How to calculate the circumference and area of a circle It's better to use a simple method. You can also help to solve some problems Brother and sister, help

How to calculate the circumference and area of a circle It's better to use a simple method. You can also help to solve some problems Brother and sister, help

Perimeter = 2 × 3.14 × radius or 3.14 × diameter
Area = 3.14 × radius 2
1. For a circle with a radius of 5cm, calculate the perimeter and area
2. Calculate the area of a circle with a circumference of 18.84

The radius of the big circle is equal to the radius of the small circle. The perimeter ratio of the two circles is () and the area ratio is ()

The radius of the big circle is equal to the radius of the small circle. The perimeter ratio of the two circles is (2:1) and the area ratio is (4:1)

The radius of a circle is 3cm, the perimeter of another circle is 1 / 3 of this circle, the perimeter of the other circle is () and the area is ()

Perimeter = 2 π r = 6 π cm
So the other circle has a circumference of 2 π cm
So the other circle has a radius of 1 cm
So area = π R ^ 2 = 1cm ^ 2

It is known that ∠ AOB = 3 ∠ BOC, if ∠ BOC = 30 °, then ∠ AOC=______ Degree

∵ - BOC = 30 °, ∵ AOB = 3 ∵ BOC, ∵ - AOB = 3 × 30 ° = 90 ° (1) when OC is on the outside of ∵ AOB, ∵ - AOC = ∵ AOB + ∵ BOC = 90 ° + 30 ° = 120 degree; (2) when OC is on the inside of ∵ AOB, ∵ AOC = ∵ AOB - ∵ BOC = 90 ° - 30 ° = 60 degree

Three non coplanar straight lines OA, OB and OC are introduced from a point O in space. If ∠ BOC = 90 °, AOB = ∠ AOC = 60 °, the angle between straight line OA and plane BOC is calculated

Take a point a on OA, make ah ⊥ plane BOC at h, connect Oh, then ⊥ AOH is the angle formed by straight line OA and plane BOC, respectively make he ⊥ ob, intersect ob at point E, HF ⊥ OC, intersect OC at point F, connect AE and AF, get AE ⊥ ob, AF ⊥ OC, △ ofH is isosceles right triangle, let of = a, then Oh = 2A, OA = 2A, cos ⊥ AOH = ohoa = 22, ⊥ AOH = 45 °

Three non coplanar straight lines OA, OB and OC are introduced from a point O in space. If ∠ BOC = 90 °, AOB = ∠ AOC = 60 °, the angle between straight line OA and plane BOC is calculated

Take a point a on OA, make ah ⊥ plane BOC at h, connect Oh, then ⊥ AOH is the angle formed by straight line OA and plane BOC, respectively make he ⊥ ob, intersect ob at point E, HF ⊥ OC, intersect OC at point F, connect AE and AF, get AE ⊥ ob, AF ⊥ OC, △ ofH is isosceles right triangle, let of = a, then Oh = 2A, OA = 2A, cos ⊥ a

As shown in the figure, it is known that ∠ AOC: ∠ BOC = 1:3, ∠ AOD: ∠ BOD = 5:7. If ∠ cod = 15 °, calculate the degree of ∠ AOB

Let ∠ AOC = x, then ∠ BOC = 3x, then ∠ AOD = x + 15 ° and ∠ BOD = 3x-15 ° according to the meaning of (x + 15 °): (3x-15 °) = 5:7, the solution is x = 22.5 ° and ∠ AOB = x + 3x = 90 °. Therefore, the degree of ∠ AOB is 90 °

The degree of ∠ BOD can be calculated by ∠ AOD = 140 °, BOC = 3 ∠ AOB, cod is 20 ° larger than AOC

∵∠COD=∠AOC+20° ∴∠AOD=2∠AOC+20°=140° ∠AOC=60°
∵∠BOC=3∠AOB ∴∠BOC=3/4∠AOC=45°
∴∠BOD=∠BOC+∠COD=45°+60°+20°=125°

As shown in the figure, ∠ AOD = ∠ BOC = 90 ° and ∠ cod = 42 ° calculate the degree of ∠ AOC and ∠ AOB

As shown in the figure, (1) ∵ AOD = 90 °, COD = 42 °, AOC = ∠ AOD + ∠ cod = 90 ° + 42 ° = 132 °; (2) ∵ AOD + ∠ cod + ∠ BOC + ∠ AOB = 360 °, AOB = 360 ° - ∠ AOD - ∠ cod - ∠ BOC, = 360 ° - 90 ° - 42 ° - 90 °, = 138 °. So the answers are 132 ° and 138 °

As shown in the figure, ∠ AOB = ∠ cod = 90 ° and ∠ AOC = 2 ∠ BOC, then ∠ AOD = how much

∵∠AOC+∠BOC＝∠AOB＝90,∠AOC＝2∠BOC
∴3∠BOC＝90
∴∠BOC＝30
∵∠COD＝90
∴∠BOD＝∠COD-∠BOC＝90-30＝60
∴∠AOD＝∠AOB+∠BOD＝90+60＝150°
The math group answered your question,