Class 6 Maths - Mensuration NCERT Solutions

Chapter 10: Mensuration (NCERT Solutions)

Exercise 10.1 (Perimeter)

Q1. Find the perimeter of each of the following figures:
(a) A square with side 5 cm (b) A rectangle with length 10 cm and breadth 4 cm
(c) An equilateral triangle with side 9 cm (d) A regular pentagon with side 7 cm

(a) Perimeter of square = 4 × side = 4 × 5 = 20 cm

(b) Perimeter of rectangle = 2(l + b) = 2(10 + 4) = 2 × 14 = 28 cm

(d) Perimeter of regular pentagon = 5 × side = 5 × 7 = 35 cm

Q2. The lid of a rectangular box of sides 40 cm by 10 cm is sealed all round with tape. What is the length of the tape required?

Length of tape = Perimeter of rectangle = 2(l + b) = 2(40 + 10) = 2 × 50 = 100 cm.

Q3. A table-top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the table-top?

Length = 2 m 25 cm = 2.25 m. Breadth = 1 m 50 cm = 1.50 m.
Perimeter = 2(2.25 + 1.50) = 2 × 3.75 = 7.50 m.

Q4. What is the length of the wooden strip required to frame a photograph of length and breadth 32 cm and 21 cm respectively?

Length of wooden strip = Perimeter = 2(32 + 21) = 2 × 53 = 106 cm.

Q5. A rectangular piece of land measures 0.7 km by 0.5 km. Each side is to be fenced with 4 rows of wires. What is the length of the wire needed?

Perimeter = 2(0.7 + 0.5) = 2 × 1.2 = 2.4 km.
Total wire needed = 4 × 2.4 = 9.6 km.

Q6. Perimeter of a regular hexagon is 88 cm. What is the length of each side?

Number of sides of hexagon = 6.
Side = Perimeter / 6 = 88 / 6 ≈ 14.67 cm (or exactly 44/3 cm).

Q7. Find the perimeter of a triangle with sides measuring 10 cm, 14 cm, and 15 cm.

Perimeter = 10 + 14 + 15 = 39 cm.

Q9. Avneet buys 9 square paving slabs, each with a side of 1/2 m. He lays them in the form of a square. What is the perimeter of the square laid?

9 slabs arranged as a 3×3 square.
Side of big square = 3 × (1/2) = 3/2 m = 1.5 m.
Perimeter = 4 × 1.5 = 6 m.

Exercise 10.2 (Area)

Q1. Find the area of the following figures by counting the squares (each square = 1 sq. cm).

Area is calculated by counting the number of complete squares enclosed within the figure, plus adding half for each half-square.

Full square counts as 1 sq. cm. Half-covered squares count as 0.5 sq. cm. Squares covered less than half are ignored; more than half count as 1.

Exercise 10.3 (Area Formulas)

Q1. Find the area of a rectangle whose: (a) length = 12 m, breadth = 21 m (b) length = 2 km, breadth = 3 km (c) length = 3 m, breadth = 70 cm

(a) Area = l × b = 12 × 21 = 252 sq. m

(b) Area = 2 × 3 = 6 sq. km

Q2. Find the area of a square park whose perimeter is 320 m.

Perimeter = 4 × side = 320 m → Side = 80 m.
Area = side² = 80 × 80 = 6400 sq. m.

Q3. Find the breadth of a rectangular plot of land, if its area is 440 sq. m and length is 22 m. Also find its perimeter.

Area = l × b → 440 = 22 × b → b = 440/22 = 20 m.
Perimeter = 2(22 + 20) = 2 × 42 = 84 m.

Q4. The perimeter of a rectangular sheet is 100 cm. If length is 35 cm, find its breadth. Also find the area.

100 = 2(35 + b) → 50 = 35 + b → b = 15 cm.
Area = 35 × 15 = 525 sq. cm.

Q5. The area of a square park is the same as that of a rectangular park. If the side of the square park is 60 m and length of the rectangular park is 90 m, find the breadth of the rectangular park.

Area of square = 60 × 60 = 3600 sq. m.
Area of rectangle = 3600 sq. m.
b = 3600 / 90 = 40 m.

Q6. 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 in the shape of a square, what will be the measure of each side?

Perimeter of rectangle = 2(40 + 22) = 2 × 62 = 124 cm.
This is the total length of wire.
Side of square = 124 / 4 = 31 cm.

Q7. The perimeter of a rectangle and a square are equal. If the perimeter is 40 cm and length of the rectangle is 14 cm, find breadth of rectangle and side of the square. Which has greater area?

Breadth of rectangle = (Perimeter/2) − length = 20 − 14 = 6 cm.
Side of square = 40/4 = 10 cm.
Area of rectangle = 14 × 6 = 84 sq. cm.
Area of square = 10 × 10 = 100 sq. cm.
Square has greater area.

Class 6 Maths - Mensuration Practice Questions

Chapter 10: Mensuration (Practice Questions)

RD Sharma / HOT Practice

Q1. A garden is 18 m long and 12 m wide. It has a path 2 m wide inside all around it. Find the area of the path.

Area of outer garden = 18 × 12 = 216 sq. m.
Inner dimensions (after removing 2 m path on each side): Length = 18 − 4 = 14 m; Breadth = 12 − 4 = 8 m.
Area of inner plot = 14 × 8 = 112 sq. m.
Area of path = 216 − 112 = 104 sq. m.

Q2. The side of a square is 6 cm. If the side is doubled, how does the area change?

Original area = 6² = 36 sq. cm.
New side = 12 cm. New area = 12² = 144 sq. cm.
Ratio = 144/36 = 4. The area becomes 4 times the original area.

Q3. How many tiles of size 50 cm × 50 cm are needed to cover a floor of size 5 m × 4 m?

Area of floor = 500 × 400 = 2,00,000 sq. cm (converting to cm).
Area of one tile = 50 × 50 = 2500 sq. cm.
Number of tiles = 2,00,000 / 2500 = 80 tiles.

Q4. The length of a rectangle is three times its breadth. If perimeter is 64 cm, find length and breadth.

Let breadth = x. Length = 3x.
2(3x + x) = 64 → 2 × 4x = 64 → 8x = 64 → x = 8.
Breadth = 8 cm. Length = 24 cm.

Q5. A rectangular field has area 630 sq. m. If its length is 35 m, find its perimeter.

Breadth = 630 / 35 = 18 m.
Perimeter = 2(35 + 18) = 2 × 53 = 106 m.

Q6. The perimeter of a square field is 360 m. A rectangular field has the same perimeter. If length of rectangle is 100 m, compare their areas.

Side of square = 360/4 = 90 m. Area of square = 90² = 8100 sq. m.
Breadth of rectangle = (360/2) − 100 = 180 − 100 = 80 m. Area of rectangle = 100 × 80 = 8000 sq. m.
Square has greater area (8100 > 8000).

Q7. A room is 15 m long and 9 m wide. How many carpet tiles of 1.5 m × 1 m are needed to cover the floor?

Area of room = 15 × 9 = 135 sq. m.
Area of one tile = 1.5 × 1 = 1.5 sq. m.
Tiles needed = 135 / 1.5 = 90 tiles.

Q8. A farmer has a square plot with side 120 m. He wants to fence it with 3 rows of barbed wire. How much wire does he need?

Perimeter = 4 × 120 = 480 m. Wire for 3 rows = 3 × 480 = 1440 m.

Q9. Find the cost of fencing a square plot of side 40 m at the rate of Rs. 10 per metre.

Perimeter = 4 × 40 = 160 m.
Cost = 160 × 10 = Rs. 1600.

Q10. A rectangular park is 60 m × 40 m. Find the cost of laying grass at Rs. 25 per sq. m.

Area = 60 × 40 = 2400 sq. m.
Cost = 2400 × 25 = Rs. 60,000.

Q11. The cost of paving a square courtyard is Rs. 45,000 at Rs. 500 per sq. m. Find the side of the courtyard.

Area = 45000 / 500 = 90 sq. m.
Side = square root of 90 ≈ 9.49 m (approx). Or if area = 81 sq. m, side = 9 m.

Q12. Two rectangles A and B have the same area. A has length 20 m and breadth 15 m. B has length 25 m. Find the breadth of B.

Area of A = 20 × 15 = 300 sq. m.
Breadth of B = 300 / 25 = 12 m.

Q13. A square floor tile has side 20 cm. How many tiles are needed to lay 400 such tiles in a row of length?

Length of row = 400 × 20 cm = 8000 cm = 80 m.

Q14. The perimeter of an isosceles triangle is 42 cm. If one of the equal sides is 15 cm, find the base.

Base = Perimeter − 2 × equal side = 42 − 2 × 15 = 42 − 30 = 12 cm.

Q15. A wire 132 cm long is bent into the shape of a rectangle with length 20 cm more than its breadth. Find dimensions and area.

Perimeter = 132. Let breadth = b, length = b + 20.
2(b + b + 20) = 132 → 2(2b + 20) = 132 → 2b + 20 = 66 → 2b = 46 → b = 23.
Breadth = 23 cm. Length = 43 cm. Area = 43 × 23 = 989 sq. cm.

Q16. A floor 25 m × 16 m is covered with square tiles of side 2 m. How many tiles are needed?

Area of floor = 25 × 16 = 400 sq. m. Area of tile = 2 × 2 = 4 sq. m.
Number of tiles = 400 / 4 = 100 tiles.

Q17. If the perimeter of each figure is 24 cm, which has the greatest area: (a) square with side 6 cm, (b) rectangle 8 cm × 4 cm?

(a) Square: side = 6 cm, area = 36 sq. cm.
(b) Rectangle: area = 8 × 4 = 32 sq. cm.
Square has greater area (36 > 32).
Key insight: For a given perimeter, the square always has the maximum area.

Q18. The area of a square is 144 sq. cm. Find its perimeter.

Side = square root of 144 = 12 cm.
Perimeter = 4 × 12 = 48 cm.

Q19. A rectangular garden 70 m × 40 m has two 2 m wide paths parallel to its sides running through the middle. Find the area occupied by the paths.

Horizontal path (along length): 70 × 2 = 140 sq. m.
Vertical path (along breadth): 40 × 2 = 80 sq. m.
Overlap (centre square): 2 × 2 = 4 sq. m.
Total path area = 140 + 80 − 4 = 216 sq. m.

Q20. Find the area of an L-shaped figure formed by two rectangles: Rectangle A is 6 cm × 4 cm and Rectangle B is 3 cm × 2 cm joined at a corner.

Area = Area of A + Area of B = (6 × 4) + (3 × 2) = 24 + 6 = 30 sq. cm.
(Divide irregular shapes into simpler rectangles and add areas.)

Class 6 Maths - Mensuration Summary

Chapter 10: Mensuration (Concepts & Summary)

1. Perimeter Formulas

Perimeter = Total length of boundary of a closed figure.

Shape	Formula
Rectangle	P = 2(l + b)
Square	P = 4 × side
Triangle	P = a + b + c (sum of all sides)
Equilateral triangle	P = 3 × side
Regular polygon (n sides)	P = n × side

2. Area Formulas

Area = Amount of surface enclosed by a closed figure. Measured in square units (sq. cm, sq. m, etc.).

Shape	Formula
Rectangle	A = l × b
Square	A = side × side = side²

3. Derived Formulas / Finding Missing Dimension

To Find	Formula
Length of rectangle	l = Area ÷ b
Breadth of rectangle	b = Area ÷ l
Side of square (from area)	side = √Area
Side of square (from perimeter)	side = P ÷ 4
Number of tiles needed	Count = Area of floor ÷ Area of one tile
Cost of fencing / laying	Cost = Perimeter (or Area) × Rate

4. Units of Measurement

1 m = 100 cm → 1 sq. m = 10,000 sq. cm
1 km = 1000 m → 1 sq. km = 10,00,000 sq. m
1 hectare = 10,000 sq. m
Always convert to same units before calculating.

5. Key Insights

Two figures can have the same perimeter but different areas.
Two figures can have the same area but different perimeters.
For a fixed perimeter, the square always has the maximum area among rectangles.
If side of a square is doubled → area becomes 4 times.
For irregular shapes: divide into simple rectangles/squares, find each area, and add.
For path problems: Path area = Outer area − Inner area.

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