Question

1. Find area of the minor
sector and major sector of
circle of radius 7cm if angle
subtended by minor arc is
60°​

Answer 1

$\underline{\sf\ \ Given:-\ \ }$

• Radius of circle=7cm
• $\sf \theta= 60^{\circ}$

$\underline{\sf\ \ To\ Find :-\ \ }$

• Area of major sector and Minor sector

$\underline{\sf\ \ SOLUTION:-\ \ }$

$\underline{\boxed{\sf\ Area\ of (Minor) \ Sector= \dfrac{\pi r^2 \theta}{360}}}$

$\dashrightarrow\sf Ar(Minor\ Sector)= \dfrac{\pi (7)^2\times \cancel{60}}{\cancel{360}}\\ \\ \\ \dashrightarrow\sf Ar(Minor\ Sector)= \dfrac{22\times \cancel{49}}{\cancel{7}\times 6}\\ \\ \\ \dashrightarrow\sf Ar(Minor\ Sector)=\dfrac{22\times 7}{6}\\ \\ \\ \dashrightarrow\sf Ar(Minor\ Sector) = \dfrac{154}{6}\\ \\ \\ \dashrightarrow\sf Ar(Minor\ Sector)= \boxed{\sf 25\dfrac{4}{6}cm^2}$

$\rule{260}{1.2}$

$\underline{\boxed{\sf\ Area\ of \ Major\ sector= Ar(Circle)- Ar(Minor\ sector)}}$

$\begin{cases}\sf{Area\ of \ circle = \pi r^2}\\ \sf{Area\ of \ minor \ sector= \dfrac{154}{6}cm^2}\end{cases}$

$\dashrightarrow\sf Area\ of \ Circle= \dfrac{22}{\not{7}}\times \not{7}\times 7\\ \\ \\ \dashrightarrow\sf Area\ of\ circle= 22\times 7\\ \\ \\ \dashrightarrow\sf Ar(Circle)= \boxed{\sf 154 cm^2}$

$\rule{260}{1.2}$

• Now Area of major sector

$\dashrightarrow\sf Ar(Major\ sector)= Ar(Circle)-Ar(Minor\ sector)\\ \\ \\ \dashrightarrow\sf Ar(Major\ sector)= 154-\dfrac{154}{6}\\ \\ \\ \dashrightarrow\sf Ar(Major\ sector)= \dfrac{924-154}{6}\\ \\ \\ \dashrightarrow\sf Ar(Major \ sector)= \dfrac{770}{6}\\ \\ \\ \dashrightarrow\sf Ar(Major\ Sector)=\boxed{\sf 128\dfrac{2}{6}cm^2}$

Answer 2

Question:-

1. Find area of the minor sector and major sector of circle of radius 7cm if angle subtended by minor arc is 60°.

Solution:-

Area of a sector of a circle of radius 'r' and angle

θ= 360/θ πr 2

Final result:-

Hence, area of the sector of the circle of radius 7 cm and angle 60° = 60/360 × 22/7 ×7×7= 77/3 cm² .