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1210
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avatar+1832 

 

How Can I Find The Area And The Parameter of the Yellow Segment ! 

 

 Dec 3, 2016

Best Answer 

 #6
avatar+33661 
+10

If the result is wanted in terms of d rather than theta then the following relationship can be used:

 

theta = cos-1(1 - 8d/D + 8d^2/D^2)

 

 

.

 Dec 4, 2016
 #1
avatar+118673 
+10

Hi 315,

I don't know what you mean by paremeter  indecision     , and I assume the black dot is the centre,   frown

 

Height of the triangle is  \(\frac{D}{2}-d = \frac{D-2d}{2}\)

 

\(sec \frac{\theta}{2}=r\div \frac{D-2d}{2}\\ ( \frac{D-2d}{2})sec \frac{\theta}{2}=r\\ r=( \frac{D-2d}{2})sec \frac{\theta}{2}\\\)

assuming that theta is in radians:

 

\(\begin{align}\\Area&=\frac{\theta}{2\pi}\times\pi r^2-\frac{1}{2}r^2sin\theta\\~\\ &=\frac{\theta r^2}{2}-\frac{1}{2}r^2sin\theta\\~\\ &=\frac{ r^2}{2}(\theta-sin\theta)\\~\\ &=\frac{\left [\frac{D-2d}{2}\;sec\frac{\theta}{2}\right]^2}{2}(\theta-sin\theta)\\~\\ &=\frac{(D-2d)^2\;sec^2\left(\frac{\theta}{2}\right)(\theta-sin\theta)}{8}\quad units^2\\~\\ \end{align}\)

 

 

*

 Dec 3, 2016
 #2
avatar+1832 
+5

Great !!!! 

 

I ment circumference by  perimeter 

and I typed  perimeter wrong blush 

sorry for that 

 Dec 3, 2016
 #3
avatar+129852 
+10

Here's an alternative approach :

 

Area of whole sector   = (1/2)r^2 * θ

 

Area of triangle within the sector   =  (1/2)r^2 sin θ

 

Yellow area    =   Area of whole secor  -  Area of triangle within the sector  =  

 

(1/2)r^2 * θ   - (1/2)r^2 sin θ   =

 

(1/2)r^2  [ θ -  sin θ ]

 

 

Perimeter  of yellow area   =

 

√ [ 2r^2  - 2r^2 cos θ]  +  r θ   =

 

r √ [2 - 2 cos θ ]  + r θ  =

 

r  [  √ [2 - 2 cos θ ]  +  θ  ]   

 

 

cool cool cool

 Dec 3, 2016
 #5
avatar+33661 
+10

The area simplifies to A = (D^2/8)(theta - sin(theta))

 

The perimeter of the yellow segment is D(theta/2 + sin(theta/2)).    The chord length can be found using either the cosine rule or the sine rule.

.

 Dec 4, 2016
 #6
avatar+33661 
+10
Best Answer

If the result is wanted in terms of d rather than theta then the following relationship can be used:

 

theta = cos-1(1 - 8d/D + 8d^2/D^2)

 

 

.

Alan  Dec 4, 2016
 #7
avatar+1832 
+5

that is amazing ! 

thank's alot CPhill & Alan 

 Dec 4, 2016

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