Home | | Maths 7th Std | Area of Pathways

# Area of Pathways

We come across pathways in different shapes. Here we restrict ourselves to two kinds of pathways namely circular and rectangular.

Area of Pathways

We come across pathways in different shapes. Here we restrict ourselves to two kinds of pathways namely circular and rectangular.

1. Circular Pathways

We observe around us circular shapes where we need to find the area of the pathway. The area of pathway is the difference between the area of outer circle and inner circle. Let ‘R’ be the radius of the outer circle and ‘r’ be the radius of inner circle. Therefore, the area of the circular pathway = πR2 πr2

= π(R2 r2 ) sq. units.

2. Rectangular Pathways

Consider a rectangular park as shown in Fig 2.23. A uniform path is to be laid outside the park. How do we find the area of the path? The uniform path including the park is also a rectangle. If we consider the path as outer rectangle, then the park will be the inner rectangle. Let l and b be the length and breadth of the park. Area of the park (inner rectangle) = l b sq.units . Let w be the width of the path. If L, B are the length and breadth of the outer rectangle, then L = l + 2w and B = b + 2w . Here, the area of the rectangular pathway

= Area of the outer rectangle – Area of the inner rectangle

= (LBlb) sq. units

Example 2.15

A park is circular in shape. The central portion has playthings for kids surrounded by a circular walking pathway. Find the walking area whose outer radius is 10 m and inner radius is 3 m.

Solution

The radius of the outer circle, R = 10 m

The radius of the inner circle, r = 3 m

The area of the circular path = Area of outer circleArea of inner circle

= π R 2 πr2

= π( R2 r2 ) sq.units

= 22/7 × (10232 )

= 22/7 × ((10 ×10)(3 × 3))

= 227 × (1009)

= 22/7 × 91

= 286 m2

Try these

(i) If the outer radius and inner radius of the circles are respectively 9 cm and 6 cm, find the width of the circular pathway.

Outer radius of the circle R = 9 cm

Inner radius r = 6 cm

The width of the pathway w = R – r

= 9 – 6 = 3 cm.

(ii) If the area of the circular pathway is 352 sq.cm and the outer radius is 16 cm, find the inner radius.

Area of the pathway = π (R2 – r2) = 352 sq.cm

Outer radius R = 16 cm

22/7 (162 – r2) = 352

256 – r2 = 352 / 22 7 = 112

– r2 = 112 – 256 = – 144

r2 = 144 = 122

r = 12 cm

(iii) If the area of the inner rectangular region is 15 sq.cm and the area covered by the outer rectangular region is 48 sq.cm, find the area of the rectangular pathway.

Area of the outer rectangular region πR2 = 48 sq.cm

Area ofthe inner rectangular region πr2 =15 sq.cm

Area of the rectangular pathway = πR2 – πr2

= (48–15) sq.cm

= 33 sq.cm

Example 2.16

The radius of a circular flower garden is 21 m. A circular path of 14 m wide is laid around the garden. Find the area of the circular path. Solution

The radius of the inner circle r = 21 m

The path is around the inner circle.

Therefore, the radius of the outer circle, R = 21 + 14 = 35 m

The area of the circular path = π ( R 2 r2 ) sq.units

= 22 /7  × (352 −212)

= 22/7 × ((35 × 35) − (21 × 21))

= 22/7 × (1225 − 441)

= 22/7 × 784

= 22 × 112 = 2464 m2

Example 2.17

The radius of a circular cricket ground is 76 m. A drainage 2 m wide has to be constructed around the cricket ground for the purpose of draining the rain water. Find the cost of constructing the drainage at the rate of ₹180/- per sq.m.

Solution

The radius of the inner circle (cricket ground), r = 76 m

A drainage is constructed around the cricket ground.

Therefore, the radius of the outer circle, R = 76 + 2 = 78 m

We have, area of the circular path = π( R 2r2 ) sq. units

= 22/7 × (782 − 762 )

= 22/7 × (6084 − 5776)

= 22/7 × 308

= 22 × 44 = 968 m2

Given, the cost of constructing the drainage per sq.m is ₹180.

Therefore, the cost of constructing the drainage = 968 × 180 = ₹1,74,240.

Example 2.18

A floor is 10 m long and 8 m wide. A carpet of size 7 m long and 5 m wide is laid on the floor. Find the area of the floor that is not covered by the carpet.

Solution

Here, L = 10 m

B = 8 m

Area of the floor = L × B

=10×8

=80 m2

Area of the carpet = l × b

= 7 × 5

=  35 m2

Therefore, the total area of the floor not covered by the carpet = 80 35

= 45 m2

Example 2.19

A picture of length 23 cm and breadth 11 cm is painted on a chart, such that there is a margin of 3 cm along each of its sides. Find the total area of the margin.

Solution

Here L = 23 cm B = 11 cm

Area of the chart = L × B

= 23 ×11

= 253 cm2

l = L 2w = 23 2(3) = 23 6 = 17 cm

b = B 2w = 11 2(3) = 11 6 = 5 cm2

Area of the picture 17× 5 = 85cm2

Therefore, the area of the marin = 235 85

= 168 cm2 .

Example 2.20

A verandah of width 3 m is constructed along the outside of a room of length 9 m and width 7 m. Find (a) the area of the verandah (b) the cost of cementing the floor of the verandah at the rate of ₹15 per sq.m.

Solution

Here, l = 9 m , b = 7 m

Area of the Room = l × b

=9×7

= 63 m2

L = l + 2w = 8 + 2(3) = 8 + 6 = 14 m

B = b + 2w = 5 + 2(3) = 5 + 6 = 11 m

Area of the room including verandah = L × B

= 14 ×11

= 154 m2

The area of the verandah = Area of the room including verandah Area of the room

=15463

= 91 m2

The cost of cementing the floor for 1sq.m = ₹15

Therfore, the cost of cementing the floor of the verandah = 91 × 15 = ₹1365.

Example 2.21

A Kho-Kho ground has dimensions 30 m × 19 m which includes a lobby on all of its sides. The dimensions of the playing area is 27 m × 16 m . Find the area of the lobby.

Solution

From the dimensions of the ground we have,

L = 30 m ; B = 19 m ; l = 27 m ; b = 16 m Area of the kho kho ground = L × B

= 30 ×19

= 570 m2

Area of the play field = l × b

= 27 ×16

= 432 m2

Area of the lobby = Area of Kho-Kho ground Area of the play field

= 570432

= 148 m2

Tags : Measurements | Term 2 Chapter 2 | 7th Maths , 7th Maths : Term 2 Unit 2 : Measurements
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
7th Maths : Term 2 Unit 2 : Measurements : Area of Pathways | Measurements | Term 2 Chapter 2 | 7th Maths