The angle of a sector of a circle radius 10.5 cm is 48\(^o\). Calculate the perimeter of the sector.
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Correct Answer: Option D
Explanation:

The lenght of Arc AB = \(\frac{Q}{360}\) 2\(\pi\)r
= \(\frac{48}{360}\) x 2\(\frac{22}{7}\) x 10.5 = \(\frac{48}{360}\) x 2\(\frac{22}{7}\) x \(\frac{21}{2}\)
= \(\frac{4 \times 22 \times 3}{30} \times \frac{88}{10}\) = 8.8cm
Perimeter = 8.8 + 2r = 8.8 + 2r
= 8.8 + 2(10.5)
= 8.8 + 21
= 29.8cm
The lenght of Arc AB = \(\frac{Q}{360}\) 2\(\pi\)r
= \(\frac{48}{360}\) x 2\(\frac{22}{7}\) x 10.5 = \(\frac{48}{360}\) x 2\(\frac{22}{7}\) x \(\frac{21}{2}\)
= \(\frac{4 \times 22 \times 3}{30} \times \frac{88}{10}\) = 8.8cm
Perimeter = 8.8 + 2r = 8.8 + 2r
= 8.8 + 2(10.5)
= 8.8 + 21
= 29.8cm