Question

Find the arc length, area and perimeter of the sector with
(i) radius $21$cm and central angle${60^ \circ }$
(ii) radius $4.9$cm and central angle${30^ \circ }$
(iii) radius $14$cm and sector angle${45^ \circ }$
(iv) radius $15$cm and sector angle${63^ \circ }$
(v) radius $21$dm and sector angle${240^ \circ }$

Answer 1

Hint: In this problem, first we will write the formula of arc length, area and perimeter of sector. Then, we will substitute given values in the formula to get the required answer.
Arc length $L$ can be obtained by using the formula $\dfrac{\theta }{{{{360}^ \circ }}} \times 2\pi r$.
Area $A$ can be obtained by using the formula $\dfrac{\theta }{{{{360}^ \circ }}} \times \pi {r^2}$.
Perimeter $P$ can be obtained by using the formula $P = L + 2r$.

Complete step-by-step answer:
First we will write the formula of arc length, area and perimeter of the sector.
Arc length is given by $L = \dfrac{\theta }{{{{360}^ \circ }}} \times 2\pi r$ where $r$ is the radius and $\theta $ is the central angle.
Area is given by $A = \dfrac{\theta }{{{{360}^ \circ }}} \times \pi {r^2}$ where $r$ is the radius and $\theta $ is central angle.
Perimeter is given by $P = L + 2r$ where $L$ is the arc length and $r$ is the radius.
(i) Given that radius $r = 21$ cm and central angle $\theta = {60^ \circ }$.
Now we will substitute values of $r$ and $\theta $ in the above formula of arc length and area of sector. Therefore, we get
$L = \dfrac{{{{60}^ \circ }}}{{{{360}^ \circ }}} \times 2 \times \dfrac{{22}}{7} \times 21 = 22$ cm
$A = \dfrac{{{{60}^ \circ }}}{{{{360}^ \circ }}} \times \pi {\left( {21} \right)^2} = \dfrac{{{{60}^ \circ }}}{{{{360}^ \circ }}} \times \dfrac{{22}}{7} \times 21 \times 21 = 231$ cm$^2$
Now we will substitute values of arc length $L$ and radius $r$ in the above formula of perimeter of sector. Therefore, we get
$P = L + 2r = 22 + 2\left( {21} \right) = 64$ cm
(ii) Given that radius $r = 4.9$ cm and central angle $\theta = {30^ \circ }$.
Now we will substitute values of $r$ and $\theta $ in the formula of arc length and area of sector. Therefore, we get
$L = \dfrac{{{{30}^ \circ }}}{{{{360}^ \circ }}} \times 2 \times \dfrac{{22}}{7} \times 4.9 \approx 2.57$ cm
$A = \dfrac{{{{30}^ \circ }}}{{{{360}^ \circ }}} \times \pi {\left( {4.9} \right)^2} = \dfrac{{{{30}^ \circ }}}{{{{360}^ \circ }}} \times \dfrac{{22}}{7} \times 4.9 \times 4.9 \approx 6.29$ cm$^2$
Now we will substitute values of arc length $L$ and radius $r$ in the formula of perimeter of sector. Therefore, we get
$P = L + 2r = 2.57 + 2\left( {4.9} \right) = 12.37$ cm
(iii) Given that radius $r = 14$ cm and sector angle $\theta = {45^ \circ }$.
Now we will substitute values of $r$ and $\theta $ in the formula of arc length and area of sector. Therefore, we get
$L = \dfrac{{{{45}^ \circ }}}{{{{360}^ \circ }}} \times 2 \times \dfrac{{22}}{7} \times 14 = 11$ cm
$A = \dfrac{{{{45}^ \circ }}}{{{{360}^ \circ }}} \times \pi {\left( {14} \right)^2} = \dfrac{{{{45}^ \circ }}}{{{{360}^ \circ }}} \times \dfrac{{22}}{7} \times 14 \times 14 = 77$ cm$^2$
Now we will substitute values of arc length $L$ and radius $r$ in the formula of perimeter of sector. Therefore, we get
$P = L + 2r = 11 + 2\left( {14} \right) = 39$ cm
(iv) Given that radius $r = 15$ cm and sector angle $\theta = {63^ \circ }$.
Now we will substitute values of $r$ and $\theta $ in the formula of arc length and area of sector. Therefore, we get
$L = \dfrac{{{{63}^ \circ }}}{{{{360}^ \circ }}} \times 2 \times \dfrac{{22}}{7} \times 15 = 16.5$ cm
$A = \dfrac{{{{63}^ \circ }}}{{{{360}^ \circ }}} \times \pi {\left( {15} \right)^2} = \dfrac{{{{63}^ \circ }}}{{{{360}^ \circ }}} \times \dfrac{{22}}{7} \times 15 \times 15 = 123.75$ cm$^2$
Now we will substitute values of arc length $L$ and radius $r$ in the formula of perimeter of sector. Therefore, we get
$P = L + 2r = 16.5 + 2\left( {15} \right) = 46.5$ cm
(v) Given that radius $r = 21$ dm and sector angle $\theta = {240^ \circ }$.
Now we will substitute values of $r$ and $\theta $ in the formula of arc length and area of sector. Therefore, we get
$L = \dfrac{{{{240}^ \circ }}}{{{{360}^ \circ }}} \times 2 \times \dfrac{{22}}{7} \times 21 = 88$ dm
$A = \dfrac{{{{240}^ \circ }}}{{{{360}^ \circ }}} \times \pi {\left( {21} \right)^2} = \dfrac{{{{240}^ \circ }}}{{{{360}^ \circ }}} \times \dfrac{{22}}{7} \times 21 \times 21 = 924$ dm$^2$
Now we will substitute values of arc length $L$ and radius $r$ in the formula of perimeter of sector. Therefore, we get
$P = L + 2r = 88 + 2\left( {21} \right) = 130$ dm

Note: The area of a circle having radius of length $r$ is given by $A = \pi {r^2}$. The circumference of a circle having radius of length $r$ is given by $C = 2\pi r$. Circumference of the circle is the length of the boundary of the circle.