Perimeter Made Simple : With Word Problems For Class 6th,7th,8th [8]

Word problems related to perimeter
Class 6th ,7th , 8th

-Q.1) The length and the breadth of rectangular field are in the ratio 7:4. The cost of fencing the field at Rs. 25 per meter is Rs. 3300. Find the dimensions of the field.

SOLUTION :- Rectangular field

Given:
Ratio of length and breadth =7:4
Assuming length = L = 7x meter
Breadth =B = 4x meter
Rate of fencing per meter = Rs.25 per m
Total cost of fencing = Rs. 3300
Find its dimensions i.e. its length and breadth=?

Meter———–Cost
1m————-Rs.25
?—————Rs.3300
Total meter (perimeter) of fencing = 3300/25 =132m.
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Q.2) The cost of putting a fence around a square field at Rs.35 per meter is is Rs.4480. Find the length of each side of field.

SOLUTION: Square Field

Given:
Rate of fencing = Rs.35 per meter
Cost of fencing = Rs. 4480
Find the length of each side = ?

I) Meter———-Cost
1m————–Rs.35
?—————Rs.4480

Hence, Total meter (perimeter) = 4480/35
Perimeter = 128m

II) Perimeter of square = 4*S
128 = 4*S
128/4 = S
32 = S

ANS: The length of each side of a square field = 32m
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Q.3) Each side of a square measures 21m. Adjacent to this field, there is a rectangular field, having its sides in the ratio 4:3. If the perimeter of both the fields are equal. Find the dimensions of rectangular field.

SOLUTION: Square Field

Given:
Side = S = 21m
Perimeter of square = Perimeter of rectangle

I) Perimeter of square = 4*S
P = 4*21
P = 84m

II) Perimeter of square = Perimeter of rectangle = 84m
Perimeter of rectangle = 2*( L + B )
84 = 2*( 4x + 3x )
84/2 = 4x + 3x
42 = 7x
42/7 = x
6= x

Hence, Length = L = 4X = 4*6 = 24m
Breadth = B = 3x = 3*6 = 18m

ANS: The dimensions of the rectangular field are 24m, 18m.
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Q.4) A rectangular piece of land measures 0.8km by 0.3km. Each side is to be fenced with 3 rows of wires. What is the length of the wire needed ?

SOLUTION: Rectangular Field

Given:
Length = L = 0.8km
Breadth = B = 0.3km
N0. of rows of fencing = 3
The length of wire needed for fencing 3 rows = ?

I) The length of the wire needed for fencing 1 row = Perimeter of land = ?
Perimeter of land = 2*( L + B )
P = 2*( 0.8 + 0.3 )
P = 2*( 1.1 )
P = 2.2km

Hence, the length of wire required for fencing 1 row = 2.2km

II) Hence, No. of row ————- Wire required
(Perimeter) 1 row ————— 2.2km
3 row ————–?

ANS: The length of wire required for 3 row = 3*2.2 = 6.6km
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Q.5) The cost of fencing a rectangular park of length 150m and breadth 100m at the rate of Rs.10 per meter.

SOLUTION: Rectangular Park

Given:
Length = L = 150m
Breadth = B = 100m
Rate of fencing = Rs.10 per meter
Cost of fencing = ?

I) Perimeter of rectangular park = 2*( L + B )
P = 2*( 150 + 100 )
P = 2*(250)
P = 500m

II) Cost of fencing = Perimeter * Rate
Cost of fencing = 500 * 10
= Rs.5000

ANS: Cost of fencing is Rs. 5000
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Q.6) What is the length of a colored tape required to stick on the sides of a a rectangular scenery. The length of the scenery is 20cm and breadth is 12cm.

SOLUTION: Rectangular Scenery

Given:
Length = L = 20cm
Breadth = B = 12cm
Length of a colored tape required = ?

Length of tape = Perimeter of the rectangular scenery

Hence, Perimeter of rectangle = 2*( L+ B )
P = 2*( 20 +12 )
P = 2*32
P = 64cm

Hence, The length of colored tape required to stick on the sides of the rectangular scenery is 64cm.
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Q.7) Two sides of a triangle are 10cm and 12cm. The perimeter of the triangle is 35cm. What is the length of third side ?

SOLUTION: Triangle

Given:
Side 1 = S1 = 10cm
Side 2 = S2 = 12cm
Side 3 = S3 = ?
Perimeter of triangle = 35cm

Perimeter of triangle = sum of length of all sides
35 = S1 + S2 + S3
35 = 10 + 12 + S3
35 = 22 +S3
35 – 22 = S3
13cm = S3

ANS: The length of the third side of the triangle is 13cm.
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