# RBSE Solutions for Class 9 Maths Chapter 11 Area of Plane Figures Ex 11.4

RBSE Solutions for Class 9 Maths Chapter 11 Area of Plane Figures Ex 11.4 is part of RBSE Solutions for Class 9 Maths. Here we have given Rajasthan Board RBSE Class 9 Maths Solutions Chapter 11 Area of Plane Figures Ex 11.4.

 Board RBSE Textbook SIERT, Rajasthan Class Class 9 Subject Maths Chapter Chapter 11 Chapter Name Area of Plane Figures Exercise Ex 11.4 Number of Questions Solved 11 Category RBSE Solutions

## Rajasthan Board RBSE Class 9 Maths Solutions Chapter 11 Area of Plane Figures Ex 11.4

Question 1.
Find the perimeter and area (RBSESolutions.com) of the following rectangles.
(i) length = 9.5 m, breadth = 7.5 m
(ii) length = 125 m, breadth = 75 m
(iii) length = 12.5 cm, breadth = 7.5 cm
Solution.
(i) Perimeter of a rectangle
= 2(9.5 + 7.5) = 2 x 17 = 34 m
Area of rectangle = 9.5 x 7.5 = 71.25 m²
(ii) Perimeter of a rectangle
= 2(125 + 75)
= 400 m
Area of rectangle = 125 x 75 = 9375 m²
(iii) Perimeter of a rectangle
= 2(12.5 + 7.5)
= 2 x 20 = 40 cm
Area of rectangle = 12.5 x 7.5 = 93.75 cm². Question 2.
Find the area and perimeter (RBSESolutions.com) of the following square whose sides are
(i) 5.3 m
(ii) 8.5 m
(iii) 9.6 m
Solution.
(i) Area of square = (side)²
= (5.3 m)² = 28.09 m²
Perimeter = 4 x one side
= 4 x 5.3 m = 21.2 m
(ii) Area of square = (8.5 m)²
= 72.25 m²
Perimeter = 4 x 8.5 = 34 m
(iii) Area of square = (9.6 m)²
= 9.6 m x 9.6 m = 92.16 m²
Perimeter = 4 x 9.6 = 38.4 m

Question 3.
The distance covered in taking 4 rounds (RBSESolutions.com) of a square ground is 4 km. Find the length and area of the ground.
Solution.
Distance covered in 4 round = 4 km
∴ Distance covered in 1 round = 1 km
= 1000 m
Hence, perimeter = 1000 m
4 x one side = 1000 m
∴ one side = 250 m
Now, area of the square field = (250 m)²
= 62500 m²

Question 4.
A rectangular plot 75 m in length and 45 m width. How (RBSESolutions.com) many flower-beds can be prepared if each bed is 5 m in length and 3 m width?
Solution.
Area of rectangular plot
= 75 m x 45 m
Area of one flower-bed
= l x b
= 5 x 3 m²
∴ No. of flower-bed Question 5.
A room is 10 m long and 5 m wide. How many squares of area 50 sq. cm are required? Also (RBSESolutions.com) find the cost of paving squares @ of Rs 20 each.
Solution.
Area of room = length x breadth
= 10 m x 5 m
= 50 sq. m
Area of one square = 50 sq. m ∵ Cost of paving 10000 squares @ Rs 20 .
= 10000 x 20
= Rs 2,00,000

Question 6.
The area of a square field is 2025 sq. m and around it, there (RBSESolutions.com) is a path way 3.5 metre wide, find the area of the pathway.
Solution.
The area of a square field = 2025 sq. m The side of the square field = √Area = √2025 = 45 m
The side of the field including the path = 45 + (3.5 + 3.5) = 52 m
∴ The area of the field including the path = 52 x 52 = 2704 m²
∴ Area of the path = (2704 – 2025) m² = 679 m² Question 7.
A road of 2.5 m wide, around (RBSESolutions.com) a rectangular room 35 metre long and 40 metre wide, is to be covered by square blocks of cement of dimensions 80 cm by 50 cm. Find the total number of such blocks.
Solution.
Area of a rectangular room
= 40 m x 35 m = 1400 m²
Area of rectangular room with road
= 45 m x 40 m = 1800 m²
Area of road = outer area – inner area
= 1800 m² -1400 m² = 400 m²
∴ Required number of square blocks Question 8.
A pathway 4 metre wide has been constructed centrally in (RBSESolutions.com) a rectangular garden parallel to its sides of length 60 metre and width 40 metre. Find the area of the pathway and find the total expenditure to cover this pathway with sand at the rate of Rs 100 per sq. metre.
Solution.
Length of the garden = 60 m
Breadth of the garden = 40 m and width of the path = 4 m
∴ Area of the path along length
= 60 x 4 = 240 m²
Area of the path along breadth
= 40 x 4 m = 160 m²
Common area = 4m x 4m = 16 m²
∴ Total area of the path
= 240 + 160 – 16 = 384 m²
∵ Expenditure of lying sand per m² Rs 100.
∴ Total expenditure of sand on 384 m² = 384 x 100 = Rs 38,400

Question 9.
Find the area and perimeter of the given (RBSESolutions.com) figure with help of the given dimension. Solution.
For finding area, we will divide the given figure, into three blocks naming A, B and C.
Area of block A = 12 m x 2 m = 24 m²
Area of block B = 5 m x 2 m = 10 m²
Area of block C = 12 m x 2 m = 24 m²
∴ Total area of the given figure = 24 m² + 10 m² + 24 m² = 58 m² and perimeter of the whole figure
= 2m + 12m + 2m + 5m + 5m + 5m + 2m + 12m + 2m + 5m + 5m + 5m
= 62 m

Question 10.
The length, breadth and height of (RBSESolutions.com) a room are 15.35 m, 4.65 m and 6.50 m respectively. If there are 4 doors each of size 1.5 m x 1.3 m and 3 windows each of size 1.5 m x 1.2 m, then find the cost to paint the four walls of room at the rate of Rs 3.40 per sq. m.
Solution.
Here l = 15.35 m, b = 4.65 m and h = 6.50 m
∴ Area of four walls
= 2 x (l + b) x h
= 2 x (15.35 + 4.65) x 6.50
= 2 x 20 x 6.50
= 260 m²
Area of four doors
= 4 x 1.5 x 1.3
= 7.8 m²
Area of 3 windows
= 3 x 1.5 x 1.2
= 5.4 m²
The area of the four walls excluding (RBSESolutions.com) door and windows
= 260 – (7.8 + 5.4)
= 260 – 13.2
= 246.8 m²
∵ Cost to paint 1 m² is Rs 3.40
∴ Cost to paint 246.8 m²
= 246.8 x 3.40
= Rs 839.12

Question 11.
A rectangular water tank is 10 metre long, 8 metre wide and 2 metre deep. Find the total cost to repair (RBSESolutions.com) its four walls and bottom at the rate of Rs 15 per square metre.
Solution.
Floor area of water tank
= length x breadth = 10 x 8 = 80 m²
Area of four walls of the water tank
= 2(l + b) x h
= 2 x (10 + 8) x 2
= 2 x 18 x 2
= 72 sq. m
Total area = (72 + 80) m²
= 152 m²
∵ Cost of repairing 1 m² is Rs 15.
∴Cost of repairing 152 m² = Rs 15 x 152
= Rs 2,280 We hope the given RBSE Solutions for Class 9 Maths Chapter 11 Area of Plane Figures Ex 11.4 will help you. If you have any query regarding RBSE Rajasthan Board Solutions for Class 9 Maths Chapter 11 Area of Plane Figures Ex 11.4, drop a comment below and we will get back to you at the earliest.