RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21D

RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21D

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 21 Concept of Perimeter and Area Ex 21D

Other Exercises

Question 1.
Solution:
(i) Length (l) = 46 cm
Breadth (b) = 25 cm
Area of rectangle = l x b
= 46 x 25 sq. cm
= 1150 sq.cm.
(ii) Length (l) = 9 m
Breadth (b) = 6 m
Area = l x b = 9 x 6
= 54 sq. metre Ans.
(iii) Length (l) = 14.5 m
Breadth (b) = 6.8 m
Area = l x b
= 14.5 x 6.8 sq. m
= 98.6 sq. m Ans.
(iv) Length (l) = 2m 5cm
= 2x 100 cm + 5 cm
= 200 cm + 5 cm
= 205 cm
Breadth = 60 cm
Area = l x b
= 205 cm x 60 cm
= 12300 cm²
(v) Length (l) = 3.5 km
Breadth (b) = 2 km
Area = l x b
= 3.5 x 2
= 7 sq. km. Ans.

Question 2.
Solution:
Side of a square plot = 14 m
Area = (Side)²
= (14)²
= 196 m²

Question 3.
Solution:
Length of top of table (l) = 2 m, 25 cm = 2.25 m
Breadth (l) = 1 m 20 cm = 1.20 m
Area of the top of the table = l x b
= (2.25 x 1.20) sq. m
= 2.7 sq. m. Ans.

Question 4.
Solution:
Length of carpet (l) = 30 m 75 cm
= 30.75 m
Breadth (b) = 80 cm = 0.80 m
Area of the carpet = l x b
= (30.75 x 0.80) sq. m
= 24.6 sq. m
Cost of one square metre = Rs. 20
Total cost = 24.6 x 150
= Rs. 3690. Ans

Question 5.
Solution:
Length of the sheet of paper 3 m 24 cm
= 300 cm + 24 cm
= 324 cm
Breadth of the sheet of the paper 1 m 72 cm
= 100 cm + 72 cm
= 172 cm
Area of the sheet of paper = (324 x 172) cm²
Also, area of the piece of paper required for an envelope = (18 x 12) cm².
Number of envelopes that can be made
= \(\\ \frac { 324\times 172 }{ 18\times 12 } \)
= 258

Question 6.
Solution:
Length of room (l) = 12.5 m
Breadth (b) = 8 m
Area = l x b
= (12.5 x 8) sq. m
= 100 sq. m
Side of square carpet (a) = 8 m
Area of carpet = a² = (8 x 8) sq. m.
= 64 square metre
Area left without carpet = 100 sq. m – 64 sq. m
= 36 sq. m Ans.

Question 7.
Solution:
Length of the lane = 150 m
Breadth of the lane = 9 m
Area of the lane = (150 x 9) m²
= 1350 m²
Area of the brick = 22.5 cm x 7.5 cm
= 168.75 cm²
RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21D Q7.1

Question 8.
Solution:
Length of room (l) = 13 m
Breadth (b) = 9 m
Area of floor or carpet = l x b
= 13 x 9
= 117 sq. m
RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21D Q8.1

Question 9.
Solution:
Let the length of the rectangular park = 5x metres
and the breadth of the rectangular park = 3x metres
RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21D Q9.1

Question 10.
Solution:
Side of the square plot = 64 m
Perimeter of the square plot = 4 x Side
= 4 x 64
= 256 m
Perimeter of the rectangular plot = Perimeter of the square plot = 256 m
Length of the rectangular plot = 70 m
Perimeter = 2 x (Length + Breadth)
256 = 2 (70 + b)
256 = 140 + 2b
=> 2b = 256 – 140
=> 2b = 116
b = \(\\ \frac { 116 }{ 2 } \) = 58 cm
Area of the rectangular plot = (length x breadth)]
= (70 x 58) m²
= 4060 m²
Area of the square plot = Side x Side
= (64 x 64) m²
= 4096 m².
Square plot has the greater area than that of the rectangular plot by
(4096 – 4060)
= 36 m².

Question 11.
Solution:
Total cost of cultivating the rectangular field = Rs. 71400
Rate of cultivating = Rs. 35 per sq. m
RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21D Q11.1

Question 12.
Solution:
Area of rectangle = 540 sq. cm
Length (l) = 36 cm
RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21D Q12.1

Question 13.
Solution:
Measure of a marble tile = 12cm x 10cm
Area of wall = 4 m x 3 m
= 12 m²
Area of one marble tile
= 12 x 10
= 120 cm²
RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21D Q13.1

Question 14.
Solution:
Area of a rectangle = 600 cm²
Breadth = 25 cm
RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21D Q14.1

Question 15.
Solution:
diagonal of square = 5 √2
RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21D Q15.1
RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21D Q15.2

Question 16.
Solution:
(i) We name the given region as shown in the figure
RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21D Q16.1
RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21D Q16.2
RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21D Q16.3
RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21D Q16.4

Question 17.
Solution:
Measures are in cm
(i) In the figure, there are three rectangles and one square
RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21D Q17.1
RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21D Q17.2
RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21D Q17.3
= 5(6 x 6) cm²
= 180 cm²

 

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