Ask your own question, for FREE!
Mathematics 21 Online
OpenStudy (gudetamas):

how would i do this problem?

OpenStudy (gudetamas):

OpenStudy (mathstudent55):

You are given a circle with an arc of length of 4.82 in. The sector which intercepts this arc has a central angle of 65 degrees. You are looking for the circumference of the circle. Is the above correct?

OpenStudy (gudetamas):

yes that is correct

OpenStudy (gudetamas):

so you would step it up like this right ? 4.82 = 65/360 and get 26.69?

OpenStudy (gudetamas):

set*

OpenStudy (mathstudent55):

The circumference of a circle is \(\large C = 2 \pi r \) This is for a full circle with a central angle of 360 degrees. For a sector of a circle, the length of an arc is a fraction of the full circumference. \(S = \dfrac{n}{360^\circ} \times 2 \pi r\) S = arc length n = central angle measure

OpenStudy (mathstudent55):

You are on the right track. 65/360 is part of it. Use 65/360 for n/360 in the formula.

OpenStudy (mathstudent55):

\(S = \dfrac{n}{360^\circ} \times 2 \pi r\) Your arc length is given as 4.82 in., so you can substitute that for S: \(4.82 ~in. = \dfrac{65^\circ}{360^\circ} \times 2 \pi r\) Now you have only one unknown, the radius. Solve the equation for r. Then just use \(C = 2 \pi r\) to find the circumference of the circle.

OpenStudy (mathstudent55):

Oh, I see what you did. That was very good. If 4.82 in. is 65/360 of the circumference, then you can solve it more easily without having to find the radius. \(4.82 ~in. = \dfrac{65}{360} C\) \(C = 4.82~in. \times \dfrac{360}{65} \) \(C = 26.70~in.\)

OpenStudy (mathstudent55):

You are correct.

OpenStudy (gudetamas):

oh yes! thanks

OpenStudy (mathstudent55):

You're welcome.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!