Exercise Solutions Available
Step-by-step answers for all textbook questions in this chapter
NCERT Solutions for Class 6 Maths Chapter 10 — Mensuration
Complete step-by-step NCERT exercise solutions for CBSE Class 6 Maths Chapter 10 Mensuration. Covers perimeter and area of rectangles, squares, triangles, and regular polygons with detailed worked examples, common mistakes, and exam tips.
Key Concepts
| Concept | Definition / Formula |
|---|---|
| Perimeter | The total distance around a closed figure; sum of all side lengths |
| Perimeter of Rectangle | P = 2 × (l + b), where l = length, b = breadth |
| Perimeter of Square | P = 4 × side |
| Perimeter of Triangle | P = a + b + c (sum of three sides) |
| Perimeter of Regular Polygon | P = number of sides × length of one side |
| Area | The amount of surface enclosed by a closed figure; measured in square units |
| Area of Rectangle | A = l × b |
| Area of Square | A = side × side = side² |
| Area of Triangle | A = (1/2) × base × height |
| Area using Unit Squares | Count the number of 1-unit squares that fit inside the figure |
Exercise 10.1 — Perimeter
Question 1
Find the perimeter of each of the following figures:
(a) A rectangle with length 12 cm and breadth 8 cm.
Perimeter of rectangle = 2 × (l + b)
= 2 × (12 + 8)
= 2 × 20
= 40 cm
(b) A square with side 9 cm.
Perimeter of square = 4 × side
= 4 × 9
= 36 cm
(c) An equilateral triangle with side 6 cm.
Perimeter of equilateral triangle = 3 × side
= 3 × 6
= 18 cm
Question 2
A rectangular field is 60 m long and 40 m wide. Find its perimeter. How much fencing is required to enclose it?
Perimeter = 2 × (l + b)
= 2 × (60 + 40)
= 2 × 100
= 200 m
Therefore, 200 m of fencing is required to enclose the field.
Question 3
Find the perimeter of a regular hexagon with each side measuring 5 cm.
A regular hexagon has 6 equal sides.
Perimeter of regular hexagon = Number of sides × Length of one side
= 6 × 5
= 30 cm
Question 4
Find the perimeter of a triangle with sides 7 cm, 9 cm, and 11 cm.
Perimeter = Sum of all three sides
= 7 + 9 + 11
= 27 cm
Question 5
Find the perimeter of a regular pentagon with each side measuring 3.5 cm.
A regular pentagon has 5 equal sides.
Perimeter = Number of sides × Length of one side
= 5 × 3.5
= 17.5 cm
Question 6
A piece of string is 48 cm long. What will be the length of each side if the string is used to form:
(a) A square
Perimeter of square = 48 cm
4 × side = 48
side = 48 ÷ 4 = 12 cm
(b) An equilateral triangle
Perimeter of equilateral triangle = 48 cm
3 × side = 48
side = 48 ÷ 3 = 16 cm
(c) A regular hexagon
Perimeter of regular hexagon = 48 cm
6 × side = 48
side = 48 ÷ 6 = 8 cm
Question 7
Two sides of a triangle are 12 cm and 14 cm. If the perimeter of the triangle is 38 cm, find the third side.
Perimeter = Side 1 + Side 2 + Side 3
38 = 12 + 14 + Side 3
38 = 26 + Side 3
Side 3 = 38 − 26
= 12 cm
Question 8
A rectangular garden is 75 m long and 45 m broad. A path of 2 m width runs all around it on the outside. Find the perimeter of the outer boundary of the path.
The path adds 2 m on each side of the garden.
Outer length = 75 + 2 + 2 = 79 m
Outer breadth = 45 + 2 + 2 = 49 m
Perimeter of outer boundary = 2 × (79 + 49)
= 2 × 128
= 256 m
Question 9
Find the cost of fencing a rectangular park of length 125 m and breadth 75 m at the rate of Rs 12 per metre.
Perimeter = 2 × (l + b) = 2 × (125 + 75) = 2 × 200 = 400 m
Cost of fencing = Perimeter × Rate
= 400 × 12
= Rs 4,800
Question 10
Sweety runs around a square field of side 75 m. Raju runs around a rectangular field with length 80 m and breadth 70 m. Who covers more distance and by how much?
Sweety's distance (square):
Perimeter of square = 4 × 75 = 300 m
Raju's distance (rectangle):
Perimeter of rectangle = 2 × (80 + 70) = 2 × 150 = 300 m
Both cover the same distance of 300 m. The difference is 0 m.
Exercise 10.2 — Area
Question 1
Find the area of a rectangle with length 14 cm and breadth 9 cm.
Area of rectangle = l × b
= 14 × 9
= 126 cm²
Question 2
Find the area of a square whose side is 11 cm.
Area of square = side × side
= 11 × 11
= 121 cm²
Question 3
The area of a rectangle is 240 cm². If its length is 20 cm, find its breadth.
Area = l × b
240 = 20 × b
b = 240 ÷ 20
= 12 cm
Question 4
The area of a square is 625 cm². Find the length of its side.
Area = side × side
side² = 625
side = √625
= 25 cm
Question 5
A floor is 5 m long and 4 m wide. A square carpet of side 3 m is laid on the floor. Find the area of the floor that is not covered by the carpet.
Area of floor = l × b = 5 × 4 = 20 m²
Area of carpet = side × side = 3 × 3 = 9 m²
Area not covered = Area of floor − Area of carpet
= 20 − 9
= 11 m²
Question 6
How many tiles whose length and breadth are 12 cm and 5 cm respectively will be needed to fit in a rectangular region whose length and breadth are respectively:
(a) 100 cm and 144 cm
Area of one tile = 12 × 5 = 60 cm²
Area of region = 100 × 144 = 14,400 cm²
Number of tiles = Area of region ÷ Area of one tile
= 14,400 ÷ 60
= 240 tiles
(b) 70 cm and 36 cm
Area of region = 70 × 36 = 2,520 cm²
Number of tiles = 2,520 ÷ 60
= 42 tiles
Question 7
A room is 6 m long and 5 m wide. A square carpet of side 4 m is laid on the floor. Find the area of the floor that is left uncovered.
Area of room = 6 × 5 = 30 m²
Area of carpet = 4 × 4 = 16 m²
Uncovered area = 30 − 16
= 14 m²
Question 8
Find the area of the following figure by counting the unit squares (each unit square has area 1 cm²):
Exercise 10.3 — Word Problems on Perimeter and Area
Question 1
Find the area of a rectangle whose length is 15 cm and breadth is 10 cm. Also find its perimeter.
Area = l × b = 15 × 10 = 150 cm²
Perimeter = 2 × (l + b) = 2 × (15 + 10) = 2 × 25 = 50 cm
Question 2
A rectangular piece of land measures 300 m by 200 m. Find:
(a) Its perimeter
Perimeter = 2 × (300 + 200) = 2 × 500 = 1,000 m
(b) Cost of fencing at Rs 8 per metre
Cost = 1,000 × 8 = Rs 8,000
(c) Cost of levelling at Rs 2 per square metre
Area = 300 × 200 = 60,000 m²
Cost = 60,000 × 2 = Rs 1,20,000
Question 3
A square playground has a perimeter of 320 m. Find its area.
Perimeter = 4 × side
320 = 4 × side
side = 320 ÷ 4 = 80 m
Area = side × side = 80 × 80 = 6,400 m²
Question 4
The perimeter of a rectangle is 130 cm. If the breadth is 30 cm, find its length and area.
Perimeter = 2 × (l + b)
130 = 2 × (l + 30)
130 = 2l + 60
2l = 130 − 60 = 70
l = 35 cm
Area = l × b = 35 × 30 = 1,050 cm²
Question 5
A verandah of width 2 m is constructed all along outside a room which is 5 m long and 4 m wide. Find the area of the verandah.
Inner dimensions: 5 m × 4 m
The verandah of 2 m width is added on all sides.
Outer length = 5 + 2 + 2 = 9 m
Outer breadth = 4 + 2 + 2 = 8 m
Area of outer rectangle = 9 × 8 = 72 m²
Area of inner rectangle = 5 × 4 = 20 m²
Area of verandah = 72 − 20 = 52 m²
Question 6
A wire is in the shape of a rectangle. Its length is 40 cm and breadth is 22 cm. If the same wire is rebent into a square, what will be the side of the square? Which encloses more area?
Perimeter of rectangle = 2 × (40 + 22) = 2 × 62 = 124 cm
When rebent into square, perimeter remains the same.
4 × side = 124
side = 31 cm
Area of rectangle = 40 × 22 = 880 cm²
Area of square = 31 × 31 = 961 cm²
The square encloses more area by 961 − 880 = 81 cm²
Question 7
The sides of a rectangular field are in the ratio 3:2. If the area of the field is 3,456 m², find the cost of fencing it at Rs 4.50 per metre.
Let length = 3x, breadth = 2x
Area = l × b
3,456 = 3x × 2x
3,456 = 6x²
x² = 576
x = 24
Length = 3 × 24 = 72 m
Breadth = 2 × 24 = 48 m
Perimeter = 2 × (72 + 48) = 2 × 120 = 240 m
Cost of fencing = 240 × 4.50 = Rs 1,080
Question 8
A farmer has a rectangular field of length 120 m and breadth 80 m. He wants to fence it with 3 rounds of wire. If the cost of wire is Rs 5 per metre, find the total cost.
Perimeter = 2 × (120 + 80) = 2 × 200 = 400 m
Length of wire for 3 rounds = 3 × 400 = 1,200 m
Total cost = 1,200 × 5 = Rs 6,000
Question 9
How many square tiles of side 25 cm will be needed to pave the floor of a room 10 m long and 8 m wide?
Convert room dimensions to cm: 10 m = 1,000 cm, 8 m = 800 cm
Area of floor = 1,000 × 800 = 8,00,000 cm²
Area of one tile = 25 × 25 = 625 cm²
Number of tiles = 8,00,000 ÷ 625
= 1,280 tiles
Question 10
The length and breadth of a playground are 80 m and 35 m respectively. Find the cost of levelling it at Rs 3.50 per m². Also find the cost of fencing it at Rs 5.25 per metre.
Levelling:
Area = 80 × 35 = 2,800 m²
Cost of levelling = 2,800 × 3.50 = Rs 9,800
Fencing:
Perimeter = 2 × (80 + 35) = 2 × 115 = 230 m
Cost of fencing = 230 × 5.25 = Rs 1,207.50
Common Mistakes
1. Confusing perimeter and area formulas: Students often write Area = 2(l + b) instead of Perimeter = 2(l + b). Remember: perimeter is the boundary (add all sides), area is the surface inside (length × breadth).
2. Forgetting to convert units: If the length is in metres and breadth is in centimetres, you must convert to the same unit before calculating. 1 m = 100 cm. Always check units first.
3. Writing area in wrong units: Area must be written in square units (cm², m²). Writing "cm" or "m" for area is incorrect. Perimeter uses linear units (cm, m).
4. Applying rectangle formula to triangles: The area of a triangle is NOT l × b. It is (1/2) × base × height. Always halve the rectangle area to get the triangle area.
5. Not reading "number of rounds" in fencing problems: If a question says "3 rounds of fencing," multiply the perimeter by 3. Many students calculate just one round and stop.
6. Incorrect side calculation from area of square: If Area = side², then side = √Area. Students sometimes divide the area by 4 (which gives the side only when area = 4 × side², which is never true). Use the square root.
Exam Checklist
- [ ] I know that perimeter = sum of all sides for any closed figure.
- [ ] I can write and apply the formula P = 2(l + b) for a rectangle.
- [ ] I can write and apply the formula P = 4 × side for a square.
- [ ] I can calculate the perimeter of a regular polygon (number of sides × side length).
- [ ] I know that area of rectangle = l × b and area of square = side².
- [ ] I always express area in square units (cm², m²) and perimeter in linear units (cm, m).
- [ ] I convert all measurements to the same unit before computing.
- [ ] I can find a missing side when given perimeter and other sides of a figure.
- [ ] I can solve word problems involving cost of fencing (perimeter) or cost of levelling/tiling (area).
- [ ] I remember to multiply by the number of rounds when fencing is done in multiple rounds.
Quick Revision Table
| Figure | Perimeter | Area |
|---|---|---|
| Rectangle | 2(l + b) | l × b |
| Square | 4 × side | side² |
| Equilateral Triangle | 3 × side | (√3/4) × side² |
| Regular Pentagon | 5 × side | — |
| Regular Hexagon | 6 × side | — |
These solutions follow the NCERT textbook pattern for Class 6 Maths Chapter 10 — Mensuration. Practice each question step-by-step and verify your answers.