As shown in Figure AB is the diameter of circle O, CD is the two points on circle O and C is the midpoint of ad arc. If ∠ bad = 20 °, calculate the degree of ∠ ACO

As shown in Figure AB is the diameter of circle O, CD is the two points on circle O and C is the midpoint of ad arc. If ∠ bad = 20 °, calculate the degree of ∠ ACO

"Lingya"
Ad arc = 180 ° - 20 ° = 160 °
Center angle AOC = 160 ° / 2 △ 2 = 40 °
AO＝CO.
∠ACO＝∠CAO＝（180°-40）÷2＝70°
Good bye
∠ACD＝180°-20°＝160°
∠CAD＝CDA＝（180°-20°）÷2＝80°
AOC = 10 °× 2 = 20 °
∠ACO＝（180°-20°）÷2＝

As shown in the figure, AB is the diameter of circle O, chord CD is perpendicular to AB, and point E is a point on arc ad. if angle BCD = 35 degrees, calculate the degree of angle AED

125 degrees
Do you want the detailed process, and it seems that I haven't seen your drawing. I don't know if I drew it, right
Connecting AC, AB is the diameter, so the angle ACB is a right angle, so the angle ACD is 55 degrees. Now there is an inscribed quadrilateral in the garden. ACDE has a property that the diagonal of the inscribed quadrilateral of a circle is complementary. Therefore, the angle AED + angle ACD is 180 ° and the angle AED is 125

As shown in the figure, AB is the diameter of circle O, e is a point on circle O, C is the midpoint of arc EB, CD is perpendicular to D. try to judge the position relationship between OC and AD People's education press page 88 page 9. Not secondary users can't transfer pictures. Sorry

Arc BC = arc CE, ﹤ EAC = ∠ cab. ∠ EAB = 2 ∠ cab
∠ cob = 2 ∠ cab (the center angle of the same arc is twice of the circumference angle)
∠EAB=∠COB
OC ∥ AE, namely OC ∥ ad

As shown in the figure, AB is known to be the diameter of ⊙ o, BC＝2 Ad, de ⊥ AB, proved: BC = 2DE

The extension de intersects the circle at point F. according to the theorem of vertical diameter, the
DF＝2
AD，
Also known
BC＝2
AD，
So,
DF＝
BC, BC = DF, so BC = 2DE

If the radius of a circular ground is increased by 5 m, the area of the site will be increased by 2 times! If the radius of a circular ground is increased by 5 m, the area of the site will be increased by 2 times

Let the radius be x meters
(X+5)*(X+5)*3.14=2X*2X*3.14
X²+10X+25=4X²
3X²-10-25=0
(X-5)(3X+5)=0
X1 = 5, X2 = - 5 / 3 (omitted)
The radius of the round field is 5 meters

As shown in the figure, increase the radius of a small circular site by 5m to get a large circular site. If the site area is doubled, the radius of the small circular site will be obtained=______ .

If the radius of the small circle is XM, then the radius of the big circle is (x + 5) M,
According to the meaning of the title: π (x + 5) 2 = 2 π x2,
X = 5 + 5
2 or x = 5-5
If you don't agree with the question, give it up
So the answer is: (5 + 5
2）m．

The area of the shadow part in the figure is 50 square centimeters. Find the area of the ring

Let the radius of the big circle and the small circle be r,
Then the area of the shadow part in the graph = R ^ 2 - R ^ 2 = 50
Therefore, the area of the ring = π (R ^ 2-r ^ 2) = 50 π = 157 square centimeter

The radius of the big and small circles are 5M and 3M respectively, and their perimeter ratio and area ratio are respectively

The ratio of perimeter is: [5:3]
The area ratio is: [25:9]

As shown in the figure, if the two equal circles with radius 1 and external tangent are both tangent to the circle with radius 3, then the perimeter of the shadow part in the figure is___ .

As shown in the figure, connecting the centers of three circles, ab = BC = AC = 2
∴∠A=∠ABC=∠ACB=60°．
∴∠DBF=∠ECF=120°．
The perimeter of the shadow part is 60 π × 3
180+120π×1
180×2=7π
3．

If the radius of a circular site is increased by 5 m, the area of the site will be increased by 2 times

Assuming that the original radius is x and PI is pi, then the following is true:
(pi*x*x)*2 = pi*(x+5)*(x+5)
So x = 5 (√ 2 + 1)