It is known that, as shown in the figure, the chords AB and CD of circle O intersect at p. it is proved that PA * Pb = PC * PD

It is known that, as shown in the figure, the chords AB and CD of circle O intersect at p. it is proved that PA * Pb = PC * PD

It is proved that triangle ACP is similar to BPD

If the two chords of a circle AB, CD intersect at point P, and PA = Pb = 4, PD = 2, then PC = 2___ .

According to the meaning of the title, Pa? Pb = PC? PD,
4 × 4 = 2pc,
So PC = 8
So the answer is 8

As shown in Fig. 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

∵ AB is the diameter, and the circular angle corresponding to arc BD is 20 degrees
The circular angle corresponding to the arc ad is 70 degrees
And C is the midpoint of the arc ad
The circular angles corresponding to arc AC and arc CD are both 35 degrees
∴∠CAD=35°,∠AOC=70°
Ao = OC
∴∠ACO=(180°-70°)÷2=55°

Known: as shown in the figure, AB is the diameter of ⊙ o, C, D are two points on ⊙ o, and C is For the midpoint of AD, if ∠ bad = 20 °, calculate the degree of ∠ ACO

∵ AB is the diameter of ⊙ o, and C is
The midpoint of AD,
∴OC⊥AD，
∵∠BAD=20°，
∴∠AOC=90°-∠BAD=70°，
∵OA=OC，
∴∠ACO=∠CAO=180°−∠AOC
2=180°−70°
2=55°．

If the radius of a circular ground is increased by 5 m, the area of the site will be increased by 2 times. Find the radius of the circular site RT. detailed process... Good bonus points~ The detailed process is more than three or four steps... Don't give me two steps is good, perfunctory I give a lot of points

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)

If the radius of the circle is increased by 5 m, the area of the site will be increased by 2 times Find the radius of a circular site Let the radius be r 2πr^2=π（r+5)^2 2πr^2-π（r^2+10r+25)=0 2πr^2-πr^2-10πr-25π=0 πr^2-10πr-25π=0 r^2-10r-25=0 The root of the equation is: R1 = 5 + 5 √ 2 R2 = 5-5 √ 2 So the radius is 5 + 5 √ 2 Why give up? I can't think clearly

A negative radius makes no sense

There is a circular figure, the inner radius is 3M, the outer inner radius is 5m, and the outer circle of shadow part: calculate the perimeter and area

Zhou Changwei
3 × 2 × π + 5 × 2 × π = 16 π = 50.24m
The area is
π (5? - 3?) = 16 π = 50.24 square meters
complete!

A circular site is shown in the figure (shadow part). If its radius is increased by 5 m, the area of the site will be increased by 2 times. The radius of the original circular site is calculated Denoted by π

The original area is π R 2
Now it's 2 π R 2
Now the radius R + 5
So π (F + 5) 2 = 2 π r
r²+10r+25=2r²
r²-10r-25=0
R>0
So r = 5 + 5 √ 2

As shown in the figure, the side length of square ABCD is 1cm, e and F are the midpoint of BC and CD respectively. Connecting BF and De, the shadow area in the figure is () cm2 A. 4 Five B. 2 Three C. 5 Six D. 3 Four

As shown in the figure, ∵ the side length of the square ABCD is 1cm, e and F are the midpoint of BC and CD respectively, ≌≌△ CBF, easy to get, △ Bge ≌ △ DGF, so s △ Bge = s △ EGC, s △ DGF = s △ CGF, so s △ Bge = s △ EGC = s △ DGF = s △ CGF, and because s △ BFC = 1 × 12 × 12 = 14cm2, so

As shown in the figure, the side length of square ABCD is 1cm, e and F are the midpoint of BC and CD respectively. Connecting BF and De, the shadow area in the figure is () cm2 A. 4 Five B. 2 Three C. 5 Six D. 3 Four

As shown in the figure, ∵ the side length of the square ABCD is 1cm, e and F are the midpoint of BC and CD respectively, ≌≌△ CBF, easy to get, △ Bge ≌ △ DGF, so s △ Bge = s △ EGC, s △ DGF = s △ CGF, so s △ Bge = s △ EGC = s △ DGF = s △ CGF, and because s △ BFC = 1 × 12 × 12 = 14cm2, so