1. Area of rectangular pathway

2. Area of rectangular pathway
Lets draw a rectangle within another rectangle with uniform width around it.
area of the pathway is the area within outer rectangle but outside the inner
rectangle.
Outer rectangle
Rectangular
pathway
Inner rectangle
Area of rectangular pathway = Area of outer rectangle – Area of inner rectangle

3. Area of uniform pathway outside the rectangle
Consider a rectangular building. A uniform flower garden is to be laid outside
the building. How do we find the area of the flower garden?
Outer rectangle
Area of flower garden
Inner rectangle
The uniform flower garden including the building is also a rectangle in shape. Let us
call it as outer rectangle. We call the building as inner rectangle.
To find area of flower garden we need to first find area of outer and inner rectangle

4. Let l and b be the length and breadth of the building. w l w w b
Breadth w
Let l and b be the length and breadth of the building.
Let w be the width of the flower garden
Area of the inner rectangle = l b sq.units.
What is the length and breadth of the outer rectangle ?
The length of the outer rectangle (L) = w + l + w = (l + 2w) units
The breadth of the outer rectangle (B) = w + b + w = (b + 2w) units

5. Area of uniform pathway
Area of the outer rectangle = L x B
Area of the outer rectangle = (l + 2w) (b + 2w) sq. units
Actually, the area of the flower garden is the pathway bounded between two
rectangles.
∴ Area of the flower garden = (Area of building and flower garden) – (Area of building)
We know that
Area of the pathway = (Area of outer rectangle) – (Area of inner rectangle)
i.e. Area of the pathway = (l + 2w) (b + 2w) – lb

6. Area of uniform pathway inside a rectangle
A swimming pool is built in the middle of a rectangular ground leaving an uniform
width all around it to maintain the lawn. What will be the area of lawn?
Outer rectangle
Inner rectangle
Area of lawn
Let l and b be the length and breadth of the outer rectangular ground.
Let w be the width of the lawn
Area of the outer rectangle = l b sq.units.

7. What is the length and breadth of the inner rectangle ?
Length = l - w - w w w
Breadth = b – w - w w
What is the length and breadth of the inner rectangle ?
The length of the inner rectangle (L) = l -w -w = (l - 2w) units
The breadth of the inner rectangle (B) = b -w -w = (b - 2w) units
Area of the outer rectangle = L x B
Area of the outer rectangle = (l - 2w) (b - 2w) sq. units

8. i.e. Area of the pathway = lb – (l - 2w) (b - 2w)
Actually, the area of the lawn is the pathway bounded between two rectangles.
Area of the flower garden = (Area of building and flower garden) – (Area of building)
We know that
Area of the pathway = (Area of outer rectangle) – (Area of inner rectangle)
i.e. Area of the pathway = lb – (l - 2w) (b - 2w)

9. Recap
Area of rectangular pathway = Area of outer rectangle – Area of inner rectangle
2) If the length and breadth of the inner rectangle is given then
area of inner and outer rectangle are as follows
Area of inner rectangle = l x b
Outer rectangle length = l + 2w, breadth= b + 2w respectively
∴ Area of outer rectangle = (l + 2w) x (b + 2w)
Where w is the width of pathway
3) Suppose the length and breadth of the outer rectangle is given then
Area of outer rectangle = l x b
Inner rectangle length = l - 2w, breadth= b - 2w respectively
∴ Area of inner rectangle = (l - 2w) x (b - 2w)

10. Example 1 : The area of outer rectangle is 360 m2
Example 1 : The area of outer rectangle is 360 m2. The area of inner rectangle is
280 m2. The two rectangles have uniform pathway between them. What is the
area of the pathway?
Solution
Given:- Area of outer rectangle = 360m2 , Area of inner rectangle = 280m2
To Find:- Area of the pathway
Area of the pathway = Area of outer rectangle – Area of inner rectangle
= – m2
Area of the pathway = 80 m2 (ans)

11. Inner rectangle (given)
Example2:- The length of a building is 20 m and its breadth is 10 m. A path of
width 1 m is made all around the building outside. Find the area of the path.
Solution :
Given:- Inner rectangle Length(l) =20m, Breadth=10m, width(w) =1 m
To Find:- Find the area of the path
Inner rectangle (given)
Outer rectangle
Since inner rectangle given
Outer rectangle length = l + 2w
breadth= b + 2w
l = 20 m , b = 10 m
Area = l × b
Area = 20 m × 10 m
= 200 m2
width(w) = 1 m
2w= 2 x w = 2x 1 =2
L = l + 2w = = 22 m
B = b + 2w = = 12 m
Area = (l + 2w) (b + 2w)
Area = 22 m x 12 m = 264 m2
Area of the path = Area of outer rectangle – Area of inner rectangle
= –
Area of the path = 64 m2 (ans)

12. Example3:-A school auditorium is 45 m long and 27 m wide
Example3:-A school auditorium is 45 m long and 27 m wide. This auditorium is
surrounded by a varandha of width 3 m on its outside. Find the area of the
varandha. Also, find the cost of laying the tiles in the varandha at the rate of Rs 100 per sq. m.
Solution :
Given:- Inner rectangle Length(l) =45m, Breadth=27m, width(w) =3 m
To Find:- Find the cost of laying the tiles in the varandha

13. Inner rectangle (given)
Step1:- Find area of varandha
Inner rectangle (given)
Outer rectangle
l = 45 m , b = 27 m
Area = l × b
Area = 45 m × 27 m
= 1215 m2
width(w) = 3 m
2w= 2 x w = 2x 3 =6
L = l + 2w = = 51 m
B = b + 2w = = 33 m
Area = (l + 2w) (b + 2w)
Area = 51 m x 33 m= 1683 m2
Since inner rectangle given
Outer rectangle length = l + 2w
breadth= b + 2w
Area of the verandha = Area of outer rectangle – Area of inner rectangle
= –
Area of the verandha = 468 m2 (Ans)

14. Step2:- Find Cost of laying tiles in verandha
Cost of laying tiles for 1 sq. m = Rs 100
Cost of laying tiles for 468 sq. m = 100 x 468 = 46800
Ans:- Cost of laying tiles in the verandha = Rs 46800

15. Example 4:- The length and breadth of a room are 8 m and 5 m respectively. A red colour border of uniform width of 0.5 m has been painted all around on its inside. Find the area of the border.
Solution :
Given:- Outer rectangle Length(l) =8m, Breadth= 5m, width(w) =0.5 m
To Find:- Find area of the border

16. Outer rectangle (given)
Inner rectangle
l = 8 m , b = 5 m
Area = l × b
Area = 8 m × 5 m
= 40 m2
width(w) = 0.5 m
2w= 2 x w = 2x 0.5 =1
L = l - 2w = = 7 m
B = b - 2w = 5 - 1= 4 m
Area = (l - 2w) (b - 2w)
Area = 7 m x 4 m= 28 m2
Since outer rectangle given
Inner rectangle length = l - 2w
breadth= b - 2w
We know that
Area of the pathway = (Area of outer rectangle) – (Area of inner rectangle)
= –
Ans:- Area of the border painted with red colour = 12 m2

17. Try these
A school play ground is rectangular in shape with length 80 m and breadth
60 m. A cemented pathway running all around it on its outside of
width 2 m is built.
A garden is in the form of a rectangle of dimension 30 m # 20 m. A path of width 1.5 m is laid all around the garden on the outside at the
rate of Rs 10 per sq. m. What is the total expense.