AREAS RELATED
TO CIRCLE
ar (A – MXN)
Sol.
B
D
C
15 m
Q. A Horse is tied to a peg at one corner of a square shaped grass
field of side 15 m by means of a 5 m long rope.
Find the area of that part of the field in which the horse can graze.
Also find the increase in the grazing area if the rope were 10m long
instead of 5m. (Use π = 3.14)
Sector
M
N
X
What is the formula to find area of sector?
Length of rope = 5 m
Radius of sector (r) = length of rope = 5 m
∴
✔
θ
360
πr2
θ = 90°
✔
?
?
ar (A – MXN) =
θ
360
×
π
r2
=
360
90
×
3.14
×
5
×
5
1
4
=
4
1
×
×
5
×
5
100
314
A
5 m
∴
Q. A Horse is tied to a peg at one corner of a square shaped grass
field of side 15 m by means of a 5 m long rope.
Find the area of that part of the field in which the horse can graze.
the increase in the grazing area if the rope were 10m long instead of 5m. (Use π = 3.14)
=
4
1
×
×
5
×
5
100
314
ar (A – MXN)
Sol.
2
157
=
2 × 100
3925
=
19.625 m2
Area of the part of the field in which horse can graze is 19.625 m2
∴
B
D
C
15 m
A
M
N
X
5 m
∴
ar (A – MXN)
=
100
1962.5
1962.5
10
×
10
×
Sol.
Q. A Horse is tied to a peg at one corner of a square shaped grass
field of side 15 m by means of a 5 m long rope.
Find the area of that part of the field in which the horse can graze.
the increase in the grazing area if the rope were 10m long instead of 5m. (Use π = 3.14)
A
B
D
C
15 m
V
S
U
M
N
X
10 m
What is the formula to find area of sector?
?
Length of rope = 10 m
Radius of sector (r) = length of rope = 10m
✔
θ
360
πr2
θ = 90°
?
✔
ar (A – SUV) =
θ
360
×
π
r2
=
360
90
3.14
×
1
4
2
5
5
=
3.14 × 5 × 5
=
78.5 m2
ar (A – SUV)
∴
Sector
Area of the part of the field in which horse can graze is 78.5 m2
∴
Sol.
Q. A Horse is tied to a peg at one corner of a square shaped grass
field of side 15 m by means of a 5 m long rope.
Find the area of that part of the field in which the horse can graze.
the increase in the grazing area if the rope were 10m long instead of 5m. (Use π = 3.14)
B
D
C
15 m
A
Increase in grazing area
–
ar(A–MXN)
10 m
ar (A – SUV) = 78.5 m2
ar (A – MXN) = 19.625 m2
= ar(A–SUV)
Increase in grazing area =
–
ar(A–MXN)
ar(A–SUV)
=
78.5
–
19.625
=
58.875 m2
Increase in the grazing area is 58.875 m2
∴
M
N
X
X
V
S
U