°，
∴∠A=360°-∠ADI-∠AFI-∠DIF=80°．
So the answer is: 80 degrees