Download free pdf of Chapter 10 **Mensuration** **Worksheets for Class 6 Maths with answers**. Students can easily get all the Class 6 Maths Worksheets and Exercises in Free Printable ( PDF Format) as per the latest syllabus and examination pattern in CBSE and NCERT pattern. Below you will find all the Mensuration related sums that are asked in 6th Grade Mathematics Exams Across Indian Schools.

## Mensuration Worksheets for Class 6 in PDF Format

- Class 6 Maths Chapter 10 Mensuration Worksheet – 4
- CBSE Class 6 Maths Worksheet Mensuration – 3
- CBSE Class 6 Maths Worksheet Mensuration – 2
- CBSE Class 6 Maths Worksheet Mensuration – 1
- Class 6 Maths Chapter 10 Mensuration Worksheet-6 with Answer
- Class 6 Maths Chapter 10 Mensuration Worksheet-5 with Answer
- Class 6 Maths Chapter 10 Mensuration Worksheet-4 with Answer
- Class 6 Maths Chapter 10 Mensuration Worksheet-3 with Answer
- Class 6 Maths Chapter 10 Mensuration Worksheet-2 with Answer
- Class 6 Maths Chapter 10 Mensuration Worksheet-1 with Answer
- Mensuration CBSE Class 6 Worksheet for Maths
- Worksheet on Mensuration for CBSE Class 6 Maths
- Maths Worksheets for CBSE Class 6 Mensuration
- Mensuration CBSE Class 6 Maths Worksheet in PDF
- Class 6 Maths Worksheet – Mensuration
- Class 6 Maths MCQ Worksheet 5 – Mensuration
- Class 6 Maths MCQ Worksheet 4 – Mensuration
- Class 6 Maths MCQ Worksheet 3 – Mensuration
- Class 6 Maths MCQ Worksheet 2 – Mensuration
- Class 6 Maths MCQ Worksheet 1 – Mensuration

### Class 6 Mensuration Worksheet with Answers

**Q.1** Find the perimeter of a regular pentagon with each side measuring 3 cm.

**Answer** : A regular pentagon has 5 equal sides. Perimeter = 5 * Side length = 5 * 3 cm = 15 cm.

**Q.2** The perimeter of a regular hexagon is 18 cm. How long is its one side?

**Answer:** A regular hexagon has 6 equal sides. Side length = Perimeter / 6 = 18 cm / 6 = 3 cm.

**Q.3** Find the perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.

**Answer** : Perimeter = Sum of all sides = 10 cm + 14 cm + 15 cm = 39 cm.

**Q.4** Find the perimeter of a regular hexagon with each side measuring 8 m.

**Answer: **A regular hexagon has 6 equal sides. Perimeter = 6 * Side length = 6 * 8 m = 48 m.

**Q.5** Find the side of the square whose perimeter is 20 m.

**Answer:** For a square, all sides are equal. Perimeter = 4 * Side length => 20 m = 4 * Side length. Solving for Side length: Side length = 20 m / 4 = 5 m.

**Q.6** The perimeter of a regular pentagon is 100 cm. How long is its each side?

**Answer :**A regular pentagon has 5 equal sides. Perimeter = 5 * Side length = 100 cm. Solving for Side length: Side length = 100 cm / 5 = 20 cm.

**Q.7** Two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is its third side?

**Answer**: Let the third side be x. The perimeter is the sum of all sides, so 12 cm + 14 cm + x = 36 cm. Solving for x: x = 36 cm – 12 cm – 14 cm = 10 cm.

**Q.8** Find the area of the rectangles whose lengths and breadths are given below :

(a) 15cm, 8 cm (b) 3 dm, 5.6 cm (c) 2 m 5 dm, 1 m 20 cm (d) 6.5 m, 4.5 m

**Answer **:

(a) 15 cm, 8 cm Area = Length × Breadth = 15 cm × 8 cm = 120 sq cm

(b) 3 dm, 5.6 cm Convert dm to cm (1 dm = 10 cm): 3 dm = 3 * 10 cm = 30 cm Area = Length × Breadth = 30 cm × 5.6 cm = 168 sq cm

(c) 2 m 5 dm, 1 m 20 cm Convert m and dm to cm: 2 m 5 dm = (2 * 100 cm) + (5 * 10 cm) = 200 cm + 50 cm = 250 cm 1 m 20 cm = (1 * 100 cm) + (20 cm) = 120 cm Area = Length × Breadth = 250 cm × 120 cm = 30000 sq cm

(d) 6.5 m, 4.5 m Area = Length × Breadth = 6.5 m × 4.5 m = 29.25 sq m

**Q.9** Find the cost of fencing a square park of side 250 m at the rate of Rs 20 per metre.

**Answer** : Cost of fencing = Perimeter * Rate per meter = 1000 m * Rs 20/m = Rs 20,000.

**Q.10** Find the cost of fencing a rectangular park of length 175 m and breadth 125 m at the rate of Rs 12 per metre.

**Answer** : Perimeter of the rectangle = 2 * (Length + Breadth) = 2 * (175 m + 125 m) = 2 * 300 m = 600 m Cost of fencing = Perimeter * Rate per meter = 600 m * Rs 12/m = Rs 7,200

**Q.11** There is a rectangular lawn 10m long and 4m wide in front of Meena’s house (see above right sided figure). It is fenced along the two smaller sides and one longer side leaving a gap of 1m for the entrance. Find the length of fencing.

**Answer **:

Perimeter of the rectangle = 2 * (Length + Breadth) = 2 * (10 m + 4 m) = 2 * 14 m = 28 m Length of the entrance gap = 1 m Length of fencing required = 28 m – 1 m = 27 m.

**Q.12** Perimeter of an isosceles triangle is 50cm. If one of the two equal sides is 18cm, find the third side.

**Answer**: Let the length of the third side be x. An isosceles triangle has two equal sides. So, the perimeter can be expressed as: 18 cm + 18 cm + x = 50 cm. Solving for x: 36 cm + x = 50 cm x = 50 cm – 36 cm x = 14 cm

So, the third side is 14 cm.

**Q.13** Find (a) area and (b) perimeter of a square whose side is 12 cm.

**Q.14** The side of a square wall is 3m 50 cm. Determine the cost of colour washing it at the rate of Rs. 2 per sq.m.

**Q.15** The perimeter of a square playground is 1200m. Find its area

**Q.16** Find the area of the squares, whose sides are given below:

(a) 7cm (b) 12 dm (c) 2 m 25 cm (d) 3.2 m

**Q.17** Find the area of square land in hectares whose side is 250 m.

**Answer:** Area of the square = Side^2 = (250 m)^2 = 62,500 sq m Since 1 hectare = 10,000 sq m, we can convert the area to hectares: Area in hectares = 62,500 sq m / 10,000 = 6.25 hectares

**Q.18** Calculate the cost of levelling a square garden of side 75 m at the rate of Rs. 8 per sq.m

**Answer**: Area of the square garden = Side^2 = (75 m)^2 = 5625 sq m Cost of leveling = Area * Rate per sq.m = 5625 sq m * Rs 8/sq.m = Rs 45,000.

**Q.19** The side of a square hall is 8 m 5 dm. Find the cost of fixing tiles on its floor at the rate of Rs. 300 per sq.m.

**Answer :**

Convert the side length to cm: 8 m 5 dm = (8 * 100 cm) + (5 * 10 cm) = 800 cm + 50 cm = 850 cm Area of the square hall = Side^2 = 850 cm * 850 cm = 722,500 sq cm Now, convert the area to sq.m by dividing by 10,000 (since 1 sq.m = 10,000 sq cm): Area in sq.m = 722,500 sq cm / 10,000 = 72.25 sq m Cost of fixing tiles = Area * Rate per sq.m = 72.25 sq m * Rs 300/sq.m = Rs 21,675.

**Q.20** Find the area of a square whose perimeter is 600 m.

**Answer:**

Perimeter of the square = 4 * Side 600 m = 4 * Side Side = 600 m / 4 = 150 m Area of the square = Side^2 = (150 m)^2 = 22,500 sq m

**Q.21** The perimeter of the floor of a square room is 22m. Find its area.

**Answer **:

Perimeter of the square room = 4 * Side 22 m = 4 * Side Side = 22 m / 4 = 5.5 m Area of the square room = Side^2 = (5.5 m)^2 = 30.25 sq m

**Q.22** Find the area and perimeter of a rectangle whose length is 2m and breadth is 70 cm.

**Answer:**

Perimeter of the square room = 4 * Side 22 m = 4 * Side Side = 22 m / 4 = 5.5 m Area of the square room = Side^2 = (5.5 m)^2 = 30.25 sq m

**Q.23** The area of a rectangular field is 3.75 hectares. If the length is 250 m find its breadth.

**Answer **:

(a) Area = Length * Breadth = 2 m * 70 cm = 2 m * 0.7 m = 1.4 sq m

(b) Perimeter = 2 * (Length + Breadth) = 2 * (2 m + 70 cm) = 2 * (2 m + 0.7 m) = 2 * 2.7 m = 5.4 m

**Q.24** What is the length of outer boundary of the park shown in the above right figure? What will be the total cost of fencing it at the rate of Rs 20 per metre? There is a rectangular flower bed in the center of the park. Find the cost of manuring the flower bed at the rate of Rs 50 per square metre.

**Answer **:

To find the length of the outer boundary, we need to add the lengths of all four sides:

Outer boundary = 150 m + 100 m + 150 m + 100 m = 500 m

Cost of fencing = Outer boundary * Rate per meter = 500 m * Rs 20/m = Rs 10,000

Now, let’s find the area of the rectangular flower bed. It is 50 m long and 30 m wide:

Area of flower bed = Length * Breadth = 50 m * 30 m = 1500 sq m

Cost of manuring the flower bed = Area * Rate per sq.m = 1500 sq m * Rs 50/sq.m = Rs 75,000.

**Q.25** The length of an aluminium strip is 40cm. If the lengths in cm are measured in natural numbers, write the measurement of all the possible rectangular frames which can be made out of it. (For example, a rectangular frame with 15cm length and 5cm breadth can be made from this strip.)

**Answer:**

The aluminum strip can be used to make rectangular frames with dimensions in cm where the sum of the length and breadth is equal to 40 cm. Here are the possible dimensions:

- Length = 1 cm, Breadth = 39 cm
- Length = 2 cm, Breadth = 38 cm
- Length = 3 cm, Breadth = 37 cm
- Length = 4 cm, Breadth = 36 cm
- Length = 5 cm, Breadth = 35 cm
- Length = 6 cm, Breadth = 34 cm …
- Length = 19 cm, Breadth = 21 cm
- Length = 20 cm, Breadth = 20 cm

These are the possible rectangular frames that can be made from the 40 cm strip.

**Q.26** In an exhibition hall, there are 24 display boards each of length 1m 50cm and breadth 1m. There is a 100m long aluminium strip, which is used to frame these boards. How many boards will be framed using this strip? Find also the length of the aluminium strip required for the remaining boards.

**Answer:**

First, let’s calculate the total length of aluminum strip required to frame all 24 boards:

Each board has a length of 1 m 50 cm and a breadth of 1 m, which means it requires aluminum framing on all four sides. Length of aluminum strip required for one board = 2 * (Length + Breadth) Length of aluminum strip required for one board = 2 * (1 m 50 cm + 1 m) = 2 * 2.5 m = 5 m

Now, for 24 boards: Total length of aluminum strip required for 24 boards = 24 * 5 m = 120 m

Since there is a 100 m long aluminum strip available, it can frame 100 m / 5 m = 20 boards.

To find the length of aluminum strip required for the remaining boards: Remaining boards = Total boards – Framed boards = 24 boards – 20 boards = 4 boards Length of aluminum strip required for remaining boards = 4 boards * 5 m/board = 20 m

So, 20 boards can be framed using the 100 m long aluminum strip, and an additional 20 m of aluminum strip is needed for the remaining 4 boards.