Class 6: Maths Chapter 21 solutions. Complete Class 6 Maths Chapter 21 Notes.
RS Aggarwal Solutions for Class 6 Maths Chapter 21–Concept of Perimeter and Area
RS Aggarwal 6th Maths Chapter 21, Class 6 Maths Chapter 21 solutions
Ex 21A Solutions
Question 1.
Solution:
(i) Length (l) = 16.8 cm
Breadth (b) = 6.2 cm
Perimeter = 2 (l + b)
= 2 (16.8 + 6.2) cm
= 2 x 23
= 46 cm
= 30m 6 dm
Question 2.
Solution:
Length of rectangular field (l) = 62 m
and breadth (b) = 33 m
Question 3.
Solution:
Perimeter of field = 128 m
Length + Breadth = 1282 = 64 m
Ratio in length and breadth = 5:3
Let length (l) = 5x
Then breadth = 3x
5x + 3x = 64
=> 5x = 64
=> x = 648 = 8
Length of the field = 5x = 5 x 8 = 40m
and breadth = 3x = 3 x 8 = 24m
Question 4.
Solution:
Cost of fencing a rectangular field
= Rs. 18 per m
Total cost = Rs. 1980
Question 5.
Solution:
Total cost of fencing a rectangular field
= Rs 3300
Rate of fencing = Rs 25 per m
Question 6.
Solution:
(i) Side of square = 3.8 cm
Perimeter = 4 x side
= 4 x 3.8 cm
= 15.2 cm
(ii) Side of a square = 4.6 m
Perimeter = 4 x side
= 4 x 4.6 m
= 18.4 m
(iii) Side of a square = 2 m 5 dm
= 2.5 m
Perimeter = 4 x side
= 4 x 2.5 m
= 10 m
Question 7.
Solution:
Total cost of fencing a square field = Rs. 4480
Rate of fencing = Rs. 35 per m
Question 8.
Solution:
Side of a square field (a) = 21 m
Perimeter = 4a = 4 x 21 = 84m
Perimeter of rectangular field = 84 m
Ratio in length and breadth = 4 : 3
Let length (l) = 4x
and breadth (b) = 3x
Perimeter = 2 (l + b)
Question 9.
Solution:
(i) Sides of a triangle are 7.8 cm, 6.5 cm and 5.9 cm
Question 10.
Solution:
(i) Each side of a regular pentagon
= 8 cm
Perimeter = 5 x Side
= 5 x 8
= 40 cm
(ii) Each side of an octagon = 4.5 cm
Perimeter = 8 x Side
= 8 x 4.5
= 36 cm
(iii) Each side of a regular decagon = 3.6 cm
Perimeter = 10 x Side
= 10 x 3.6
= 36 cm
Question 11.
Solution:
We know that perimeter of a closed figure or a polygon = Sum of its sides
(i) In the figure, sides of a quadrilateral are 45 cm, 35 cm, 27 cm, 35 cm
Its perimeter = Sum of its sides
= (45 + 35 + 27 + 35) cm
= 142 cm
(ii) Sides of a quadrilateral (rhombus) are 18 cm, 18 cm, 18 cm, 18 cm
i.e., each side = 18 cm Perimeter
= 4 x Side
= 4 x 18
= 72 cm
(iii) Sides of the polygon given are 16 cm, 4 cm, 12 cm, 12 cm, 4 cm, 16 cm and 8 cm
Its perimeter = Sum of its sides
= (16 + 4 + 12 + 12 + 4 + 16 + 8) cm
= 72 cm
Ex 21B Solutions
Question 1.
Solution:
(i) Radius of the circle (r) = 28 cm
Circumference = 2 πr
= 2 x 227 x 28 cm
= 176 cm Ans.
Question 2.
Solution:
(i) Diameter of the circle (d) = 14 cm
Circumference = πd
= 227 x 14
= 44 cm
Question 3.
Solution:
Circumference of the circle = 176 cm
Let r be the radius, then
Question 4.
Solution:
Circumference of a wheel = 264 cm
Let d be its diameter, then
πd = 264
=> 227d
= 264
Question 5.
Solution:
Diameter of the wheel (d) = 77 cm
Circumference = πd
Question 6.
Solution:
Diameter of the wheel = 70 cm
circumference = πd
Ex 21C Solutions
Question 1.
Solution:
1. In the given figure, there are 12 complete squares, so its area is 12 cm².
2. In the given figure, there are 18 complete squares, so its area is 18 cm².
3. In the given figure, there are 14.5 complete squares, so its area is 14.5 cm² .
4. In the given figure, there are 6 complete squares and four half parts of a square, the area of the figure is 8 cm².
5. The given figure contains 9 complete squares and 6 half parts of a square. So the area of the figure is
(9+62) = 9 + 3 = 12 cm².
6. The given figure contains 16 complete squares. So, its area is 16 cm².
7. The given figure contains 4 complete squares, 8 more than half parts and 4 less than half parts of a square. Neglecting the less than half parts and considering the more than half parts as complete squares, the approximate area of the figure is 12 cm².
8. The given figure contains 7 complete squares and 5 more than half parts and some less than half parts of a square. Neglecting the less than half parts and considering more than half parts as complete square, the area of the figure is 12 cm² approximately.
9. The given figure contains 14 complete squares and four half parts of a square. So, the area of the figure is
(14+42) cm² = (14 + 2) cm² = 16 cm².
Ex 21D Solutions
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
= 324×17218×12
= 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
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
Question 9.
Solution:
Let the length of the rectangular park = 5x metres
and the breadth of the rectangular park = 3x metres
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 = 1162 = 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
Question 12.
Solution:
Area of rectangle = 540 sq. cm
Length (l) = 36 cm
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
Question 14.
Solution:
Area of a rectangle = 600 cm²
Breadth = 25 cm
Question 15.
Solution:
diagonal of square = 5 √2
Question 16.
Solution:
(i) We name the given region as shown in the figure
Question 17.
Solution:
Measures are in cm
(i) In the figure, there are three rectangles and one square
= 5(6 x 6) cm²
= 180 cm²
Ex 21E Solutions
Objective questions
Mark against the correct answer in each of the following :
Question 1.
Solution:
Ratio in the sides of a rectangle = 7 : 5
and perimeter = 96 cm
Question 2.
Solution:
Area of a rectangle = 650 cm²
and breadth (b) = 13 cm
Question 3.
Solution:
Length of a rectangular field (l) = 34 m
and breadth (b) = 18 m
Circumference = 2 (l + b)
= 2 (34 + 18)m
= 2 x 52
= 104 m
Rate of fencing = Rs. 22.50 per m
Total cost = Rs. 22.50 x 104
= Rs. 2340 (b)
Question 4.
Solution:
Total cost of fencing = Rs. 2400
Rate = Rs. 30 per m
Perimeter of the rectangular field
Question 5.
Solution:
Area of rectangular carpet =120 cm²
Perimeter = 46 m
Now 2 (l + b)
= 46 m
Question 6.
Solution:
Let width of a rectangle = x
Then length = 3x
and diagonal = 6√10 cm
Question 7.
Solution:
Ratio in length and perimeter of a rectangle = 1 : 3
Let length = x,
then perimeter = 3x
Question 8.
Solution:
Length of diagonal of a square = 20 cm
Question 9.
Solution:
Total cost of fencing around a square field = Rs. 2000
and rate = Rs. 25 per metre
Question 10.
Solution:
Circumference = πd
= 227 x 7
= 22 cm (b)
Question 11.
Solution:
(a) Diameter = circumferenceπ
= 88×722
= 28 cm
Question 12.
Solution:
Circumference = πd
= 227 x 70
= 220 cm
Question 13.
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
Question 14.
Solution:
Length of a rectangular room (l) = 5 m 40 cm = 5.4 m
and breadth (b) = 4 m 50 cm
= 4.5 m
Area = l x b = 5.4 x 4.5 m²
= 24.3 m² (b)
Question 15.
Solution:
Length of a sheet (l) = 72 cm
and breadth (b) = 48 cm
Area = l x b = 72 x 48 cm²
Area of paper for one envelope = 18 x 12 cm²
No. of envelopes = 72×4818×12 =16 (d)
RS Aggarwal Solutions for Class 6 Maths Chapter 21: Download PDF
RS Aggarwal Solutions for Class 6 Maths Chapter 21–Concept of Perimeter and Area
Download PDF: RS Aggarwal Solutions for Class 6 Maths Chapter 21–Concept of Perimeter and Area PDF
Chapterwise RS Aggarwal Solutions for Class 6 Maths :
- Chapter 1–Number System
- Chapter 2–Factors and Multiples
- Chapter 3–Whole Numbers
- Chapter 4–Integers
- Chapter 5–Fractions
- Chapter 6–Simplification
- Chapter 7–Decimals
- Chapter 8–Algebraic Expressions
- Chapter 9–Linear Equations in One Variable
- Chapter 10–Ratio, Proportion and Unitary Method
- Chapter 11–Line Segment, Ray and Line
- Chapter 12–Parallel Lines
- Chapter 13–Angles and Their Measurement
- Chapter 14–Constructions (Using Ruler and a Pairs of Compasses)
- Chapter 15–Polygons
- Chapter 16–Triangles
- Chapter 17–Quadrilaterals
- Chapter 18–Circles
- Chapter 19–Three-Dimensional Shapes
- Chapter 20–Two-Dimensional Reflection Symmetry (Linear Symmetry)
- Chapter 21–Concept of Perimeter and Area
- Chapter 22–Data Handling
- Chapter 23–Pictograph
- Chapter 24–Bar Graph
About RS Aggarwal Class 6 Book
Investing in an R.S. Aggarwal book will never be of waste since you can use the book to prepare for various competitive exams as well. RS Aggarwal is one of the most prominent books with an endless number of problems. R.S. Aggarwal's book very neatly explains every derivation, formula, and question in a very consolidated manner. It has tonnes of examples, practice questions, and solutions even for the NCERT questions.
He was born on January 2, 1946 in a village of Delhi. He graduated from Kirori Mal College, University of Delhi. After completing his M.Sc. in Mathematics in 1969, he joined N.A.S. College, Meerut, as a lecturer. In 1976, he was awarded a fellowship for 3 years and joined the University of Delhi for his Ph.D. Thereafter, he was promoted as a reader in N.A.S. College, Meerut. In 1999, he joined M.M.H. College, Ghaziabad, as a reader and took voluntary retirement in 2003. He has authored more than 75 titles ranging from Nursery to M. Sc. He has also written books for competitive examinations right from the clerical grade to the I.A.S. level.
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