Mensuration Exercise 20.2 Class 6 Notes | EduRev

Mathematics (Maths) Class 6

Class 6 : Mensuration Exercise 20.2 Class 6 Notes | EduRev

 Page 1


 
 
 
 
 
 
                                                                           
1. Find the perimeters of the rectangles whose lengths and breadths are given below: 
(i) 7 cm, 5 cm 
(ii) 5 cm, 4 cm 
(iii) 7.5 cm, 4.5 cm 
Solution: 
 
(i) We know that the perimeter of a rectangle = 2 (L + B) 
It is given that L = 7 cm and B = 5 cm 
So the perimeter of a rectangle = 2 (7 + 5) = 2 × 12 = 24 cm 
 
(ii) We know that the perimeter of a rectangle = 2 (L + B) 
It is given that L = 5 cm and B = 4 cm 
So the perimeter of a rectangle = 2 (5 + 4) = 2 × 9 = 18 cm 
 
(iii) We know that the perimeter of a rectangle = 2 (L + B) 
It is given that L = 7.5 cm and B = 4.5 cm 
So the perimeter of a rectangle = 2 (7.5 + 4.5) = 2 × 12 = 24 cm 
 
2. Find the perimeters of the squares whose sides are given below: 
(i) 10 cm 
(ii) 5 m 
(iii) 115.5 cm 
Solution: 
 
(i) We know that the perimeter of a square = 4 × Length of one side 
It is given that L = 10 cm 
So the perimeter of a square = 4 × 10 = 40 cm 
 
(ii) We know that the perimeter of a square = 4 × Length of one side 
It is given that L = 5 m 
So the perimeter of a square = 4 × 5 = 20 m 
 
(iii) We know that the perimeter of a square = 4 × Length of one side 
It is given that L = 115.5 cm 
So the perimeter of a square = 4 × 115.5 = 462 cm 
 
3. Find the side of the square whose perimeter is: 
(i) 16 m 
(ii) 40 cm 
(iii) 22 cm 
Solution: 
 
(i) We know that side of a square = perimeter/ 4 
It is given that perimeter = 16 m 
So the side of the square = 16/4 = 4 m 
 
(ii) We know that side of a square = perimeter/ 4 
It is given that perimeter = 40 cm 
Page 2


 
 
 
 
 
 
                                                                           
1. Find the perimeters of the rectangles whose lengths and breadths are given below: 
(i) 7 cm, 5 cm 
(ii) 5 cm, 4 cm 
(iii) 7.5 cm, 4.5 cm 
Solution: 
 
(i) We know that the perimeter of a rectangle = 2 (L + B) 
It is given that L = 7 cm and B = 5 cm 
So the perimeter of a rectangle = 2 (7 + 5) = 2 × 12 = 24 cm 
 
(ii) We know that the perimeter of a rectangle = 2 (L + B) 
It is given that L = 5 cm and B = 4 cm 
So the perimeter of a rectangle = 2 (5 + 4) = 2 × 9 = 18 cm 
 
(iii) We know that the perimeter of a rectangle = 2 (L + B) 
It is given that L = 7.5 cm and B = 4.5 cm 
So the perimeter of a rectangle = 2 (7.5 + 4.5) = 2 × 12 = 24 cm 
 
2. Find the perimeters of the squares whose sides are given below: 
(i) 10 cm 
(ii) 5 m 
(iii) 115.5 cm 
Solution: 
 
(i) We know that the perimeter of a square = 4 × Length of one side 
It is given that L = 10 cm 
So the perimeter of a square = 4 × 10 = 40 cm 
 
(ii) We know that the perimeter of a square = 4 × Length of one side 
It is given that L = 5 m 
So the perimeter of a square = 4 × 5 = 20 m 
 
(iii) We know that the perimeter of a square = 4 × Length of one side 
It is given that L = 115.5 cm 
So the perimeter of a square = 4 × 115.5 = 462 cm 
 
3. Find the side of the square whose perimeter is: 
(i) 16 m 
(ii) 40 cm 
(iii) 22 cm 
Solution: 
 
(i) We know that side of a square = perimeter/ 4 
It is given that perimeter = 16 m 
So the side of the square = 16/4 = 4 m 
 
(ii) We know that side of a square = perimeter/ 4 
It is given that perimeter = 40 cm 
 
 
 
 
 
 
So the side of the square = 40/4 = 10 cm 
 
(iii) We know that side of a square = perimeter/ 4 
It is given that perimeter = 22 cm 
So the side of the square = 22/4 = 5.5 cm 
 
4. Find the breadth of the rectangle whose perimeter is 360 cm and whose length is 
(i) 116 cm 
(ii) 140 cm 
(iii) 102 cm 
Solution: 
 
We know that the perimeter of a rectangle = 2 (L + B) 
So the breadth of the rectangle = perimeter/2 – length 
 
(i) It is given that perimeter = 360 cm and length = 116 cm 
So the breadth of the rectangle = 360/2 – 116 = 180 – 116 = 64 cm 
 
(ii) It is given that perimeter = 360 cm and length = 140 cm 
So the breadth of the rectangle = 360/2 – 140 = 180 – 140 = 40 cm 
 
(iii) It is given that perimeter = 360 cm and length = 102 cm 
So the breadth of the rectangle = 360/2 – 102 = 180 – 102 = 78 cm 
 
5. A rectangular piece of lawn is 55 m wide and 98 m long. Find the length of the fence around it. 
Solution: 
 
The dimensions of lawn are 
Breadth = 55 m 
Length = 98 m 
We know that  
Perimeter of lawn = 2 (L + B) 
By substituting the values 
Perimeter of lawn = 2 (98 +55)  
So we get 
Perimeter of lawn = 2 × 153 = 306 m 
 
Hence, the length of the fence around the lawn is 306 m. 
 
6. The side of a square field is 65 m. What is the length of the fence required all around it? 
Solution: 
 
It is given that 
Side of a square field = 65 m 
So the perimeter of square field = 4 × side of the square 
By substituting the values 
Perimeter of square field = 4 × 65 = 260 m 
 
Hence, the length of the fence required all around the square field is 260 m. 
 
Page 3


 
 
 
 
 
 
                                                                           
1. Find the perimeters of the rectangles whose lengths and breadths are given below: 
(i) 7 cm, 5 cm 
(ii) 5 cm, 4 cm 
(iii) 7.5 cm, 4.5 cm 
Solution: 
 
(i) We know that the perimeter of a rectangle = 2 (L + B) 
It is given that L = 7 cm and B = 5 cm 
So the perimeter of a rectangle = 2 (7 + 5) = 2 × 12 = 24 cm 
 
(ii) We know that the perimeter of a rectangle = 2 (L + B) 
It is given that L = 5 cm and B = 4 cm 
So the perimeter of a rectangle = 2 (5 + 4) = 2 × 9 = 18 cm 
 
(iii) We know that the perimeter of a rectangle = 2 (L + B) 
It is given that L = 7.5 cm and B = 4.5 cm 
So the perimeter of a rectangle = 2 (7.5 + 4.5) = 2 × 12 = 24 cm 
 
2. Find the perimeters of the squares whose sides are given below: 
(i) 10 cm 
(ii) 5 m 
(iii) 115.5 cm 
Solution: 
 
(i) We know that the perimeter of a square = 4 × Length of one side 
It is given that L = 10 cm 
So the perimeter of a square = 4 × 10 = 40 cm 
 
(ii) We know that the perimeter of a square = 4 × Length of one side 
It is given that L = 5 m 
So the perimeter of a square = 4 × 5 = 20 m 
 
(iii) We know that the perimeter of a square = 4 × Length of one side 
It is given that L = 115.5 cm 
So the perimeter of a square = 4 × 115.5 = 462 cm 
 
3. Find the side of the square whose perimeter is: 
(i) 16 m 
(ii) 40 cm 
(iii) 22 cm 
Solution: 
 
(i) We know that side of a square = perimeter/ 4 
It is given that perimeter = 16 m 
So the side of the square = 16/4 = 4 m 
 
(ii) We know that side of a square = perimeter/ 4 
It is given that perimeter = 40 cm 
 
 
 
 
 
 
So the side of the square = 40/4 = 10 cm 
 
(iii) We know that side of a square = perimeter/ 4 
It is given that perimeter = 22 cm 
So the side of the square = 22/4 = 5.5 cm 
 
4. Find the breadth of the rectangle whose perimeter is 360 cm and whose length is 
(i) 116 cm 
(ii) 140 cm 
(iii) 102 cm 
Solution: 
 
We know that the perimeter of a rectangle = 2 (L + B) 
So the breadth of the rectangle = perimeter/2 – length 
 
(i) It is given that perimeter = 360 cm and length = 116 cm 
So the breadth of the rectangle = 360/2 – 116 = 180 – 116 = 64 cm 
 
(ii) It is given that perimeter = 360 cm and length = 140 cm 
So the breadth of the rectangle = 360/2 – 140 = 180 – 140 = 40 cm 
 
(iii) It is given that perimeter = 360 cm and length = 102 cm 
So the breadth of the rectangle = 360/2 – 102 = 180 – 102 = 78 cm 
 
5. A rectangular piece of lawn is 55 m wide and 98 m long. Find the length of the fence around it. 
Solution: 
 
The dimensions of lawn are 
Breadth = 55 m 
Length = 98 m 
We know that  
Perimeter of lawn = 2 (L + B) 
By substituting the values 
Perimeter of lawn = 2 (98 +55)  
So we get 
Perimeter of lawn = 2 × 153 = 306 m 
 
Hence, the length of the fence around the lawn is 306 m. 
 
6. The side of a square field is 65 m. What is the length of the fence required all around it? 
Solution: 
 
It is given that 
Side of a square field = 65 m 
So the perimeter of square field = 4 × side of the square 
By substituting the values 
Perimeter of square field = 4 × 65 = 260 m 
 
Hence, the length of the fence required all around the square field is 260 m. 
 
 
 
 
 
 
 
7. Two sides of a triangle are 15 cm and 20 cm. The perimeter of the triangle is 50 cm. What is the third 
side? 
Solution: 
 
It is given that 
First side of triangle = 15 cm 
Second side of triangle = 20 cm 
In order to find the length of third side 
We know that perimeter of a triangle is the sum of all three sides of a triangle 
So the length of third side = perimeter of triangle – sum of length of other two sides 
By substituting the values 
Length of third side = 50 – (15 + 20) = 15 cm. 
 
Hence, the length of third side is 15 cm. 
 
8. A wire of length 20 m is to be folded in the form of a rectangle. How many rectangles can be formed by 
folding the wire if the sides are positive integers in metres? 
Solution: 
 
Given: 
Length of wire 20 m is folded in the form of rectangle 
So the perimeter = 20 m 
It can be written as 
2 (L + B) = 20 m 
On further calculation 
L + B = 10 m 
If the sides are positive integers in metres the possible dimensions are (1m, 9m), (2m, 8m), (3m, 7m), (4m, 6m) 
and (5m, 5m) 
Hence, five rectangles can be formed using the given wire. 
 
9. A square piece of land has each side equal to 100 m. If 3 layers of metal wire has to be used to fence it, 
what is the length of the wire needed? 
Solution: 
 
It is given that 
Each side of a square field = 100 m 
We can find the wire required to fence the square field by determining the perimeter = 4 × each side of a square 
field 
By substituting the values 
Perimeter of the square field = 4 × 100 = 400 m 
So the length of wire which is required to fence three layers is = 3 × 400 = 1200 m 
 
Hence, the length of wire needed to fence 3 layers is 1200 m. 
 
10. Shikha runs around a square of side 75 m. Priya runs around a rectangle with length 60 m and breadth 
45 m. Who covers the smaller distance? 
Solution: 
 
It is given that  
Shikha runs around a square of side = 75 m 
Page 4


 
 
 
 
 
 
                                                                           
1. Find the perimeters of the rectangles whose lengths and breadths are given below: 
(i) 7 cm, 5 cm 
(ii) 5 cm, 4 cm 
(iii) 7.5 cm, 4.5 cm 
Solution: 
 
(i) We know that the perimeter of a rectangle = 2 (L + B) 
It is given that L = 7 cm and B = 5 cm 
So the perimeter of a rectangle = 2 (7 + 5) = 2 × 12 = 24 cm 
 
(ii) We know that the perimeter of a rectangle = 2 (L + B) 
It is given that L = 5 cm and B = 4 cm 
So the perimeter of a rectangle = 2 (5 + 4) = 2 × 9 = 18 cm 
 
(iii) We know that the perimeter of a rectangle = 2 (L + B) 
It is given that L = 7.5 cm and B = 4.5 cm 
So the perimeter of a rectangle = 2 (7.5 + 4.5) = 2 × 12 = 24 cm 
 
2. Find the perimeters of the squares whose sides are given below: 
(i) 10 cm 
(ii) 5 m 
(iii) 115.5 cm 
Solution: 
 
(i) We know that the perimeter of a square = 4 × Length of one side 
It is given that L = 10 cm 
So the perimeter of a square = 4 × 10 = 40 cm 
 
(ii) We know that the perimeter of a square = 4 × Length of one side 
It is given that L = 5 m 
So the perimeter of a square = 4 × 5 = 20 m 
 
(iii) We know that the perimeter of a square = 4 × Length of one side 
It is given that L = 115.5 cm 
So the perimeter of a square = 4 × 115.5 = 462 cm 
 
3. Find the side of the square whose perimeter is: 
(i) 16 m 
(ii) 40 cm 
(iii) 22 cm 
Solution: 
 
(i) We know that side of a square = perimeter/ 4 
It is given that perimeter = 16 m 
So the side of the square = 16/4 = 4 m 
 
(ii) We know that side of a square = perimeter/ 4 
It is given that perimeter = 40 cm 
 
 
 
 
 
 
So the side of the square = 40/4 = 10 cm 
 
(iii) We know that side of a square = perimeter/ 4 
It is given that perimeter = 22 cm 
So the side of the square = 22/4 = 5.5 cm 
 
4. Find the breadth of the rectangle whose perimeter is 360 cm and whose length is 
(i) 116 cm 
(ii) 140 cm 
(iii) 102 cm 
Solution: 
 
We know that the perimeter of a rectangle = 2 (L + B) 
So the breadth of the rectangle = perimeter/2 – length 
 
(i) It is given that perimeter = 360 cm and length = 116 cm 
So the breadth of the rectangle = 360/2 – 116 = 180 – 116 = 64 cm 
 
(ii) It is given that perimeter = 360 cm and length = 140 cm 
So the breadth of the rectangle = 360/2 – 140 = 180 – 140 = 40 cm 
 
(iii) It is given that perimeter = 360 cm and length = 102 cm 
So the breadth of the rectangle = 360/2 – 102 = 180 – 102 = 78 cm 
 
5. A rectangular piece of lawn is 55 m wide and 98 m long. Find the length of the fence around it. 
Solution: 
 
The dimensions of lawn are 
Breadth = 55 m 
Length = 98 m 
We know that  
Perimeter of lawn = 2 (L + B) 
By substituting the values 
Perimeter of lawn = 2 (98 +55)  
So we get 
Perimeter of lawn = 2 × 153 = 306 m 
 
Hence, the length of the fence around the lawn is 306 m. 
 
6. The side of a square field is 65 m. What is the length of the fence required all around it? 
Solution: 
 
It is given that 
Side of a square field = 65 m 
So the perimeter of square field = 4 × side of the square 
By substituting the values 
Perimeter of square field = 4 × 65 = 260 m 
 
Hence, the length of the fence required all around the square field is 260 m. 
 
 
 
 
 
 
 
7. Two sides of a triangle are 15 cm and 20 cm. The perimeter of the triangle is 50 cm. What is the third 
side? 
Solution: 
 
It is given that 
First side of triangle = 15 cm 
Second side of triangle = 20 cm 
In order to find the length of third side 
We know that perimeter of a triangle is the sum of all three sides of a triangle 
So the length of third side = perimeter of triangle – sum of length of other two sides 
By substituting the values 
Length of third side = 50 – (15 + 20) = 15 cm. 
 
Hence, the length of third side is 15 cm. 
 
8. A wire of length 20 m is to be folded in the form of a rectangle. How many rectangles can be formed by 
folding the wire if the sides are positive integers in metres? 
Solution: 
 
Given: 
Length of wire 20 m is folded in the form of rectangle 
So the perimeter = 20 m 
It can be written as 
2 (L + B) = 20 m 
On further calculation 
L + B = 10 m 
If the sides are positive integers in metres the possible dimensions are (1m, 9m), (2m, 8m), (3m, 7m), (4m, 6m) 
and (5m, 5m) 
Hence, five rectangles can be formed using the given wire. 
 
9. A square piece of land has each side equal to 100 m. If 3 layers of metal wire has to be used to fence it, 
what is the length of the wire needed? 
Solution: 
 
It is given that 
Each side of a square field = 100 m 
We can find the wire required to fence the square field by determining the perimeter = 4 × each side of a square 
field 
By substituting the values 
Perimeter of the square field = 4 × 100 = 400 m 
So the length of wire which is required to fence three layers is = 3 × 400 = 1200 m 
 
Hence, the length of wire needed to fence 3 layers is 1200 m. 
 
10. Shikha runs around a square of side 75 m. Priya runs around a rectangle with length 60 m and breadth 
45 m. Who covers the smaller distance? 
Solution: 
 
It is given that  
Shikha runs around a square of side = 75 m 
 
 
 
 
 
 
So the perimeter = 4 × 75 = 300 m 
Priya runs around a rectangle having 
Length = 60 m 
Breadth = 45 m 
So the distance covered can be found from the perimeter = 2 (L + B) 
By substituting the values 
Perimeter = 2 (60 + 45) = 2 × 105 = 210 m 
 
Hence, Priya covers the smaller distance of 210 m. 
 
11. The dimensions of a photographs are 30 cm × 20 cm. What length of wooden frame is needed to frame 
the picture? 
Solution: 
 
It is given that 
Dimensions of a photographs = 30 cm × 20 cm 
So the required length of the wooden frame can be determined from the perimeter of the photograph = 2 (L + B) 
By substituting the values = 2 (30 + 20) = 2 × 50 = 100 cm 
 
Hence, the length of the wooden frame required to frame the picture is 100 cm. 
 
12. The length of a rectangular field is 100 m. If the perimeter is 300 m, what is its breadth? 
Solution: 
 
The dimensions of rectangular field are 
Length = 100 m 
Perimeter = 300 m 
We know that perimeter = 2 (L + B)  
It can be written as 
Breadth = perimeter/2 – length 
By substituting the values 
Breadth = (300-200)/2 = 100/2 = 50 m 
 
Hence, the breadth of the rectangular field is 50 m. 
 
13. To fix fence wires in a garden, 70 m long and 50 m wide, Arvind bought metal pipes for posts. He fixed 
a post every 5 metres apart. Each post was 2 m long. What is the total length of the pipes he bought for the 
posts? 
Solution: 
 
The dimensions of garden are 
Length = 70 m 
Breadth = 50 m 
So the perimeter = 2 (L + B) 
By substituting the values 
Perimeter = 2 (70 + 50) = 2 × 120 = 240 m 
 
Given: 
Arvind fixes a post every 5 metres apart 
No. of posts required = 240/5 = 48 
Page 5


 
 
 
 
 
 
                                                                           
1. Find the perimeters of the rectangles whose lengths and breadths are given below: 
(i) 7 cm, 5 cm 
(ii) 5 cm, 4 cm 
(iii) 7.5 cm, 4.5 cm 
Solution: 
 
(i) We know that the perimeter of a rectangle = 2 (L + B) 
It is given that L = 7 cm and B = 5 cm 
So the perimeter of a rectangle = 2 (7 + 5) = 2 × 12 = 24 cm 
 
(ii) We know that the perimeter of a rectangle = 2 (L + B) 
It is given that L = 5 cm and B = 4 cm 
So the perimeter of a rectangle = 2 (5 + 4) = 2 × 9 = 18 cm 
 
(iii) We know that the perimeter of a rectangle = 2 (L + B) 
It is given that L = 7.5 cm and B = 4.5 cm 
So the perimeter of a rectangle = 2 (7.5 + 4.5) = 2 × 12 = 24 cm 
 
2. Find the perimeters of the squares whose sides are given below: 
(i) 10 cm 
(ii) 5 m 
(iii) 115.5 cm 
Solution: 
 
(i) We know that the perimeter of a square = 4 × Length of one side 
It is given that L = 10 cm 
So the perimeter of a square = 4 × 10 = 40 cm 
 
(ii) We know that the perimeter of a square = 4 × Length of one side 
It is given that L = 5 m 
So the perimeter of a square = 4 × 5 = 20 m 
 
(iii) We know that the perimeter of a square = 4 × Length of one side 
It is given that L = 115.5 cm 
So the perimeter of a square = 4 × 115.5 = 462 cm 
 
3. Find the side of the square whose perimeter is: 
(i) 16 m 
(ii) 40 cm 
(iii) 22 cm 
Solution: 
 
(i) We know that side of a square = perimeter/ 4 
It is given that perimeter = 16 m 
So the side of the square = 16/4 = 4 m 
 
(ii) We know that side of a square = perimeter/ 4 
It is given that perimeter = 40 cm 
 
 
 
 
 
 
So the side of the square = 40/4 = 10 cm 
 
(iii) We know that side of a square = perimeter/ 4 
It is given that perimeter = 22 cm 
So the side of the square = 22/4 = 5.5 cm 
 
4. Find the breadth of the rectangle whose perimeter is 360 cm and whose length is 
(i) 116 cm 
(ii) 140 cm 
(iii) 102 cm 
Solution: 
 
We know that the perimeter of a rectangle = 2 (L + B) 
So the breadth of the rectangle = perimeter/2 – length 
 
(i) It is given that perimeter = 360 cm and length = 116 cm 
So the breadth of the rectangle = 360/2 – 116 = 180 – 116 = 64 cm 
 
(ii) It is given that perimeter = 360 cm and length = 140 cm 
So the breadth of the rectangle = 360/2 – 140 = 180 – 140 = 40 cm 
 
(iii) It is given that perimeter = 360 cm and length = 102 cm 
So the breadth of the rectangle = 360/2 – 102 = 180 – 102 = 78 cm 
 
5. A rectangular piece of lawn is 55 m wide and 98 m long. Find the length of the fence around it. 
Solution: 
 
The dimensions of lawn are 
Breadth = 55 m 
Length = 98 m 
We know that  
Perimeter of lawn = 2 (L + B) 
By substituting the values 
Perimeter of lawn = 2 (98 +55)  
So we get 
Perimeter of lawn = 2 × 153 = 306 m 
 
Hence, the length of the fence around the lawn is 306 m. 
 
6. The side of a square field is 65 m. What is the length of the fence required all around it? 
Solution: 
 
It is given that 
Side of a square field = 65 m 
So the perimeter of square field = 4 × side of the square 
By substituting the values 
Perimeter of square field = 4 × 65 = 260 m 
 
Hence, the length of the fence required all around the square field is 260 m. 
 
 
 
 
 
 
 
7. Two sides of a triangle are 15 cm and 20 cm. The perimeter of the triangle is 50 cm. What is the third 
side? 
Solution: 
 
It is given that 
First side of triangle = 15 cm 
Second side of triangle = 20 cm 
In order to find the length of third side 
We know that perimeter of a triangle is the sum of all three sides of a triangle 
So the length of third side = perimeter of triangle – sum of length of other two sides 
By substituting the values 
Length of third side = 50 – (15 + 20) = 15 cm. 
 
Hence, the length of third side is 15 cm. 
 
8. A wire of length 20 m is to be folded in the form of a rectangle. How many rectangles can be formed by 
folding the wire if the sides are positive integers in metres? 
Solution: 
 
Given: 
Length of wire 20 m is folded in the form of rectangle 
So the perimeter = 20 m 
It can be written as 
2 (L + B) = 20 m 
On further calculation 
L + B = 10 m 
If the sides are positive integers in metres the possible dimensions are (1m, 9m), (2m, 8m), (3m, 7m), (4m, 6m) 
and (5m, 5m) 
Hence, five rectangles can be formed using the given wire. 
 
9. A square piece of land has each side equal to 100 m. If 3 layers of metal wire has to be used to fence it, 
what is the length of the wire needed? 
Solution: 
 
It is given that 
Each side of a square field = 100 m 
We can find the wire required to fence the square field by determining the perimeter = 4 × each side of a square 
field 
By substituting the values 
Perimeter of the square field = 4 × 100 = 400 m 
So the length of wire which is required to fence three layers is = 3 × 400 = 1200 m 
 
Hence, the length of wire needed to fence 3 layers is 1200 m. 
 
10. Shikha runs around a square of side 75 m. Priya runs around a rectangle with length 60 m and breadth 
45 m. Who covers the smaller distance? 
Solution: 
 
It is given that  
Shikha runs around a square of side = 75 m 
 
 
 
 
 
 
So the perimeter = 4 × 75 = 300 m 
Priya runs around a rectangle having 
Length = 60 m 
Breadth = 45 m 
So the distance covered can be found from the perimeter = 2 (L + B) 
By substituting the values 
Perimeter = 2 (60 + 45) = 2 × 105 = 210 m 
 
Hence, Priya covers the smaller distance of 210 m. 
 
11. The dimensions of a photographs are 30 cm × 20 cm. What length of wooden frame is needed to frame 
the picture? 
Solution: 
 
It is given that 
Dimensions of a photographs = 30 cm × 20 cm 
So the required length of the wooden frame can be determined from the perimeter of the photograph = 2 (L + B) 
By substituting the values = 2 (30 + 20) = 2 × 50 = 100 cm 
 
Hence, the length of the wooden frame required to frame the picture is 100 cm. 
 
12. The length of a rectangular field is 100 m. If the perimeter is 300 m, what is its breadth? 
Solution: 
 
The dimensions of rectangular field are 
Length = 100 m 
Perimeter = 300 m 
We know that perimeter = 2 (L + B)  
It can be written as 
Breadth = perimeter/2 – length 
By substituting the values 
Breadth = (300-200)/2 = 100/2 = 50 m 
 
Hence, the breadth of the rectangular field is 50 m. 
 
13. To fix fence wires in a garden, 70 m long and 50 m wide, Arvind bought metal pipes for posts. He fixed 
a post every 5 metres apart. Each post was 2 m long. What is the total length of the pipes he bought for the 
posts? 
Solution: 
 
The dimensions of garden are 
Length = 70 m 
Breadth = 50 m 
So the perimeter = 2 (L + B) 
By substituting the values 
Perimeter = 2 (70 + 50) = 2 × 120 = 240 m 
 
Given: 
Arvind fixes a post every 5 metres apart 
No. of posts required = 240/5 = 48 
 
 
 
 
 
 
The length of each post = 2 m 
So the total length of the pipe required = 48 × 2 = 96 m 
Hence, the total length of the pipes he bought for the posts is 96 m. 
 
14. Find the cost of fencing a rectangular park of length 175 m and breadth 125 m at the rate of Rs 12 per 
meter. 
Solution: 
 
The dimensions of the rectangular park are 
Length = 175 m 
Breadth = 125 m 
So the perimeter = 2 (L + B) 
By substituting the values 
Perimeter = 2 (175 + 125) = 2 × 300 = 600 m 
It is given that the cost of fencing = Rs 12 per meter 
So the total cost of fencing = 12 × 600 = Rs 7200 
 
Hence, the cost of fencing a rectangular park is Rs 7200. 
 
15. The perimeter of a regular pentagon is 100 cm. How long is each side? 
Solution: 
 
We know that a regular pentagon is a closed polygon having 5 sides of same length. 
It is given that 
Perimeter of a regular pentagon = 100 cm 
It can be written as 
Perimeter = 5 × side of the regular pentagon  
So we get 
Side of the regular pentagon = Perimeter/5 
By substituting the values 
Side of the regular pentagon = 100/5 = 20 cm 
 
Hence, the side of the regular pentagon measures 20 cm. 
 
16. Find the perimeter of a regular hexagon with each side measuring 8 m. 
Solution: 
 
We know that a regular hexagon is a closed polygon which has six sides of same length. 
It is given that 
Side of the regular hexagon = 8 m 
So we get  
Perimeter = 6 × side of the regular hexagon 
By substituting the values 
Perimeter = 6 × 8 = 48 m 
 
Hence, the perimeter of a regular hexagon is 48 m. 
 
17. A rectangular piece of land measure 0.7 km by 0.5 km. Each side is to be fenced with four rows of wires. 
What length of the wire is needed? 
Solution: 
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