Factor 15x+35 using the GCF
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Edited 2018-03-09 07:49
Given unit circle, so radius=1.
We calculate lengths of vertical segments, with the help of Pythagoras Theorem, based on a right triangle radiating from circle centre O, and hypotenuse from O to a point on the circumference.
AO=1 (given unit circle)
BB'=sqrt(1^2-0.25^2)=0.968246
CC'=sqrt(1^2-0.5^2)=0.866025
DD'=sqrt(1^2-0.75^2)=0.661438
EE'=0
Now we proceed to calculate the segments approximating the arc. Again, we use a right triangle in which the hypotenuse is the segment joining two points on the circumference. The height is the difference between the two vertical segments, and the base is 0.25 for all four segments.
AB=sqrt((AO-BB)^2+0.25^2)=0.252009BC=sqrt((BB-CC)^2+0.25^2)=0.270091CD=sqrt((CC-DD)^2+0.25^2)=0.323042DE=sqrt((DD-0)^2+0.25^2)=0.7071068
giving a total estimation of the arc length
approximation of arc=AB+BC+CD+DE=1.55225
Given unit circle, so radius=1.
We calculate lengths of vertical segments, with the help of Pythagoras Theorem, based on a right triangle radiating from circle centre O, and hypotenuse from O to a point on the circumference.
AO=1 (given unit circle)
BB'=sqrt(1^2-0.25^2)=0.968246
CC'=sqrt(1^2-0.5^2)=0.866025
DD'=sqrt(1^2-0.75^2)=0.661438
EE'=0
Now we proceed to calculate the segments approximating the arc. Again, we use a right triangle in which the hypotenuse is the segment joining two points on the circumference. The height is the difference between the two vertical segments, and the base is 0.25 for all four segments.
AB=sqrt((AO-BB)^2+0.25^2)=0.252009BC=sqrt((BB-CC)^2+0.25^2)=0.270091CD=sqrt((CC-DD)^2+0.25^2)=0.323042DE=sqrt((DD-0)^2+0.25^2)=0.7071068
giving a total estimation of the arc length
approximation of arc=AB+BC+CD+DE=1.55225