WORD PROBLEMS RELATED TO PERIMETER
CLASS 6th, 7th, 8th & 9th
Q.1) Aditya runs around a square park of side 50m three times. Shweta runs twice around a rectangular park of length 100m and breadth 60m. Who covers more distance?
SOLUTION:
Given:
Aditya (Square Park)
Sides = S = 50m
No. of rounds = 3 times
Perimeter of square park = P = ?
Perimeter of square = 4*S
P = 4*50
P = 200m
ANS: Distance covered by Aditya is 200m in
1 round.
Hence, Distance covered by Aditya in 3 rounds = 200*3
= 600m
Given:
Shweta (Rectangular Park)
Length = L = 100m
Breadth = B = 60m
No. of rounds = 2
Perimeter of rectangle = P = ?
Perimeter of rectangle = 2*( L+B )
P = 2*( 100 + 60 )
P = 2*( 160 )
P = 320m
ANS: Distance covered by Shweta is 320m in 1 round.
Hence, Distance covered in 2 rounds
= 320*2
= 640m
Clearly, proved that Shweta covers more distance than Aditya.
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Q.2) Ranjan walks around a rectangular park of length 60m and breadth 40m and Shreya walks around a square park of side 55m. Both take 3 rounds each. Who walks more distance and by how much?
SOLUTION:
Given:
Ranjan (Rectangular Park )
Length = L = 60m
Breadth = B = 40m
No. of rounds = 3
Perimeter of rectangle = P = ?
Distance covered in 1 round = P of rectangle
Perimeter of rectangle = 2*(L + B)
P = 2*( 60 + 40)
P = 2*( 100 )
P = 200m
Distance covered in 1 round = 200m
Distance covered in 3 rounds = ?
= 200*3
= 600m
Given:
Shreya ( Square Park)
Sides = S = 55m
No. of rounds = 3
Perimeter of square = P = ?
Distance covered in 1 round = P of square
Perimeter of square = 4*S
P = 4*55
P = 220m
Distance covered in 1 round = 220m
Distance covered in 3 rounds = ?
= 3*220
= 660m
Hence, clearly proved Shreya walks more distance by = 660 – 600
= 60m
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Q.3) A rectangular field is 15m long and 10m wide. Another rectangular field having the same perimeter has it’s sides in the ratio 4:1 . Find the dimensions of the rectangular field ?
SOLUTION : 1st Rectangular Field
Given : Length =L = 15m
Breadth = B = 10m
Perimeter of 1st rectangle = 2 * ( L + B)
P = 2 * ( 15 + 10 )
P = 2 * 25
P = 50m
SOLUTION : 2nd Rectangular Field
Given : Ratio of length and breadth = 4:1
Assuming Length = L = 4x m
Breadth = B = 1x m
Perimeter of the 2nd rectangle = Perimeter of the 1st rectangle = 50 m
Perimeter of the 2nd rectangle = 2 * ( L + B )
50 = 2* ( 4x + 1x )
50 = 2* 5x
50 = 10x
50/10 = x
5m = x
Ans : Dimensions : Length = 4x = 4*x =4*5 = 20m
Breadth = 1x = 1*x = 1*5 = 5m
Q.4) A rectangular field of plot measures 0.6Km by 0.5Km. Each side is to be fenced with 5 rows of
wires. What is the length of wire needed ? If the length of wire cost Rs. 9 per m, what is the cost of wire required for fencing ?
SOLUTION : Rectangular Plot
Given : Length = L = 0.6 Km.
Breadth = B = 0.5 Km
No. of rows of wires = 5 rows
Rate of wire = Rs. 9 per m
Find the length of wire needed for 5 rows =?
Cost of wire =?
Step 1 : Length of wire needed for 1 row = Perimeter of the rectangle
Perimeter of the rectangle = 2 * ( L + B )
P = 2 * ( 0.6 + 0.5 )
P = 2 * 1.1
P = 2.2 Km.
Step 2 : Length of wire needed for 1 row = 2.2 km
. Length of wire needed for 5 rows = 2.2 * 5
= 11 km.
= 11 * 1000m ( 1km = 1000m)
= 11000m
Step 3 : Rate of per m or wire is = Rs. 9
Cost of 11000m of wire = 11000 * 9 = Rs. 99000.
Q.5) A rectangular piece of land is to be fenced for an exhibition. The length of the exhibition land is 250m and it’s breadth is 200m. We need to keep an entrance of 10m and an emergency exit of 5m. If the cost of this special fencing is Rs. 175 per m, what is the total cost of fencing ?
SOLUTION : Rectangular land
Given : Length = L = 250m
Breadth = B= 200m
Length of entrance = 10m Length of exit = 5m
Rate of fencing = Rs. 175 per m
Find the total cost of fencing = ?
Step 1 : Perimeter of the rectangle = 2 * ( L + B )
P = 2 * ( 250 + 200 )
P = 2 * 450
P = 900m
Step 2 : Perimeter without entrance and exit = 900 – (10 + 5 )
= 900 – 15
= 885m
Step 3 : cost of fencing per m = Rs. 175
Total cost of fencing 885m = 885 * 175
= Rs. 1,54,875
Ans : The total cost of fencing is Rs. 1,54,875