Question

For the sectors with given measures, find the length of the arc, area and perimeter, (π = 3.14)

★ central angle 60°, r = 14 cm

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Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

• Radius of sector, r = 14 cm

• Central angle of sector, x = 60°

We know that,

Area of sector of radius r and central angle x° is

$\purple{\rm :\longmapsto\:\boxed{\tt{ Area_{(sector)} \: = \: \pi \: {r}^{2} \: \frac{x}{360} \: }}}$

So, on substituting the values, we get

$\rm :\longmapsto\:Area_{(sector)} = 3.14 \times 14 \times 14 \times \dfrac{60}{360}$

$\bf\implies \:Area_{(sector)} \: = \: 102.57 \: {cm}^{2} \: approx$

We know that,

Length of arc of sector of radius r and central angle x is

$\purple{\boxed{\tt{ Length \: of \: arc_{(sector)} = \: \pi \: r \: \frac{x}{180} \: }}}$

$\rm :\longmapsto\:Length \: of \: arc_{(sector)} = 3.14 \times 14 \times \dfrac{60}{180}$

$\bf\implies \:Length \: of \: arc_{(sector)} \: = \: 14.65 \: cm$

Now, We know that,

Perimeter of sector of radius r and Length of arc, l is given by

$\purple{\rm :\longmapsto\:\boxed{\tt{ Perimeter_{(sector)} = 2r + l}}}$

So, on substituting the values, we get

$\rm :\longmapsto\:Perimeter_{(sector)} = 2 \times 14 + 14.65$

$\bf\implies \:\:Perimeter_{(sector)} = 42.65 \: cm$

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$\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Breadth\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}$

Answer 2

Step-by-step explanation:

$\large\underline{\sf{Solution-}}$

Given that,

Radius of sector, r = 14 cm

Central angle of sector, x = 60°

We know that,

Area of sector of radius r and central angle x° is

$\purple{\rm :\longmapsto\:\boxed{\tt{ Area_{(sector)} \: = \: \pi \: {r}^{2} \: \frac{x}{360} \: }}}$

So, on substituting the values, we get

$\rm :\longmapsto\:Area_{(sector)} = 3.14 \times 14 \times 14 \times \dfrac{60}{360}$

$\bf\implies \:Area_{(sector)} \: = \: 102.57 \: {cm}^{2} \: approx$

We know that,

Length of arc of sector of radius r and central angle x is

$\purple{\boxed{\tt{ Length \: of \: arc_{(sector)} = \: \pi \: r \: \frac{x}{180} \: }}}$

$\rm :\longmapsto\:Length \: of \: arc_{(sector)} = 3.14 \times 14 \times \dfrac{60}{180}$

$\bf\implies \:Length \: of \: arc_{(sector)} \: = \: 14.65 \: cm$

Now, We know that,

Perimeter of sector of radius r and Length of arc, l is given by

$\purple{\rm :\longmapsto\:\boxed{\tt{ Perimeter_{(sector)} = 2r + l}}}$

So, on substituting the values, we get

$\rm :\longmapsto\:Perimeter_{(sector)} = 2 \times 14 + 14.65$

$\bf\implies \:\:Perimeter_{(sector)} = 42.65 \: cm$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

$\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Breadth\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}$