The perimeter of a sector of a circle of radius r cm and of central angle θ (in degrees) is:

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C = 2 r (\frac{πθ}{360} + 1)

C = \frac{q}{360} \times 4 pr + 2 r

C = \frac{q}{360} \times 2 pr + 6 r

C = \frac{q}{360} \times 2 pr + 5 r

answer is A.

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Detailed Solution

Given,
AOB is a sector of radius=r and the central angle=θ.
Now, the arc length =

\frac{θ}{360 °} \times 2 \times π \times r

.
So, t

he arc length AB = \frac{θ}{360 °} \times 2 πr

.
Now the perimeter of the sector = OA+OB+ arc AB.
The circumference of a sector of angle θ of a circle with radius r is,

C = r + r + \frac{θ}{360} \times 2 πr

\Rightarrow C = 2 r + \frac{θ}{360} \times 2 πr

\Rightarrow C = 2 r (\frac{θπ}{360} + 1)

Thus, the perimeter of a sector of a circle of radius r cm and of central angle θ (in degrees) is

2 r (\frac{πθ}{360} + 1)

.
Therefore, required answer is 1.

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