MENSURATION-sk
CIRCLE-LENGTH OF THE ARC & AREA OF A SECTOR
STD X
MAHARASHTRA STATE BOARD OF EDUCATION,
MUMBAI
CIRCLE-LENGTH OF THE ARC & AREA OF A
SECTOR
Circleis a set of all points in the plane which are equidistant
from a given point, called the centre of a circle.
CIRCLE-LENGTH OF THE ARC & AREA OF A
SECTOR
An arc of a circle is a "portion" of
the circumference of the circle.
In a circle, the length of an arc is
the length of the portion of the
circumference.
ARC OF THE CIRCLE
CIRCLE-LENGTH OF THE ARC & AREA OF A
SECTOR
In a circle, the degree measure of
an arc is equal to the measure of
the central angle that intercepts
the arc.
DEGREE MEASURE OF AN
ARC
m∠ AOB = 50o ∴ m(arc AB) = 50o
m∠ COB = 80o ∴ m(arc BC) = 80o
m∠ DOC = 50o ∴ m(arc DC) = 50o
CIRCLE-LENGTH OF THE ARC & AREA OF A SECTOR
LENGTH OF THE ARC
Measure of
the arc (𝜃)
𝜃
360
Length of the
arc
360o 360
360
=1 1 x 2𝜋r
180o 180
360
=
1
2
1
2
x 2𝜋r
900 90
360
=
1
4
1
4
x 2𝜋r
60o 60
360
=
1
6
1
6
x 2𝜋r
45O 45
360
=
1
8
1
8
x 2𝜋r
𝜃o 𝜃
360
𝜃
360
x 2𝜋r
LENGTH (ARC) =
𝜃
360
x 2𝜋r
CIRCLE-LENGTH OF THE ARC & AREA OF A
SECTOR
A circular sector or circle sector, is the
portion of a circle enclosed by two radii
and an arc.
SECTOR of a circle
CIRCLE-LENGTH OF THE ARC & AREA OF A
SECTOR
Measure of
the arc (𝜃)
𝜃
360
Area of the
sector
360o 360
360
=1 1 x 𝜋r2
180o 180
360
=
1
2
1
2
x 𝜋r2
900 90
360
=
1
4
1
4
x 𝜋r2
60o 60
360
=
1
6
1
6
x 𝜋r2
45O 45
360
=
1
8
1
8
x 𝜋r2
𝜃o 𝜃
360
𝜃
360
x 𝜋r2
A (sector) =
𝜃
360
x 𝜋r2
AREA OF A SECTOR
CIRCLE-LENGTH OF THE ARC &
AREA OF A SECTOR
A (sector) =
𝜃
360
x 𝜋r2
LENGTH (ARC) =
𝜃
360
x 2𝜋r
A(SECTOR) = l(ARC) X
𝑅
2
APPLICATION
Radius of a circle is 10 cm. Measure of an arc
of the circle is 54o. Find the l(arc) and area of
the sector.
R = 10cm, 𝜃 =54o
l (arc)=
𝜃
360
x 2𝜋r
=
54
360
x 2 x 3.14 x 10 = 9.42 cm
Radius of a circle is 10 cm, Area of a sector is 100 cm2. Find
the area of its corresponding major sector.
APPLICATION
ALSO FIND:
If AO = 7 cm, 𝜃 = 60o then find 1) area of the circle, 2) A(O-ADC)
and 3) A(O-ABC)
HINT:
step 1: Find the area of a circle
Step 2: subtract area of the sector from the
area of the circle
APPLICATION
R = 3.4 cm, perimeter of the sector O-ADC = 12.8 cm.
Find A(O-ADC)
perimeter of the sector O-ADC = OA +OC + l(arc ADC)
12.8 = 3.4 + 3.4 +l(arc ADC)
12.8 – 6.8 = l(arc ADC)
6 = l(arc ADC)
A(O-ADC) = l(arc ADC) x
3.4
2
= 6 x
3.4
2
= 10. 2 sq cm
APPLICATION
PQ = 14cm, QR = 21 cm. PQRS is a rectangle
Find area of X,Y and Z regions.
Hint: step 1: find a(rectangle)
step 2: Find radius of sector X and Y region.
step 3: find area of sectors X and Y
step 4: subtract the sum of the area of
sectors X and Y from area of rectangle.
APPLICATION
LMN is an equilateral triangle, LM = 14 cm.
Three sectors are drawn with vertices as
centres and radius 7 cm.
Find
A( triangle LMN)
Area of any one sector
Total area of all the three sectors
Area of the shaded region
A(equilateral triangle) =
3
4
x side2,
( 3 =1.73)
APPLICATION
A (circle) = 𝜋r2= 36 𝜋 sq cm
A(sector) =
𝜃
360
x 𝜋r2
15𝜋 x 360= 𝜃 x 36 𝜋
𝜃 =1500
R = 6cm, A(sector) = 15𝜋 sq cm. Find: measure of the
arc(𝜃) and length of the arc corresponding to the sector
A(sector) = l(arc) x
𝑟
2
15𝜋 = l(arc) x 3
5𝜋 cm = l(arc)
APPLICATION
Square ABCD is inscribed in the sector A-PCQ.
The radius of sector C-BXD is 20 cm. Find the
area of shaded region
Side of the square = radius of sector C-BXD
AC = 20 2cm
Area of shaded region = A(A-PCQ) – A(C-BXD)
=
90
360
x 𝜋(20 2)2 -
90
360
x 𝜋(20)2
=
1
4
x 3-14 x 400 (2-1)
= 3.14 x 100
=314 sq cm
THANK YOU

Circle - arc sector

  • 1.
    MENSURATION-sk CIRCLE-LENGTH OF THEARC & AREA OF A SECTOR STD X MAHARASHTRA STATE BOARD OF EDUCATION, MUMBAI
  • 2.
    CIRCLE-LENGTH OF THEARC & AREA OF A SECTOR Circleis a set of all points in the plane which are equidistant from a given point, called the centre of a circle.
  • 3.
    CIRCLE-LENGTH OF THEARC & AREA OF A SECTOR An arc of a circle is a "portion" of the circumference of the circle. In a circle, the length of an arc is the length of the portion of the circumference. ARC OF THE CIRCLE
  • 4.
    CIRCLE-LENGTH OF THEARC & AREA OF A SECTOR In a circle, the degree measure of an arc is equal to the measure of the central angle that intercepts the arc. DEGREE MEASURE OF AN ARC m∠ AOB = 50o ∴ m(arc AB) = 50o m∠ COB = 80o ∴ m(arc BC) = 80o m∠ DOC = 50o ∴ m(arc DC) = 50o
  • 5.
    CIRCLE-LENGTH OF THEARC & AREA OF A SECTOR LENGTH OF THE ARC Measure of the arc (𝜃) 𝜃 360 Length of the arc 360o 360 360 =1 1 x 2𝜋r 180o 180 360 = 1 2 1 2 x 2𝜋r 900 90 360 = 1 4 1 4 x 2𝜋r 60o 60 360 = 1 6 1 6 x 2𝜋r 45O 45 360 = 1 8 1 8 x 2𝜋r 𝜃o 𝜃 360 𝜃 360 x 2𝜋r LENGTH (ARC) = 𝜃 360 x 2𝜋r
  • 6.
    CIRCLE-LENGTH OF THEARC & AREA OF A SECTOR A circular sector or circle sector, is the portion of a circle enclosed by two radii and an arc. SECTOR of a circle
  • 7.
    CIRCLE-LENGTH OF THEARC & AREA OF A SECTOR Measure of the arc (𝜃) 𝜃 360 Area of the sector 360o 360 360 =1 1 x 𝜋r2 180o 180 360 = 1 2 1 2 x 𝜋r2 900 90 360 = 1 4 1 4 x 𝜋r2 60o 60 360 = 1 6 1 6 x 𝜋r2 45O 45 360 = 1 8 1 8 x 𝜋r2 𝜃o 𝜃 360 𝜃 360 x 𝜋r2 A (sector) = 𝜃 360 x 𝜋r2 AREA OF A SECTOR
  • 8.
    CIRCLE-LENGTH OF THEARC & AREA OF A SECTOR A (sector) = 𝜃 360 x 𝜋r2 LENGTH (ARC) = 𝜃 360 x 2𝜋r A(SECTOR) = l(ARC) X 𝑅 2
  • 9.
    APPLICATION Radius of acircle is 10 cm. Measure of an arc of the circle is 54o. Find the l(arc) and area of the sector. R = 10cm, 𝜃 =54o l (arc)= 𝜃 360 x 2𝜋r = 54 360 x 2 x 3.14 x 10 = 9.42 cm
  • 10.
    Radius of acircle is 10 cm, Area of a sector is 100 cm2. Find the area of its corresponding major sector. APPLICATION ALSO FIND: If AO = 7 cm, 𝜃 = 60o then find 1) area of the circle, 2) A(O-ADC) and 3) A(O-ABC) HINT: step 1: Find the area of a circle Step 2: subtract area of the sector from the area of the circle
  • 11.
    APPLICATION R = 3.4cm, perimeter of the sector O-ADC = 12.8 cm. Find A(O-ADC) perimeter of the sector O-ADC = OA +OC + l(arc ADC) 12.8 = 3.4 + 3.4 +l(arc ADC) 12.8 – 6.8 = l(arc ADC) 6 = l(arc ADC) A(O-ADC) = l(arc ADC) x 3.4 2 = 6 x 3.4 2 = 10. 2 sq cm
  • 12.
    APPLICATION PQ = 14cm,QR = 21 cm. PQRS is a rectangle Find area of X,Y and Z regions. Hint: step 1: find a(rectangle) step 2: Find radius of sector X and Y region. step 3: find area of sectors X and Y step 4: subtract the sum of the area of sectors X and Y from area of rectangle.
  • 13.
    APPLICATION LMN is anequilateral triangle, LM = 14 cm. Three sectors are drawn with vertices as centres and radius 7 cm. Find A( triangle LMN) Area of any one sector Total area of all the three sectors Area of the shaded region A(equilateral triangle) = 3 4 x side2, ( 3 =1.73)
  • 14.
    APPLICATION A (circle) =𝜋r2= 36 𝜋 sq cm A(sector) = 𝜃 360 x 𝜋r2 15𝜋 x 360= 𝜃 x 36 𝜋 𝜃 =1500 R = 6cm, A(sector) = 15𝜋 sq cm. Find: measure of the arc(𝜃) and length of the arc corresponding to the sector A(sector) = l(arc) x 𝑟 2 15𝜋 = l(arc) x 3 5𝜋 cm = l(arc)
  • 15.
    APPLICATION Square ABCD isinscribed in the sector A-PCQ. The radius of sector C-BXD is 20 cm. Find the area of shaded region Side of the square = radius of sector C-BXD AC = 20 2cm Area of shaded region = A(A-PCQ) – A(C-BXD) = 90 360 x 𝜋(20 2)2 - 90 360 x 𝜋(20)2 = 1 4 x 3-14 x 400 (2-1) = 3.14 x 100 =314 sq cm
  • 16.