RS Aggarwal Solutions: Concept of Perimeter and Area Class 6 Notes | EduRev

Mathematics (Maths) Class 6

Class 6 : RS Aggarwal Solutions: Concept of Perimeter and Area Class 6 Notes | EduRev

 Page 1


Points to Remember :
Region. A plane figure together with its interior is called the region enclosed by the plane figure.
Area. The measurement of the region encloed by a plane figure is called the area of the plane
figure. The standard unit of area is cm
2
.
Source Formulae :
1. Area of rectangle = (length × breadth) Sq. units.
2. Length of a rectangle 
F
H
G
I
K
J
Area
Breadth
units
3. Breadth of a rectangle 
F
H
G
I
K
J
Area
Length
units
4. Perimeter of a rectangle = 2 × (length + breadth) units.
5. Area of a square = (Side)
2
 sq. units.
6. Perimeter of a square = (4 × side) units.
Units of Area. The standard unit of area is cm
2
.
For measuring areas of lands and fields, we use larger units such as m
2
, areas and hectares etc.
Conversion of Units :
Units of Length Units of Area
  1 cm = 10 mm  1 cm
2
 = (10 × 10) mm
2
 = 100 mm
2
    1 m = 100 cm.    1 m
2
 = (100 × 100) cm
2
 = 10000 cm
2
  1 dm = 10 cm  1 dm
2
 = (10 × 10) cm
2
 = 100 cm
2
1 dam = 10 m 1 dam
2
 = (10 × 10) m
2
 = 100 m
2
 1 km = 10 hm  1 km
2
 = (10 × 10) hm
2
 = 100 hm
2
 1 km = 1000 m 1 km
2
 = (1000 × 1000) m
2
 = 1000000 m
2
Page 2


Points to Remember :
Region. A plane figure together with its interior is called the region enclosed by the plane figure.
Area. The measurement of the region encloed by a plane figure is called the area of the plane
figure. The standard unit of area is cm
2
.
Source Formulae :
1. Area of rectangle = (length × breadth) Sq. units.
2. Length of a rectangle 
F
H
G
I
K
J
Area
Breadth
units
3. Breadth of a rectangle 
F
H
G
I
K
J
Area
Length
units
4. Perimeter of a rectangle = 2 × (length + breadth) units.
5. Area of a square = (Side)
2
 sq. units.
6. Perimeter of a square = (4 × side) units.
Units of Area. The standard unit of area is cm
2
.
For measuring areas of lands and fields, we use larger units such as m
2
, areas and hectares etc.
Conversion of Units :
Units of Length Units of Area
  1 cm = 10 mm  1 cm
2
 = (10 × 10) mm
2
 = 100 mm
2
    1 m = 100 cm.    1 m
2
 = (100 × 100) cm
2
 = 10000 cm
2
  1 dm = 10 cm  1 dm
2
 = (10 × 10) cm
2
 = 100 cm
2
1 dam = 10 m 1 dam
2
 = (10 × 10) m
2
 = 100 m
2
 1 km = 10 hm  1 km
2
 = (10 × 10) hm
2
 = 100 hm
2
 1 km = 1000 m 1 km
2
 = (1000 × 1000) m
2
 = 1000000 m
2
           Q. 1. Find the perimeter of a rectangle in which :
(i) length = 16·8 cm and breadth = 6·2 cm
(ii) length = 2 m 25 cm and breadth = 1 m
50 cm
(iii) length = 8 m 5 dm and breadth = 6 m
8 dm
Sol. (i) Length (l) = 16·8 cm
   Breadth (b) = 6·2 cm
  Perimeter = 2 (l + b)
         = 2 (16·8 + 6·2) cm
         = 2 × 23 = 46 cm   Ans.
(ii) Length of rectanlge (l) = 2 m, 25 cm
= 2.25 m
and breadth (b) = 1 m 50 cm = 1.50
m
D C
A B 2m 25cm
1m 50cm
Perimeter = 2 (l + b)
= 2 (2.25 + 1.50) m
= 2 × 3.75 m = 7.50 m
= 7 m 50 cm
(iii) Length of a rectangle (l) = 8 m 5 dm
= 8.5 m
and breadth (b) = 6 m 8 dm = 6.8 m
D C
A B 8m 5dm
6m 8dm
Perimeter = 2 (l + b)
= 2 (8.5 + 6.8) m
= 2 × 15.3 m = 30.6 m
= 30 m 6 dm
Q.2. Find the cost of fencing a rectangular field
62 m long and 33 m wide at Rs. 16 per
metre.
Sol. Length of rectangular field (l) = 62 m
and breadth (b) = 33 m
C D
B A 33 m
62 m
Perimeter = 2 (l + b)
= 2 (62 + 33) = 2 × 95 m = 190 m
Rate of fencing = Rs. 16 per m
Total cost of fencing the field
= Rs. 16 × 190 = Rs. 3040
Q.3. The length and the breadth of a rectangular
field are in the ratio 5 : 3. If its perimeter
is 128 m, find the dimensions of the field.
Sol. Perimeter of field = 128 m
Length + Breadth = 
2
128
 = 64 m
Ratio in length and breadth = 5 : 3
Let length (l) = 5x
Then breadth = 3x
5x + 3x = 64  8x = 64
x = 
8
64
 = 8
Length of the field = 5x = 5 × 8 = 40 m
and breadth = 3x = 3 × 8 = 24 m
Q.4. The cost of fencing a rectangular field at
Rs. 18 per metre is Rs. 1980. If the
width of the field is 23 m, find its length.
Page 3


Points to Remember :
Region. A plane figure together with its interior is called the region enclosed by the plane figure.
Area. The measurement of the region encloed by a plane figure is called the area of the plane
figure. The standard unit of area is cm
2
.
Source Formulae :
1. Area of rectangle = (length × breadth) Sq. units.
2. Length of a rectangle 
F
H
G
I
K
J
Area
Breadth
units
3. Breadth of a rectangle 
F
H
G
I
K
J
Area
Length
units
4. Perimeter of a rectangle = 2 × (length + breadth) units.
5. Area of a square = (Side)
2
 sq. units.
6. Perimeter of a square = (4 × side) units.
Units of Area. The standard unit of area is cm
2
.
For measuring areas of lands and fields, we use larger units such as m
2
, areas and hectares etc.
Conversion of Units :
Units of Length Units of Area
  1 cm = 10 mm  1 cm
2
 = (10 × 10) mm
2
 = 100 mm
2
    1 m = 100 cm.    1 m
2
 = (100 × 100) cm
2
 = 10000 cm
2
  1 dm = 10 cm  1 dm
2
 = (10 × 10) cm
2
 = 100 cm
2
1 dam = 10 m 1 dam
2
 = (10 × 10) m
2
 = 100 m
2
 1 km = 10 hm  1 km
2
 = (10 × 10) hm
2
 = 100 hm
2
 1 km = 1000 m 1 km
2
 = (1000 × 1000) m
2
 = 1000000 m
2
           Q. 1. Find the perimeter of a rectangle in which :
(i) length = 16·8 cm and breadth = 6·2 cm
(ii) length = 2 m 25 cm and breadth = 1 m
50 cm
(iii) length = 8 m 5 dm and breadth = 6 m
8 dm
Sol. (i) Length (l) = 16·8 cm
   Breadth (b) = 6·2 cm
  Perimeter = 2 (l + b)
         = 2 (16·8 + 6·2) cm
         = 2 × 23 = 46 cm   Ans.
(ii) Length of rectanlge (l) = 2 m, 25 cm
= 2.25 m
and breadth (b) = 1 m 50 cm = 1.50
m
D C
A B 2m 25cm
1m 50cm
Perimeter = 2 (l + b)
= 2 (2.25 + 1.50) m
= 2 × 3.75 m = 7.50 m
= 7 m 50 cm
(iii) Length of a rectangle (l) = 8 m 5 dm
= 8.5 m
and breadth (b) = 6 m 8 dm = 6.8 m
D C
A B 8m 5dm
6m 8dm
Perimeter = 2 (l + b)
= 2 (8.5 + 6.8) m
= 2 × 15.3 m = 30.6 m
= 30 m 6 dm
Q.2. Find the cost of fencing a rectangular field
62 m long and 33 m wide at Rs. 16 per
metre.
Sol. Length of rectangular field (l) = 62 m
and breadth (b) = 33 m
C D
B A 33 m
62 m
Perimeter = 2 (l + b)
= 2 (62 + 33) = 2 × 95 m = 190 m
Rate of fencing = Rs. 16 per m
Total cost of fencing the field
= Rs. 16 × 190 = Rs. 3040
Q.3. The length and the breadth of a rectangular
field are in the ratio 5 : 3. If its perimeter
is 128 m, find the dimensions of the field.
Sol. Perimeter of field = 128 m
Length + Breadth = 
2
128
 = 64 m
Ratio in length and breadth = 5 : 3
Let length (l) = 5x
Then breadth = 3x
5x + 3x = 64  8x = 64
x = 
8
64
 = 8
Length of the field = 5x = 5 × 8 = 40 m
and breadth = 3x = 3 × 8 = 24 m
Q.4. The cost of fencing a rectangular field at
Rs. 18 per metre is Rs. 1980. If the
width of the field is 23 m, find its length.
Sol. Cost of fencing a rectangular field
= Rs. 18 per m
Total cost = Rs. 1980
Perimeter of the field = 
Rate
cost Total
= 
18
1980
 = 110 m
and length + breadth = 
2
110
 = 55 m
Width of the field = 23 m
Length = 55 – 23 = 32 m
Q.5. The length and the breadth of a rectangular
field are in the ratio 7 : 4. The cost of
fencing the field at Rs. 25 per metre is
Rs. 3300. Find the dimensions of the field.
Sol. Total cost of fencing a rectangular field
= Rs. 3300
Rate of fencing = Rs. 25 per m
Perimeter = 
Rate
cost Total
 = 
25
3300
= 132 m
and length + breadth = 
2
Perimeter
= 
2
132
 = 66 m
Now ratio in length and breadth = 7 : 4
Let length = 7x
Then breadth = 4x
7x + 4x = 66  11x = 66
x = 
11
66
 = 6
Length of the field = 7x = 7 × 6 = 42 m
and breadth = 4x = 4 × 6 = 24 m
Q.6. Find the perimeter of a square, each of
whose sides measures :
(i) 3.8 cm (ii) 4.6 m (iii) 2 m 5 dm
Sol. (i) Side of square = 3.8 cm
Perimeter = 4 × side
= 4 × 3.8 cm = 15.2 cm
(ii) Side of a square = 4.6 m
Perimeter = 4 × side = 4 × 4.6 m
= 18.4 m
(iii) Side of a square = 2 m 5 dm = 2.5 m
Perimeter = 4 × side
= 4 × 2.5 m = 10 m
Q.7. The cost of putting a fence around a
square field at Rs. 35 per metre is Rs.
4480. Find the length of each side of the
field.
Sol. Total cost of fencing a square field
= Rs. 4480
Rate of fencing = Rs. 35 per m
Perimeter of square field = 
Rate
cost Total
= 
35
4480
 = 128 m
Side of the square = 
4
Perimeter
= 
4
128
 = 32 m
Q. 8. Each side of a square field measures 21
m, Adjacent to this field, there is a
rectangular field having its sides in the
ratio 4 : 3. If the perimeters of both of
the fields are equal, find the dimensions
of the rectangular field.
Sol. Side of a square field (a) = 21 m
 Perimeter = 4 a = 4 × 21 = 84 m
 Perimeter of rectangular field = 84 m
Ratio in length and breadth = 4 : 3
Let     length (l) = 4 x
and breadth (b) = 3 x
 Perimeter = 2 (l + b)
Page 4


Points to Remember :
Region. A plane figure together with its interior is called the region enclosed by the plane figure.
Area. The measurement of the region encloed by a plane figure is called the area of the plane
figure. The standard unit of area is cm
2
.
Source Formulae :
1. Area of rectangle = (length × breadth) Sq. units.
2. Length of a rectangle 
F
H
G
I
K
J
Area
Breadth
units
3. Breadth of a rectangle 
F
H
G
I
K
J
Area
Length
units
4. Perimeter of a rectangle = 2 × (length + breadth) units.
5. Area of a square = (Side)
2
 sq. units.
6. Perimeter of a square = (4 × side) units.
Units of Area. The standard unit of area is cm
2
.
For measuring areas of lands and fields, we use larger units such as m
2
, areas and hectares etc.
Conversion of Units :
Units of Length Units of Area
  1 cm = 10 mm  1 cm
2
 = (10 × 10) mm
2
 = 100 mm
2
    1 m = 100 cm.    1 m
2
 = (100 × 100) cm
2
 = 10000 cm
2
  1 dm = 10 cm  1 dm
2
 = (10 × 10) cm
2
 = 100 cm
2
1 dam = 10 m 1 dam
2
 = (10 × 10) m
2
 = 100 m
2
 1 km = 10 hm  1 km
2
 = (10 × 10) hm
2
 = 100 hm
2
 1 km = 1000 m 1 km
2
 = (1000 × 1000) m
2
 = 1000000 m
2
           Q. 1. Find the perimeter of a rectangle in which :
(i) length = 16·8 cm and breadth = 6·2 cm
(ii) length = 2 m 25 cm and breadth = 1 m
50 cm
(iii) length = 8 m 5 dm and breadth = 6 m
8 dm
Sol. (i) Length (l) = 16·8 cm
   Breadth (b) = 6·2 cm
  Perimeter = 2 (l + b)
         = 2 (16·8 + 6·2) cm
         = 2 × 23 = 46 cm   Ans.
(ii) Length of rectanlge (l) = 2 m, 25 cm
= 2.25 m
and breadth (b) = 1 m 50 cm = 1.50
m
D C
A B 2m 25cm
1m 50cm
Perimeter = 2 (l + b)
= 2 (2.25 + 1.50) m
= 2 × 3.75 m = 7.50 m
= 7 m 50 cm
(iii) Length of a rectangle (l) = 8 m 5 dm
= 8.5 m
and breadth (b) = 6 m 8 dm = 6.8 m
D C
A B 8m 5dm
6m 8dm
Perimeter = 2 (l + b)
= 2 (8.5 + 6.8) m
= 2 × 15.3 m = 30.6 m
= 30 m 6 dm
Q.2. Find the cost of fencing a rectangular field
62 m long and 33 m wide at Rs. 16 per
metre.
Sol. Length of rectangular field (l) = 62 m
and breadth (b) = 33 m
C D
B A 33 m
62 m
Perimeter = 2 (l + b)
= 2 (62 + 33) = 2 × 95 m = 190 m
Rate of fencing = Rs. 16 per m
Total cost of fencing the field
= Rs. 16 × 190 = Rs. 3040
Q.3. The length and the breadth of a rectangular
field are in the ratio 5 : 3. If its perimeter
is 128 m, find the dimensions of the field.
Sol. Perimeter of field = 128 m
Length + Breadth = 
2
128
 = 64 m
Ratio in length and breadth = 5 : 3
Let length (l) = 5x
Then breadth = 3x
5x + 3x = 64  8x = 64
x = 
8
64
 = 8
Length of the field = 5x = 5 × 8 = 40 m
and breadth = 3x = 3 × 8 = 24 m
Q.4. The cost of fencing a rectangular field at
Rs. 18 per metre is Rs. 1980. If the
width of the field is 23 m, find its length.
Sol. Cost of fencing a rectangular field
= Rs. 18 per m
Total cost = Rs. 1980
Perimeter of the field = 
Rate
cost Total
= 
18
1980
 = 110 m
and length + breadth = 
2
110
 = 55 m
Width of the field = 23 m
Length = 55 – 23 = 32 m
Q.5. The length and the breadth of a rectangular
field are in the ratio 7 : 4. The cost of
fencing the field at Rs. 25 per metre is
Rs. 3300. Find the dimensions of the field.
Sol. Total cost of fencing a rectangular field
= Rs. 3300
Rate of fencing = Rs. 25 per m
Perimeter = 
Rate
cost Total
 = 
25
3300
= 132 m
and length + breadth = 
2
Perimeter
= 
2
132
 = 66 m
Now ratio in length and breadth = 7 : 4
Let length = 7x
Then breadth = 4x
7x + 4x = 66  11x = 66
x = 
11
66
 = 6
Length of the field = 7x = 7 × 6 = 42 m
and breadth = 4x = 4 × 6 = 24 m
Q.6. Find the perimeter of a square, each of
whose sides measures :
(i) 3.8 cm (ii) 4.6 m (iii) 2 m 5 dm
Sol. (i) Side of square = 3.8 cm
Perimeter = 4 × side
= 4 × 3.8 cm = 15.2 cm
(ii) Side of a square = 4.6 m
Perimeter = 4 × side = 4 × 4.6 m
= 18.4 m
(iii) Side of a square = 2 m 5 dm = 2.5 m
Perimeter = 4 × side
= 4 × 2.5 m = 10 m
Q.7. The cost of putting a fence around a
square field at Rs. 35 per metre is Rs.
4480. Find the length of each side of the
field.
Sol. Total cost of fencing a square field
= Rs. 4480
Rate of fencing = Rs. 35 per m
Perimeter of square field = 
Rate
cost Total
= 
35
4480
 = 128 m
Side of the square = 
4
Perimeter
= 
4
128
 = 32 m
Q. 8. Each side of a square field measures 21
m, Adjacent to this field, there is a
rectangular field having its sides in the
ratio 4 : 3. If the perimeters of both of
the fields are equal, find the dimensions
of the rectangular field.
Sol. Side of a square field (a) = 21 m
 Perimeter = 4 a = 4 × 21 = 84 m
 Perimeter of rectangular field = 84 m
Ratio in length and breadth = 4 : 3
Let     length (l) = 4 x
and breadth (b) = 3 x
 Perimeter = 2 (l + b)
  84 = 2 (4 x + 3 x) = 2 × 7 x = 14 x
    x
84
14
6
   Length = 4 x = 4 × 6
 = 24 metre
and breadth = 3 x = 3 × 6
  = 18 metre   Ans.
Q.9. Find the perimeter of
(i) a triangle of sides 7.8 cm, 6.5 cm and
5.9 cm,
(ii) an equilateral triangle of side 9.4 cm,
(iii) an isosocels triangle with equal sides 8.5
cm each and third side 7 cm.
Sol. (i) Sides of a triangle are 7.8 cm, 6.5 cm
and 5.9 cm
A
B C 7.8 cm
6.5 cm
5.9 cm
Perimeter = Sum of sides
= (7.8 + 6.5 + 5.9) cm = 20.2 cm
(ii) Each side of an equilateral triangles
= 9.4 cm
A
B C
9.4 cm
9.4 cm
9.4 cm
Perimeter = 3 × side
= 3 × 9.4 = 28.2 cm
(iii) Each of equal sides = 8.5 cm
and third side = 7 cm
A
C B
8.5 cm
7 cm
8.5 cm
Perimeter = Sum of sides
= (8.5 + 8.5 + 7) cm = 24 cm
Q.10. Find the perimeter of
(i) a regular pentagon of side 8 cm,
(ii) a regular octagon of side 4.5 cm,
(iii) a regular decagon of side 3.6 cm.
Sol. (i) Each side of a regular pentagon
= 8 cm
Perimter = 5 × Side
= 5 × 8 = 40 cm
(ii) Each side of an octagon = 4.5 cm
Perimeter = 8 × Side
= 8 × 4.5 = 36 cm
(iii) Each side of a regular decagon = 3.6 cm
Perimeter = 10 × Side
= 10 × 3.6 = 36 cm
Q.11. Find the perimeter of each of the
following figures :
35cm
45 cm
27 cm
35cm
18 cm
18 cm
18 cm
18 cm
4 cm
16 cm
16 cm
8 cm
4 cm
12 cm
12 cm
Page 5


Points to Remember :
Region. A plane figure together with its interior is called the region enclosed by the plane figure.
Area. The measurement of the region encloed by a plane figure is called the area of the plane
figure. The standard unit of area is cm
2
.
Source Formulae :
1. Area of rectangle = (length × breadth) Sq. units.
2. Length of a rectangle 
F
H
G
I
K
J
Area
Breadth
units
3. Breadth of a rectangle 
F
H
G
I
K
J
Area
Length
units
4. Perimeter of a rectangle = 2 × (length + breadth) units.
5. Area of a square = (Side)
2
 sq. units.
6. Perimeter of a square = (4 × side) units.
Units of Area. The standard unit of area is cm
2
.
For measuring areas of lands and fields, we use larger units such as m
2
, areas and hectares etc.
Conversion of Units :
Units of Length Units of Area
  1 cm = 10 mm  1 cm
2
 = (10 × 10) mm
2
 = 100 mm
2
    1 m = 100 cm.    1 m
2
 = (100 × 100) cm
2
 = 10000 cm
2
  1 dm = 10 cm  1 dm
2
 = (10 × 10) cm
2
 = 100 cm
2
1 dam = 10 m 1 dam
2
 = (10 × 10) m
2
 = 100 m
2
 1 km = 10 hm  1 km
2
 = (10 × 10) hm
2
 = 100 hm
2
 1 km = 1000 m 1 km
2
 = (1000 × 1000) m
2
 = 1000000 m
2
           Q. 1. Find the perimeter of a rectangle in which :
(i) length = 16·8 cm and breadth = 6·2 cm
(ii) length = 2 m 25 cm and breadth = 1 m
50 cm
(iii) length = 8 m 5 dm and breadth = 6 m
8 dm
Sol. (i) Length (l) = 16·8 cm
   Breadth (b) = 6·2 cm
  Perimeter = 2 (l + b)
         = 2 (16·8 + 6·2) cm
         = 2 × 23 = 46 cm   Ans.
(ii) Length of rectanlge (l) = 2 m, 25 cm
= 2.25 m
and breadth (b) = 1 m 50 cm = 1.50
m
D C
A B 2m 25cm
1m 50cm
Perimeter = 2 (l + b)
= 2 (2.25 + 1.50) m
= 2 × 3.75 m = 7.50 m
= 7 m 50 cm
(iii) Length of a rectangle (l) = 8 m 5 dm
= 8.5 m
and breadth (b) = 6 m 8 dm = 6.8 m
D C
A B 8m 5dm
6m 8dm
Perimeter = 2 (l + b)
= 2 (8.5 + 6.8) m
= 2 × 15.3 m = 30.6 m
= 30 m 6 dm
Q.2. Find the cost of fencing a rectangular field
62 m long and 33 m wide at Rs. 16 per
metre.
Sol. Length of rectangular field (l) = 62 m
and breadth (b) = 33 m
C D
B A 33 m
62 m
Perimeter = 2 (l + b)
= 2 (62 + 33) = 2 × 95 m = 190 m
Rate of fencing = Rs. 16 per m
Total cost of fencing the field
= Rs. 16 × 190 = Rs. 3040
Q.3. The length and the breadth of a rectangular
field are in the ratio 5 : 3. If its perimeter
is 128 m, find the dimensions of the field.
Sol. Perimeter of field = 128 m
Length + Breadth = 
2
128
 = 64 m
Ratio in length and breadth = 5 : 3
Let length (l) = 5x
Then breadth = 3x
5x + 3x = 64  8x = 64
x = 
8
64
 = 8
Length of the field = 5x = 5 × 8 = 40 m
and breadth = 3x = 3 × 8 = 24 m
Q.4. The cost of fencing a rectangular field at
Rs. 18 per metre is Rs. 1980. If the
width of the field is 23 m, find its length.
Sol. Cost of fencing a rectangular field
= Rs. 18 per m
Total cost = Rs. 1980
Perimeter of the field = 
Rate
cost Total
= 
18
1980
 = 110 m
and length + breadth = 
2
110
 = 55 m
Width of the field = 23 m
Length = 55 – 23 = 32 m
Q.5. The length and the breadth of a rectangular
field are in the ratio 7 : 4. The cost of
fencing the field at Rs. 25 per metre is
Rs. 3300. Find the dimensions of the field.
Sol. Total cost of fencing a rectangular field
= Rs. 3300
Rate of fencing = Rs. 25 per m
Perimeter = 
Rate
cost Total
 = 
25
3300
= 132 m
and length + breadth = 
2
Perimeter
= 
2
132
 = 66 m
Now ratio in length and breadth = 7 : 4
Let length = 7x
Then breadth = 4x
7x + 4x = 66  11x = 66
x = 
11
66
 = 6
Length of the field = 7x = 7 × 6 = 42 m
and breadth = 4x = 4 × 6 = 24 m
Q.6. Find the perimeter of a square, each of
whose sides measures :
(i) 3.8 cm (ii) 4.6 m (iii) 2 m 5 dm
Sol. (i) Side of square = 3.8 cm
Perimeter = 4 × side
= 4 × 3.8 cm = 15.2 cm
(ii) Side of a square = 4.6 m
Perimeter = 4 × side = 4 × 4.6 m
= 18.4 m
(iii) Side of a square = 2 m 5 dm = 2.5 m
Perimeter = 4 × side
= 4 × 2.5 m = 10 m
Q.7. The cost of putting a fence around a
square field at Rs. 35 per metre is Rs.
4480. Find the length of each side of the
field.
Sol. Total cost of fencing a square field
= Rs. 4480
Rate of fencing = Rs. 35 per m
Perimeter of square field = 
Rate
cost Total
= 
35
4480
 = 128 m
Side of the square = 
4
Perimeter
= 
4
128
 = 32 m
Q. 8. Each side of a square field measures 21
m, Adjacent to this field, there is a
rectangular field having its sides in the
ratio 4 : 3. If the perimeters of both of
the fields are equal, find the dimensions
of the rectangular field.
Sol. Side of a square field (a) = 21 m
 Perimeter = 4 a = 4 × 21 = 84 m
 Perimeter of rectangular field = 84 m
Ratio in length and breadth = 4 : 3
Let     length (l) = 4 x
and breadth (b) = 3 x
 Perimeter = 2 (l + b)
  84 = 2 (4 x + 3 x) = 2 × 7 x = 14 x
    x
84
14
6
   Length = 4 x = 4 × 6
 = 24 metre
and breadth = 3 x = 3 × 6
  = 18 metre   Ans.
Q.9. Find the perimeter of
(i) a triangle of sides 7.8 cm, 6.5 cm and
5.9 cm,
(ii) an equilateral triangle of side 9.4 cm,
(iii) an isosocels triangle with equal sides 8.5
cm each and third side 7 cm.
Sol. (i) Sides of a triangle are 7.8 cm, 6.5 cm
and 5.9 cm
A
B C 7.8 cm
6.5 cm
5.9 cm
Perimeter = Sum of sides
= (7.8 + 6.5 + 5.9) cm = 20.2 cm
(ii) Each side of an equilateral triangles
= 9.4 cm
A
B C
9.4 cm
9.4 cm
9.4 cm
Perimeter = 3 × side
= 3 × 9.4 = 28.2 cm
(iii) Each of equal sides = 8.5 cm
and third side = 7 cm
A
C B
8.5 cm
7 cm
8.5 cm
Perimeter = Sum of sides
= (8.5 + 8.5 + 7) cm = 24 cm
Q.10. Find the perimeter of
(i) a regular pentagon of side 8 cm,
(ii) a regular octagon of side 4.5 cm,
(iii) a regular decagon of side 3.6 cm.
Sol. (i) Each side of a regular pentagon
= 8 cm
Perimter = 5 × Side
= 5 × 8 = 40 cm
(ii) Each side of an octagon = 4.5 cm
Perimeter = 8 × Side
= 8 × 4.5 = 36 cm
(iii) Each side of a regular decagon = 3.6 cm
Perimeter = 10 × Side
= 10 × 3.6 = 36 cm
Q.11. Find the perimeter of each of the
following figures :
35cm
45 cm
27 cm
35cm
18 cm
18 cm
18 cm
18 cm
4 cm
16 cm
16 cm
8 cm
4 cm
12 cm
12 cm
Sol. We know that perimeter of a closed
figure or a polygon = Sum of its sides
(i) In the figure, sides of a quadrilateral are
45 cm, 35 cm, 27 cm, 35 cm
Its perimeter = Sum of its sides
= (45 + 35 + 27 + 35) cm = 142 cm
(ii) Sides of a quadrilateral (rhombus) are
18 cm, 18 cm, 18 cm, 18 cm
i.e., each side = 18 cm
Perimeter = 4 × Side
= 4 × 18 = 72 cm
(iii) Sides of the polgyon given are 16 cm,
4 cm, 12 cm, 12 cm, 4 cm, 16 cm and
8 cm
Its perimeter = Sum of its sides
= (16 + 4 + 12 + 12 + 4 + 16 + 8) cm
= 72 cm
           Q. 1. Find the circumference of a circle whose
radius is
(i) 28 cm (ii) 10·5 cm
(iii) 3.5 m
Sol. (i) Radius of the circle (r) = 28 cm
 Circumference = 2 r 2
22
7
28 cm
= 176 cm    Ans.
(ii) Radius of the circle (r) = 10·5 cm
 Circumference = 2 r
2
22
7
10 5
.
cm
= 44 × 1·5 = 66 cm   Ans.
(iii) Radius of a circle (r) = 3.5 m
Circumference = 2 r
= 2 × 
7
22
 × 3.5 m
= 
10 7
35 22 2
 = 22 m  Ans.
Q. 2. Find the circumference of a circle whose
diameter is
(i) 14 cm (ii) 35 cm
(iii) 10.5 m
Sol. (i) Diameter of the circle (d) = 14 cm
 Circumference = d
22
7
14 44 cm   Ans.
(ii) Diameter of the circle (d) = 35 cm
 Circumference = d
22
7
35 110cm   Ans.
(iii) Diameter of a circle (d) = 10.5 m
Circumference = d = 
7
22
 × 10.5 m
= 33 m   Ans.
Q. 3. Find the radius of a circle whose
circumference is 176 cm.
Sol. Circumference of the circle = 176 cm
Let r be the radius, then
r
Circumference
2
   
176 7
2 22
4 7 28 cm
 Radius = 28 cm   Ans.
Q. 4. Find the diameter of a wheel whose
circumference is 75
3
7
cm.
Sol. Find the diameter of a wheel whose
circumference is 264 cm.
Sol. Circumference of a wheel = 264 cm
Let d be its diameter, then
d = 264  
7
22
d = 264
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