{"id":792210,"date":"2024-02-19T16:25:14","date_gmt":"2024-02-19T10:55:14","guid":{"rendered":"https:\/\/leverageedu.com\/discover\/?p=792210"},"modified":"2024-02-19T16:25:14","modified_gmt":"2024-02-19T10:55:14","slug":"cbse-ncert-class-6-maths-chapter-10","status":"publish","type":"post","link":"https:\/\/leverageedu.com\/discover\/school-education\/cbse-ncert-class-6-maths-chapter-10\/","title":{"rendered":"NCERT Solutions and Notes for Class 6 Maths Chapter 10: Mensuration- Free PDF Download!"},"content":{"rendered":"\n

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.<\/p>\n\n\n\n

Click here to download Mensuration Class 6 Maths Chapter 10 Notes and Exercise Solutions PDF<\/strong><\/a><\/p>\n\n\n\n

Explore all the Chapters of Class 6 Mathematics:-<\/strong><\/p>\n\n\n\n

Chapter 1<\/strong><\/a><\/td>Chapter 2<\/strong><\/a><\/td>Chapter 3<\/strong><\/a><\/td>Chapter 4<\/strong><\/a><\/td><\/tr>
Chapter 5<\/strong><\/a><\/td>Chapter 6<\/strong><\/a><\/td>Chapter 7<\/strong><\/a><\/td>Chapter 8<\/strong><\/a><\/td><\/tr>
Chapter 9<\/strong><\/a><\/td>Chapter 10<\/td>Chapter 11<\/strong><\/td>Chapter 12<\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n\n\n\n

Also Read:<\/strong> Multiplication and Division Word Problems<\/strong><\/a><\/p>\n\n\n\n

Mensuration: Class 6 Maths Chapter 10 Notes – PDF Download<\/h2>\n\n\n\n

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. <\/p>\n\n\n\n

Topic 1: Perimeter<\/h3>\n\n\n\n

Perimeter is the distance covered along the boundary forming a closed figure when you go around the figure once.<\/p>\n\n\n\n

Topic 2: Perimeter Formulas: Rectangle, Square, Triangles<\/h3>\n\n\n\n

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. <\/p>\n\n\n\n

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  1. Perimeter of Rectangle Formula: <\/strong>The rectangle has 4 sides, having opposite sides of equal lengths.<\/li>\n<\/ol>\n\n\n\n
    Perimeter of Rectangle = 2 \u00d7 (length + breadth)<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
      \n
    1. Perimeter of Square Formula: <\/strong>A square has sides of equal lengths.<\/li>\n<\/ol>\n\n\n\n
      Perimeter of Square = 4 \u00d7 Length of a Side<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
        \n
      1. Perimeter of Triangle Formula: <\/strong>Here, we will determine the perimeter of an equilateral triangle. An equilateral triangle has all sides of equal lengths.<\/li>\n<\/ol>\n\n\n\n
        Perimeter of an Equilateral Triangle = 3 \u00d7 Length of a Side<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

        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.<\/p>\n\n\n\n

        Perimeter of a Regular Pentagon = 5 \u00d7 Length of a Side<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

        Topic 3: Area<\/h3>\n\n\n\n

        The amount of surface enclosed by a closed figure is called its area. <\/p>\n\n\n\n

        Topic 4: Area Formulas<\/h3>\n\n\n\n

        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. <\/p>\n\n\n\n

          \n
        1. Area of Rectangle Formula: <\/strong>The area of a rectangle can be calculated by multiplying its length and breadth.<\/li>\n<\/ol>\n\n\n\n
          Area of a Rectangle = Length \u00d7 Breadth <\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
            \n
          1. Area of a Square: <\/strong>The area of a square can be calculated by multiplying its sides.<\/li>\n<\/ol>\n\n\n\n
            Area of a Rectangle = Side \u00d7 Side = (Side)2<\/sup> <\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

            Also Read:<\/strong> BODMAS Questions<\/strong><\/a><\/p>\n\n\n\n

            Click here to download Mensuration Class 6 Maths Chapter 10 Notes and Exercise Solutions PDF<\/strong><\/a><\/p>\n\n\n\n

            Explore all the Chapters of Class 6 Mathematics:-<\/strong><\/p>\n\n\n\n

            Chapter 1<\/strong><\/a><\/td>Chapter 2<\/strong><\/a><\/td>Chapter 3<\/strong><\/a><\/td>Chapter 4<\/strong><\/a><\/td><\/tr>
            Chapter 5<\/strong><\/a><\/td>Chapter 6<\/strong><\/a><\/td>Chapter 7<\/strong><\/a><\/td>Chapter 8<\/strong><\/a><\/td><\/tr>
            Chapter 9<\/strong><\/a><\/td>Chapter 10<\/td>Chapter 11<\/strong><\/td>Chapter 12<\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

            Important Questions in NCERT Class 6 Maths Chapter 10: PDF Download<\/h2>\n\n\n\n

            Below we have provided some important exercise questions and their solutions from Chapter 10 of CBSE NCERT Maths for Class 6. <\/p>\n\n\n\n

            Exercise 10.1<\/h3>\n\n\n\n

            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?<\/strong><\/p>\n\n\n\n

            Ans.<\/strong> The length of the tape required to seal all the sides of the rectangular box\u2019s lid will be equal to the perimeter of the rectangular box. <\/p>\n\n\n\n

            We know, Perimeter of a Rectangle = 2 \u00d7 (length + breadth)<\/p>\n\n\n\n

            And, the length of the tape required = Perimeter of the rectangular box\u2019s lid.<\/p>\n\n\n\n

            Given: Length of the rectangular box\u2019s lid = 40 cm<\/p>\n\n\n\n

            The breadth of the rectangular box\u2019s lid = 10 cm<\/p>\n\n\n\n

            \u2234 Length of tape required = 2 \u00d7 (40 + 10) cm = 2 \u00d7 50 cm = 100 cm<\/p>\n\n\n\n

            Q 2.  A table-top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the tabletop?<\/strong><\/p>\n\n\n\n

            Ans.<\/strong> To find the perimeter of the tabletop, let\u2019s find its measures in a single unit of length.<\/p>\n\n\n\n

            Given: Length = 2 m 25cm and Breadth = 1 m 50 cm<\/p>\n\n\n\n

            We know, 1 cm = 0.01 m, then, 25 cm = 0.25 m and 50 cm = 0.50 m<\/p>\n\n\n\n

            Now, Length = 2 m + 0.25 m = 2.25 m<\/p>\n\n\n\n

            And Breadth = 1 m + 0.50 m = 1.50 m <\/p>\n\n\n\n

            Perimeter of tabletop = 2 \u00d7 (2.25 + 1.50) m = 2 \u00d7 3.75 m = 7.5 m<\/p>\n\n\n\n

            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?<\/strong><\/p>\n\n\n\n

            Ans. <\/strong>The length of the wooden strip required must be equal to the perimeter of the photograph.<\/p>\n\n\n\n

            \u2234 length of wooden strip = 2 \u00d7 (32 + 21) cm = 2 \u00d7 53 cm = 106 cm<\/p>\n\n\n\n

            Q 4. Find the perimeter of a regular hexagon with each side measuring 8 m.<\/strong><\/p>\n\n\n\n

            Ans.<\/strong> We know that regular shapes have all sides of equal lengths. <\/p>\n\n\n\n

            Therefore, the given regular hexagon has 6 equal sides of 8 m. <\/p>\n\n\n\n

            Perimeter of a Regular Hexagon = 6 \u00d7 Length of Each Side = 6 \u00d7 8 m = 48 m<\/p>\n\n\n\n

            \u2234 The perimeter of the given regular hexagon is 48 m. <\/p>\n\n\n\n

            Q 5. The perimeter of a regular pentagon is 100 cm. How long is its each side?<\/strong><\/p>\n\n\n\n

            Ans. <\/strong>From the above question, we know that the length of each side in a regular figure is equal. <\/p>\n\n\n\n

            \u2234 The perimeter of a regular pentagon = 5 \u00d7 Length of Each Side <\/p>\n\n\n\n

            Given: Perimeter of the regular pentagon = 100 cm<\/p>\n\n\n\n

            \"\"<\/figure>\n\n\n\n

            \u2234 the length of each side of the given regular pentagon is 20 cm.<\/p>\n\n\n\n

            Q 6. Find the cost of fencing a square park of side 250 m at the rate of \u20b9 20 per metre.<\/strong><\/p>\n\n\n\n

            Ans. <\/strong>To determine the cost of fencing, we must first calculate the perimeter of the square park.<\/p>\n\n\n\n

            Given: Length of each side of the square park = 250 m<\/p>\n\n\n\n

            Perimeter of the park = 4 \u00d7 250 m = 1000 m<\/p>\n\n\n\n

            Now, the cost of fencing for 1 metre = \u20b9 20<\/p>\n\n\n\n

            \u2234 cost of fencing 1000 m = \u20b9 20 \u00d7 1000 = \u20b9 20,000<\/p>\n\n\n\n

            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.<\/strong><\/p>\n\n\n\n

              \n
            1. What is the perimeter of his arrangement?<\/strong><\/li>\n\n\n\n
            2. Shari does not like his arrangement. She gets him to lay them out like a cross. What is the perimeter of her arrangement?<\/strong><\/li>\n\n\n\n
            3. Which has a greater perimeter?<\/strong><\/li>\n<\/ol>\n\n\n\n
              \"\"<\/figure>\n\n\n\n

              Ans. <\/strong>The answers to each question are given below.<\/p>\n\n\n\n

                \n
              1. Length of each side of the square paving slabs = 0.5 m <\/li>\n<\/ol>\n\n\n\n

                In the square arrangement, Avneet placed 3 square paving slabs in each row. <\/p>\n\n\n\n

                \u2234 Length of each side of the square = 3 \u00d7 0.5 m = 1.5 m<\/p>\n\n\n\n

                Thus, the perimeter of the arrangement = 4 \u00d7 1.5 m = 6.0 m<\/p>\n\n\n\n

                  \n
                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.<\/li>\n<\/ol>\n\n\n\n

                  \u2234 The perimeter of the arrangement made by Shari = 8 \u00d7 (0.5 m + 0.5 m) + 4 \u00d7 0.5 m<\/p>\n\n\n\n

                  \u21d2 8 \u00d7 1 m + 2 m = 10 m<\/p>\n\n\n\n

                  Thus, the perimeter of the arrangement made by Shari is 10 m.<\/p>\n\n\n\n

                    \n
                  1. The arrangement made by Shari has a greater perimeter. <\/li>\n<\/ol>\n\n\n\n

                    Exercise 10.3<\/h3>\n\n\n\n

                    Q 1. What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the rate of \u20b9 8 per hundred sq m?<\/strong><\/p>\n\n\n\n

                    Ans. <\/strong>Let us find the area of the rectangular plot to determine the cost of tiling it.  <\/p>\n\n\n\n

                    We know that the area of a rectangle = Length \u00d7 Breadth<\/p>\n\n\n\n

                    So, area = 500 m \u00d7 200 m = 1,00,000 sq m<\/p>\n\n\n\n

                    Given, the cost of tiling 100 sq m of land = \u20b9 8<\/p>\n\n\n\n

                    \u2234 the cost of tiling 1,00,000 sq m of plot = \u20b9 8\/100 \u00d7 1,00,000 = \u20b9 8000<\/p>\n\n\n\n

                    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? <\/strong><\/p>\n\n\n\n

                    Ans. <\/strong>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.<\/p>\n\n\n\n

                    \u2234 Area of a tile = 12 cm \u00d7 5 cm = 60 sq cm<\/p>\n\n\n\n

                    Area of the rectangular region = 100 cm \u00d7 144 cm = 14,400 sq cm<\/p>\n\n\n\n

                    So, the number of tiles that fit the rectangular region:<\/p>\n\n\n\n

                    \"\"<\/figure>\n\n\n\n

                    So, 240 tiles will be needed to fit in a rectangular region.<\/p>\n\n\n\n

                    \n