Question

What is the perimeter of the sector with radius 10.5cm and sector angle 60?

Answer 1

Perimeter of a sector, P = 2r + l
where, r is radius and l is length of arc.
also, we know,
l = rn
where, n id angle of sector.

So, P = 2r + r(pi/3)
P = (2+ pi/3)r

Substitute the required values to get numerical answer.

Answer 2

[ Figure in attachment ]

Given :- Radius ( r ) = 10.5 cm

Angle of the sector = 60

$angle \: of \: sector \: i.e.theta = \alpha$

Solution :-

Perimeter of Sector = length of Arc + 2radius.

Perimeter of Sector =
$\frac{ \alpha }{360} \times 2\pi \: r + 2r$

$= 2r( \frac{60}{360} \times \frac{22}{7} + 1)$


$= 2 \times 10.5( \frac{1}{6} \times \frac{22}{7} + 1)$

$= 21 \times ( \frac{22 + 42}{42} )$


$= \frac{64}{2}$

$= 32cm$


Perimeter of sector ( ACB sector ) = 32cm