Area of a sector is the region bounded by the bounding radii and the arc of the sector.
Generally,we are having two formulas to find the area of any sector.
Area of the sector = (θ/360) x Π r ² square units
(or) Area of the sector = (l r/2) square units
θ - central angle formed by the sector
L - length of arc
r - radius of the sector
We can use the first formula if the central angle(θ) formed by the sector and radius given. If the length of arc(L) is given we have to use the second formula.
Question 1 :
Find the area of the sector whose radius and central angle are 42 cm and 60° respectively.
Solution :
Area of the sector = (θ/360) x Π r²
r = 42 cm , θ = 60°
by applying those values in the above formula we get,
= (60/360) x (22/7) x 42 x 42
= (1/6) x 22 x 6 x 42 ==> 924 cm²
Hence, area of sector is 924 cm²
Let us see the next question on "area of a sector"
Question 2 :
Find the area of the sector whose radius and central angle are 21 cm and 60° respectively.
Solution:
Area of the sector = (θ/360) x Π r²
r = 21 cm , θ = 60°
by applying those values in the above formula we get,
= (60/360) x (22/7) x 21 x 21
= (1/6) x 22 x 3 x 21 ==> 231 cm²
Hence, area of sector is 231 cm²
Let us see the next question on "area of a sector"
Question 3 :
Find the area of the sector whose radius and central angle are 4.9 cm and 30° respectively.
Solution:
Area of the sector = (θ/360) x Π r²
r = 4.9 cm , θ = 30°
by applying those values in the above formula we get,
= (30/360) x (22/7) x 4.9 x 4.9
= (1/12) x 22 x 0.7 x 4.9 ==> 6.3 cm²
Hence, area of sector is 6.3 cm²
Let us see the next question on "area of a sector"
Question 4 :
Find
the area of the sector and also find the central angle formed by the
sector whose radius is 21 cm and length of arc is 66 cm.
Solution :
First,let us find the area of sector using length of an arc and radius.
Area of the sector = (l r / 2) square units
L = 66 cm
r = 21 cm
= (1/2) x 66 x 21 ==> 33 x 21 ==> 693 square units
Area of the sector = 693 square units
(θ/360) x Π r ² = 693
(θ/360) x (22/7) x (21)² = 693
θ = (693 x 7 x 360) / (22 x 21 x 21)
θ = (693 x 360) / (22 x 3 x 21)
θ = (231 x 180) / (11 x 21)
θ = (21 x 180) / 21
θ = 180°
Hence, area of sector and central angle are 693 square units and 180° respectively.
Let us see the next question on "area of a sector"
Question 5 :
Find the area of the sector whose radius and length of arc are 6 cm and 20 cm.
Solution :
Area of the sector = (Lr/2) square units
L = 20 cm
r = 6 cm
= (20 x 6)/2 ==> 10 x 6 ==> 60 cm²
Hence, area of sector is 60 cm²
Let us see the next question on "area of a sector"
Question 6 :
Find the area of the sector whose diameter and length of arc are 10 cm and 40 cm.
Solution :
Area of the sector = (Lr/2) square units
L = 40 cm
diameter = 10 ==> r = 5 cm
= (40 x 5)/2 ==> 20 x 5 ==> 100 cm²
Hence, area of sector is 100 cm²
Question 7 :
Find the area of the sector whose radius is 35 cm and perimeter is 147 cm.
Solution :
radius = 35 cm
Perimeter of sector = 147 cm
L + 2r = 147 ==> L + 2 (35) = 147 ==> L + 70 = 147 ==> L = 77 cm
Now we have the length of an arc and radius. So we have to use the second formula to find the area of the given sector.
Area of the sector = (1/2) x l r square units
= (1/2) x 77 x 35 ==> 38.5 x 35 ==> 1347.5 square units
Let us see the next question on "area of a sector"
Question 8 :
Find the area of the sector whose radius is 20 cm and perimeter is 110 cm.
Solution :
radius = 20 cm
Perimeter of sector = 110 cm
L + 2r = 110 ==> L + 2 (20) = 110 ==> L + 40 = 110 ==> L = 70 cm
Now we have the length of an arc and radius. So we have to use the second formula to find the area of the given sector.
Area of the sector = (l r/2) square units
= (70 x 20) / 2 ==> 70 x 10 ==> 700 square units
Square Parallelogram |
Rectangle Rhombus |
Traingle Quadrilateral Sector
Hollow cylinder Sphere Area around circle Area of combined figures |
Trapezium Circle Semicircle Quadrant Cyclinder Cone Hemisphere Path ways |
WORD PROBLEMS
Word problems on simple equations
Word problems on linear equations
Word problems on quadratic equations
Area and perimeter word problems
Word problems on direct variation and inverse variation
Word problems on comparing rates
Converting customary units word problems
Converting metric units word problems
Word problems on simple interest
Word problems on compound interest
Word problems on types of angles
Complementary and supplementary angles word problems
Markup and markdown word problems
Word problems on mixed fractrions
One step equation word problems
Linear inequalities word problems
Ratio and proportion word problems
Word problems on sets and venn diagrams
Pythagorean theorem word problems
Percent of a number word problems
Word problems on constant speed
Word problems on average speed
Word problems on sum of the angles of a triangle is 180 degree
OTHER TOPICS
Time, speed and distance shortcuts
Ratio and proportion shortcuts
Domain and range of rational functions
Domain and range of rational functions with holes
Graphing rational functions with holes
Converting repeating decimals in to fractions
Decimal representation of rational numbers
Finding square root using long division
L.C.M method to solve time and work problems
Translating the word problems in to algebraic expressions
Remainder when 2 power 256 is divided by 17
Remainder when 17 power 23 is divided by 16
Sum of all three digit numbers divisible by 6
Sum of all three digit numbers divisible by 7
Sum of all three digit numbers divisible by 8
Sum of all three digit numbers formed using 1, 3, 4
Sum of all three four digit numbers formed with non zero digits