If the circumference of a sector with a radius of 4cm is equal to the length of the semicircle where the arc is located, then the area of the sector is () The answer is 8p-16,

If the circumference of a sector with a radius of 4cm is equal to the length of the semicircle where the arc is located, then the area of the sector is () The answer is 8p-16,

The length of the semicircle where the arc is located = π r = 4 π
For a sector with a radius of 4cm, its circumference = arc length + 2R = 8 + arc length = 4 π
Arc length = 4 π - 8
Circumference = 8 π
The area of sector = π R & sup2; × (4 π - 8) △ 8 π = (8 π - 16) CM & sup2;

If the circumference of a sector of radius 4 is equal to the circumference of the semicircle in which the arc is located, find the area of the sector
For a sector of radius 4, if its perimeter is equal to the perimeter of the semicircle where the arc is located, the area of the sector

Radius r = 8cm, it can be assumed that the central angle of the sector is θ, then the arc length L = θ R, the semicircle circumference is π R (circumference C = 8 π R), that is: 8cm + L + 8cm = π R, substituting R, 8cm + L = 8 π That's why you asked. Sector area s = θ R & amp; sup8 / 8 = θ R · R / 8 = LR / 8

As shown in the figure, the section diameter of the horizontal cylindrical drainage pipe is 1 m, and the width ab of the water surface is 0.8 m, then the depth of the water in the drainage pipe is 1 m___  m．

Through o as OC ⊥ AB, intersecting AB at point C, we can get AC = BC = 12ab = 0.4m, from the diameter is 1m, we can know that the radius is 0.5m, in RT △ AOC, according to the Pythagorean theorem: OC = ao2-ac2 = 0.52-0.42 = 0.3 (m), then the depth of water in the drainage pipe is 0.5-0.3 = 0.2 (m), so the answer is: 0.2

The cylindrical drain pipe with an inner diameter of 1m is placed horizontally. From the section direction, the width of the water surface in the pipe is 0.6m, so what is the depth of the water? There is an analysis

If the radius is 0.5m, the distance from the water surface to the center of the circle is 0.4m (hook three strands, four strings and five strings), the water depth is 0.1M

If the radius of ⊙ o is 2cm and the center angle of the circle ∠ AOB = 120 °, then the length of AB is () cm and the chord center distance of chord AB is () cm

Root 3cm, 1cm

Given that the radius of a circle is 5cm and the length of a string is 6cm, then the chord center distance of the string is 5cm______ cm．

Through o as OC ⊥ AB in C, connect OA, ∵ OC ⊥ AB, OC through O, ∵ AC = BC = 12ab = 3cm, in RT △ OAC, from Pythagorean theorem: OC = oa2 − ac2 = 52 − 32 = 4 (CM), so the answer is: 4

In a circle with a radius of 5cm, there is a string of 8cm in length. The distance between the centers of the strings is______ cm．

According to the vertical diameter theorem, half of the chord is 4cm. According to the Pythagorean theorem, the chord center distance is 3cm

In ⊙ o, if the diameter is 25cm, the chord center distance of chord AB is 10cm, the length of chord AB is calculated

As shown in the figure, OC is the chord center distance of AB, connecting OA, ∵ OC is the chord center distance of AB, that is, OC ⊥ AB, ∵ AC = BC, in RT △ OAC, ∵ OA = 252, OC = 10, ∵ AC = oa2 − oc2 = 152, ∵ AB = 2Ac = 15 (CM)

If the diameter of ⊙ o is 10 & nbsp; cm and the chord center distance of chord AB is 3cm, then the perimeter of ⊙ o inscribed rectangle with chord AB as one side is ()
A. 14cmB. 28cmC. 48cmD. 20cm

As shown in the figure, the diameter of ∵ o is 10 & nbsp; cm, the chord center distance of string AB is 3cm, and ∵ connects OA, OA = 5cm. In RT △ AOG, OA = 5cm, og = 3cm, so Ag = oa2 − og2 = 52 − 32 = 4. According to the vertical diameter theorem, ab = 2ag = 2 × 4 = 8cm

In ⊙ o, if the diameter is 25cm, the chord center distance of chord AB is 10cm, the length of chord AB is calculated

As shown in the figure, OC is the chord center distance of AB, connecting OA, ∵ OC is the chord center distance of AB, that is, OC ⊥ AB, ∵ AC = BC, in RT △ OAC, ∵ OA = 252, OC = 10, ∵ AC = oa2 − oc2 = 152, ∵ AB = 2Ac = 15 (CM)