Question

Find the peri-meter of a sector which makes an angle 45° at the centre of a circle of radius
7 cm.​

Answer 1

SoluTiση :-

Radius of circle = 7 cm

Perimeter = (2 × π × r) cm

Perimeter = (2 × π × 7) cm

Perimeter = (14 × π) cm

Perimeter = 44 cm

Thus,

$\rm {\because \ Full \ circumference = 360^{\circ}}\\\\\\\rm {\therefore \ \dfrac{360^{\circ}}{8}=45^{\circ}}$

Now,

$\sf {AB = \dfrac{44}{8} =5.5}$

Therefore,

Perimeter of sector which makes an angle of 45°

= (7 + 7 + 5.5) cm

= 19.5 cm

❝Answer = 19.5 cm❞

Answer 2

Given ,

Radius of circle = 7 cm

Centre angle = 45

We know that , the length of arc of a sector is given by

$\sf \fbox{Length \: of \: arc \: of \:a \: sector = \frac{ \theta}{360} \times 2\pi r}$

Thus ,

$\sf \Rightarrow Length = \frac{45}{360} \times 2 \times \frac{22}{7} \times 7 \\ \\ \sf \Rightarrow Length = \frac{45 \times 44}{360} \\ \\\sf \Rightarrow Length = \frac{1980}{360} \\ \\\sf \Rightarrow Length = 5.5 \: cm$

Now ,

$\sf \fbox{Perimeter \: of \: sector = 2r + Length \: of \: arc \: of \: sector }$

Thus ,

$\sf \mapsto Perimeter \: of \: sector = 2 \times 7 + 5.5 \\ \\ \sf \mapsto Perimeter \: of \: sector = 19.5$

$\therefore \sf \underline{The \: perimeter \: of \: sector \: is \: 19.5 \: cm}$