# NCERT Solutions and Notes for Class 6 Maths Chapter 10: Mensuration- Free PDF Download!

In Chapter 10, Mensuration – Class 6, students will learn to analyze and compare the area of flat figures, i.e. figures on a 2D plane. The chapter is very important is it lays the foundation for understanding the measurements which will be useful for them even in their higher studies such as engineering, Maths, or any other core fields of study. In the Mensuration chapter of Class 6, students will learn to calculate the perimeter, perimeter formulas, area, and area formulas of various shapes. After completing the chapter Mensuration, Class 6, students will be able to differentiate between the area and perimeters of various shapes. Read through for CBSE NCERT Class 6 Maths Chapter 10 Mensuration Notes and Solutions.

Below we have given topic-wise notes for Chapter 10, Mensuration, Class 6. We have also provided a downloadable free PDF at the end of these notes so you can download and take a printout to study later when you need quick revision before going to the exam hall.

### Topic 1: Perimeter

Perimeter is the distance covered along the boundary forming a closed figure when you go around the figure once.

### Topic 2: Perimeter Formulas: Rectangle, Square, Triangles

Below we have given the various perimeter formulas of some basic shapes. In the exercises, we will learn to use these perimeter formulas to calculate the perimeter of complex figures.

1. Perimeter of Rectangle Formula: The rectangle has 4 sides, having opposite sides of equal lengths.
1. Perimeter of Square Formula: A square has sides of equal lengths.
1. Perimeter of Triangle Formula: Here, we will determine the perimeter of an equilateral triangle. An equilateral triangle has all sides of equal lengths.

We can determine the length of any regular figure in a similar way if we know its number of sides. Therefore, we can determine the perimeter of a regular pentagon in a similar way.

### Topic 3: Area

The amount of surface enclosed by a closed figure is called its area.

### Topic 4: Area Formulas

Below we have given the various area formulas of some basic shapes. In the exercises, we will learn to use these area formulas to calculate the area of complex figures.

1. Area of Rectangle Formula: The area of a rectangle can be calculated by multiplying its length and breadth.
1. Area of a Square: The area of a square can be calculated by multiplying its sides.

## Important Questions in NCERT Class 6 Maths Chapter 10: PDF Download

Below we have provided some important exercise questions and their solutions from Chapter 10 of CBSE NCERT Maths for Class 6.

### Exercise 10.1

Q 1.  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?

Ans. The length of the tape required to seal all the sides of the rectangular box’s lid will be equal to the perimeter of the rectangular box.

We know, Perimeter of a Rectangle = 2 × (length + breadth)

And, the length of the tape required = Perimeter of the rectangular box’s lid.

Given: Length of the rectangular box’s lid = 40 cm

The breadth of the rectangular box’s lid = 10 cm

∴ Length of tape required = 2 × (40 + 10) cm = 2 × 50 cm = 100 cm

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

Ans. To find the perimeter of the tabletop, let’s find its measures in a single unit of length.

Given: Length = 2 m 25cm and Breadth = 1 m 50 cm

We know, 1 cm = 0.01 m, then, 25 cm = 0.25 m and 50 cm = 0.50 m

Now, Length = 2 m + 0.25 m = 2.25 m

And Breadth = 1 m + 0.50 m = 1.50 m

Perimeter of tabletop = 2 × (2.25 + 1.50) m = 2 × 3.75 m = 7.5 m

Q 3. What is the length of the wooden strip required to frame a photograph of length and breadth of 32 cm and 21 cm respectively?

Ans. The length of the wooden strip required must be equal to the perimeter of the photograph.

∴ length of wooden strip = 2 × (32 + 21) cm = 2 × 53 cm = 106 cm

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

Ans. We know that regular shapes have all sides of equal lengths.

Therefore, the given regular hexagon has 6 equal sides of 8 m.

Perimeter of a Regular Hexagon = 6 × Length of Each Side = 6 × 8 m = 48 m

∴ The perimeter of the given regular hexagon is 48 m.

Q 5. The perimeter of a regular pentagon is 100 cm. How long is its each side?

Ans. From the above question, we know that the length of each side in a regular figure is equal.

∴ The perimeter of a regular pentagon = 5 × Length of Each Side

Given: Perimeter of the regular pentagon = 100 cm

∴ the length of each side of the given regular pentagon is 20 cm.

Q 6. Find the cost of fencing a square park of side 250 m at the rate of ₹ 20 per metre.

Ans. To determine the cost of fencing, we must first calculate the perimeter of the square park.

Given: Length of each side of the square park = 250 m

Perimeter of the park = 4 × 250 m = 1000 m

Now, the cost of fencing for 1 metre = ₹ 20

∴ cost of fencing 1000 m = ₹ 20 × 1000 = ₹ 20,000

Q 7. Avneet buys 9 square paving slabs, each with a side of 1/2 m. He lays them in the form of a square.

1. What is the perimeter of his arrangement?
2. Shari does not like his arrangement. She gets him to lay them out like a cross. What is the perimeter of her arrangement?
3. Which has a greater perimeter?

Ans. The answers to each question are given below.

1. Length of each side of the square paving slabs = 0.5 m

In the square arrangement, Avneet placed 3 square paving slabs in each row.

∴ Length of each side of the square = 3 × 0.5 m = 1.5 m

Thus, the perimeter of the arrangement = 4 × 1.5 m = 6.0 m

1. Shari makes an arrangement in which 8 sides are made by placing two square paving slabs side to side and the remaining 4 sides are equal in length to 1 square paving slab.

∴ The perimeter of the arrangement made by Shari = 8 × (0.5 m + 0.5 m) + 4 × 0.5 m

⇒ 8 × 1 m + 2 m = 10 m

Thus, the perimeter of the arrangement made by Shari is 10 m.

1. The arrangement made by Shari has a greater perimeter.

### Exercise 10.3

Q 1. What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the rate of ₹ 8 per hundred sq m?

Ans. Let us find the area of the rectangular plot to determine the cost of tiling it.

We know that the area of a rectangle = Length × Breadth

So, area = 500 m × 200 m = 1,00,000 sq m

Given, the cost of tiling 100 sq m of land = ₹ 8

∴ the cost of tiling 1,00,000 sq m of plot = ₹ 8/100 × 1,00,000 = ₹ 8000

Q 2. 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 100 cm and 144 cm?

Ans. We must divide the area of the rectangular region by the area of a tile to find the number of tiles that will fit in the given rectangular region.

∴ Area of a tile = 12 cm × 5 cm = 60 sq cm

Area of the rectangular region = 100 cm × 144 cm = 14,400 sq cm

So, the number of tiles that fit the rectangular region:

So, 240 tiles will be needed to fit in a rectangular region.

Also See:

## FAQs

Q.1. What are regular closed figures?

Ans: All the regular closed figures have all sides of equal lengths and all the angles of equal measure.

Q.2. How to find the perimeter of a rectangle?

Ans: The perimeter of a rectangle can be calculated by the formula: Perimeter =  2 × (length + breadth).

Q.3. How to find the perimeter of a circle?

Ans: The perimeter of a circle is referred to as its circumference. The circumference of a circle can be calculated by the formula: Circumference = 2πr.

Q.4. How to find the perimeter of a regular triangle?

Ans: The perimeter of a regular triangle can be determined by the formula: Perimeter = 3 × Length of each side.

Q.5. How to find the area of a rectangle?

Ans: The area of a rectangle can be found by multiplying each side, i.e. Length × Breadth.

This was all about Mensuration Class 6 Maths in which we studied the perimeter, perimeter formulas, and area and area formulas of various shapes. Follow the CBSE Class 6 Maths Notes for more such chapter notes and important questions and answers for preparation for CBSE Class 6 Maths.