The area of a semicircle is 50.24. To find the circumference of the semicircle, we must have a complete formula, not just words and a few numbers,

The area of a semicircle is 50.24. To find the circumference of the semicircle, we must have a complete formula, not just words and a few numbers,

Let R be the radius,
RxRx3.14÷2=50.24
RxR=16
RxR=4x4
R=4
The circumference of this semicircle = diameter x3.14 △ 2 + diameter = 2x4x3.14 △ 2 + 4 + 4 = 20.56

Finding the circumference and area of a circle in sixth grade mathematics
Find the perimeter and area below
1.r=4.2m 2.d=15dm 3.r=14cm
Finding the area of a ring
1. R inner = 12cm r outer = 20cm2. C inner = 31.4dm D outer = 18dm

Find the perimeter and area below
1.r=4.2m
4.2*2*3.14= 26.376
3.14*4.2*4.2 = 55.3896
2.d=15dm
3.14*15= 47.1
3.148*15/2*15/2 = 176.625
3.r=14cm
3.14*14*2=87.92
3.14*14*14=615.44
Finding the area of a ring
1. R inner = 12cm r outer = 20cm
3.14*(20*20-12*12)=175.84
2. C = 31.4dm, d = 18dm
31.4/3.14/2= 5
18/2 = 9
3.14*(9*9-5*5)=175.84

It is known that in the orthorhombic o-abcd, the bottom quadrilateral ABCD is a diamond, M is the midpoint of OA, and N is the midpoint of BC

In the orthorhombic o-abcd, the bottom quadrilateral ABCD is rhombic, M is the midpoint of OA, n is the midpoint of BC. To prove: Mn is parallel to OCD. Prove: take od midpoint e, connect EM and CE ∵ m as the midpoint of OA, n is the midpoint of BC, that is, EM is the median line of △ oad, EM / / and = (1 / 2) adcn = (1 / 2) BC and ∵ bottom quadrilateral

Given rectangular ABCD, OA is perpendicular to the plane ABCD, OA = 1, the angle between od and the bottom ABCD is 30 degrees, and the angle between OB and CD is 45 degrees
Find: (1) the size of dihedral angle o-cd-a
(2) The dihedral angle between planar OBD and planar ABCD
Please be specific,

(1) When ad is perpendicular to CDDO and CD is perpendicular, the angle ADO is equal to 30 degrees. (2) OA = 1, OD = 2ad = root, 3ob and CD45 degrees, AB CD is parallel to ab = 1, AE is perpendicular to BD and intersects with BD at e0.5ad * AB = 0.5bd * aeae = 1, then the dihedral angle between planar OBD and planar ABCD = arctan1 is 45

It is known that as shown in the figure, ∠ AOB = 90 °, C and D are the three equal points of arc AB, AB intersects OC and OD at points E and f respectively

In other words, arc ACG = arc BDG is half of arc ab. so their corresponding center angles are equal. The same arc CG and arc DG are equal. The center angles are equal. In addition, M has two right angles perpendicular to the foot and a common edge OM

As shown in the figure, ab = CD, AE ⊥ BC, DF ⊥ BC, CE = BF

It is proved that: ∵ AE ⊥ BC, DF ⊥ BC, ∵ DFC = ∠ AEB = 90 ° and ∵ CE = BF, ∵ ce-ef = bf-ef, that is, CF = be, ∵ AB = CD, ≌ RT △ DFC ≌ RT △ AEB (HL), ≌ AE = DF

As shown in the figure, ab = CD, AE ⊥ BC, DF ⊥ BC, CE = BF

It is proved that: ∵ AE ⊥ BC, DF ⊥ BC, ∵ DFC = ∠ AEB = 90 ° and ∵ CE = BF, ∵ ce-ef = bf-ef, that is, CF = be, ∵ AB = CD, ≌ RT △ DFC ≌ RT △ AEB (HL), ≌ AE = DF

As shown in the figure, ab = CD, AE ⊥ BC at point E, DF ⊥ BC at point F, AE = DF, CE = BF

Certification:
∵AE⊥BC,DF⊥BC
∴∠AEB=∠DFC=90º
And ∵ AB = CD, AE = DF
∴Rt⊿ABE≌Rt⊿DCF（HL）
∴BE=CF
I don't know the figure
If CE > CF
CF + EF = be + EF, that is CE = BF
If CE ∠ CF
Cf-ef = be-ef, that is CE = BF

As shown in the figure, ab = CD, AE = DF, CE = BF

∵AB=CD
∴AB+BC=CD+BC
That is, AC = BD
In △ ace and △ DBF
AE=DF
AC=BD
CE=BF
△ACE≌△DBF（SSS）
∴∠ECA=∠FBD
‖ EC ‖ BF (equal internal stagger angle, two lines parallel)

It is known that ∠ BOC is 20 degrees larger than ∠ AOC, and OA is outside of ∠ BOC. Od bisects ∠ AOD and calculates the degree of ∠ doc

It is known that ∠ BOC - ∠ AOC = 20 ° and OA is outside ∠ BOC,
It can be concluded that: ∠ BOC + ∠ AOC = ∠ AOB,
Therefore, ∠ BOC = (1 / 2) ∠ AOB + 10 °,
It is known that OD bisects ∠ AOB
It can be concluded that: ∠ doc = ∠ BOC - ∠ BOD = (1 / 2) ∠ AOB + 10 ° - (1 / 2) ∠ AOB = 10 °