+0  
 
+1
181
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avatar+541 

What is the area under a chord with a central/arc angle of 125 and a radius of 8?

 

EDIT: I tried a few different methods and ended up with negative numbers. I really don't know what to do...indecision

After reply 2:

EDIT: To clarify, I'm looking for are that is not a part of the traingle, but still part of the sector. Thank you Pavlov!

helperid1839321  Feb 13, 2018
edited by helperid1839321  Feb 13, 2018
edited by helperid1839321  Feb 13, 2018
 #1
avatar+12565 
+1

The chord line and the the two  radii of 8  form a triangle of which one angle is 125 degrees and the other two are

27.5 degrees   (must add to 180 degrees)

  Draw a line of length 'h' perpindicular to the chord which bisects the 125 degree angle .

Now you have two equal triangles with angles of 90  125/2   and 27.5 degree angles

Now use law of sines to find 'h'  

 

sin 90 /8  = sin 27.5 /h = sin 62.5/(1/2chord)         

h = 3.694    and chord = 14.192

Then area of the triangle = 1/2 b h      where b is the chord

Area = 1/2 (14.192)(3.694) = 26.22 units^2

ElectricPavlov  Feb 13, 2018
 #2
avatar+541 
0

Sorry if I didn't clarify. I wasn't looking for the area of the triangle, I was looking for the area of the part of the section of the circle that wasn't a part of the triangle, but was ont the other side of the chord with the sector.

 

Thanks though! smiley

helperid1839321  Feb 13, 2018
 #3
avatar+12565 
+1

Then just find the area of the sector and subtract the area of the triangle

 

sector area     =   125/360 x  pi x 8^2 = 69.813 sq units

Now subtract the triangle for the area of the segment.....

ElectricPavlov  Feb 13, 2018
edited by ElectricPavlov  Feb 13, 2018
 #4
avatar+87556 
+1

 

Area of sector - Area of triangle  = Desired Area

 

The "formula" for the area is :

 

 r^2  [ (pi * 125/360)  -  (1/2) sin (125°)  ]  =

 

8^2 [  (pi * 125/360)  - (1/2)sin (125° ) ]  ≈  43.6  units^2

 

 

cool cool cool

CPhill  Feb 13, 2018

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