{"id":164055,"date":"2021-02-24T09:06:54","date_gmt":"2021-02-24T09:06:54","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=164055"},"modified":"2023-09-14T05:44:21","modified_gmt":"2023-09-14T05:44:21","slug":"ncert-solutions-for-6th-class-maths-chapter-10-mensuration","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-6th-class-maths-chapter-10-mensuration\/","title":{"rendered":"NCERT Solutions for 6th Class Maths: Chapter 10-Mensuration"},"content":{"rendered":"\n<p>Class 6: Maths Chapter 10 solutions. Complete Class 6 Maths Chapter 10 Notes.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-ncert-solutions-for-6th-class-maths-chapter-10-mensuration\"><strong>NCERT Solutions for 6th Class Maths: Chapter 10-Mensuration<\/strong><\/h2>\n\n\n\n<p>NCERT 6th Maths Chapter 10, class 6 Maths Chapter 10 solutions<\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-exercise-10-1\">Exercise 10.1<\/h2>\n\n\n\n<p><strong>1.&nbsp;Find the perimeter of each of the following figures:<br><\/strong><img loading=\"lazy\" decoding=\"async\" width=\"276\" height=\"320\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/06_math_ncert_ch10_01.png\"><br><br><strong>Answer<\/strong><br><\/p>\n\n\n\n<p>(a) Perimeter = Sum of all the sides<\/p>\n\n\n\n<p>= 4 cm + 2 cm + 1 cm + 5 cm&nbsp;<\/p>\n\n\n\n<p>= 12 cm<\/p>\n\n\n\n<p>(b) Perimeter = Sum of all the sides<\/p>\n\n\n\n<p>= 23 cm + 35 cm + 40 cm + 35 cm&nbsp;<\/p>\n\n\n\n<p>= 133 cm<\/p>\n\n\n\n<p>(c) Perimeter = Sum of all the sides<\/p>\n\n\n\n<p>= 15 cm + 15 cm + 15 cm + 15 cm&nbsp;<\/p>\n\n\n\n<p>= 60 cm<\/p>\n\n\n\n<p>(d) Perimeter = Sum of all the sides<\/p>\n\n\n\n<p>= 4 cm + 4 cm + 4 cm + 4 cm + 4 cm&nbsp;<\/p>\n\n\n\n<p>= 20 cm<\/p>\n\n\n\n<p>(e) Perimeter = Sum of all the sides<\/p>\n\n\n\n<p>= 1 cm + 4 cm + 0.5 cm + 2.5 cm + 2.5 cm + 0.5 cm + 4 cm&nbsp;<\/p>\n\n\n\n<p>= 15 cm<\/p>\n\n\n\n<p>(f) Perimeter = Sum of all the sides<\/p>\n\n\n\n<p>= 4 cm + 1 cm + 3 cm + 2 cm + 3 cm + 4 cm + 1 cm + 3 cm + 2 cm + 3 cm + 4 cm +&nbsp;<\/p>\n\n\n\n<p>1 cm + 3 cm + 2 cm + 3 cm + 4 cm + 1 cm + 3 cm + 2 cm + 3 cm&nbsp;<\/p>\n\n\n\n<p>= 52 cm<\/p>\n\n\n\n<p><strong>2.&nbsp;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<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Total length of tape required = Perimeter of rectangle<br><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/06_math_ncert_ch10_02.png\"><\/p>\n\n\n\n<p>= 2 ( length + breadth)<\/p>\n\n\n\n<p>= 2 ( 40 + 10)<\/p>\n\n\n\n<p>= 2\u00d750<\/p>\n\n\n\n<p>= 100 cm = 1 m<\/p>\n\n\n\n<p>Thus, the total length of tape required is 100 cm or 1 m.<\/p>\n\n\n\n<p><strong>3.&nbsp;A table-top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the table-top?<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Length of table top = 2 m 25 cm = 2.25 m<\/p>\n\n\n\n<p>Breadth of table top = 1 m 50 cm = 1.50 m<\/p>\n\n\n\n<p>Perimeter of table top = 2 x (length + breadth)<\/p>\n\n\n\n<p>= 2 \u00d7&nbsp;(2.25 + 1.50)<\/p>\n\n\n\n<p>= 2 \u00d7&nbsp;3.75 = 7.50 m<\/p>\n\n\n\n<p>Thus, perimeter of table top is 7.5 m.<\/p>\n\n\n\n<p><strong>4.&nbsp;What is the length of the wooden strip required to frame a photograph of length 32 cm and breadth 21 cm respectively?<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Length of wooden strip = Perimeter of photograph<\/p>\n\n\n\n<p>Perimeter of photograph = 2 \u00d7&nbsp;(length + breadth)<\/p>\n\n\n\n<p>= 2 (32 + 21)<\/p>\n\n\n\n<p>= 2 \u00d7&nbsp;53 cm = 106 cm<\/p>\n\n\n\n<p>Thus, the length of the wooden strip required is 106 cm.<\/p>\n\n\n\n<p><strong>5.&nbsp;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?<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Since the 4 rows of wires are needed. Therefore the total length of wires is equal to 4 times the perimeter of rectangle.<\/p>\n\n\n\n<p>Perimeter of rectangular piece of land = 2 \u00d7&nbsp;(length + breadth)<\/p>\n\n\n\n<p>= 2 \u00d7&nbsp;(0.7 + 0.5)&nbsp;<\/p>\n\n\n\n<p>= 2 \u00d7&nbsp;1.2&nbsp;<\/p>\n\n\n\n<p>= 2.4 km<\/p>\n\n\n\n<p>= 2.4 \u00d7&nbsp;1000 m&nbsp;<\/p>\n\n\n\n<p>= 2400 m<\/p>\n\n\n\n<p>Thus, the length of wire = 4 \u00d7&nbsp;2400 = 9600 m = 9.6 km<\/p>\n\n\n\n<p><strong>6. Find the perimeter of each of the following shapes:<\/strong><\/p>\n\n\n\n<p><strong>(a) A triangle of sides 3 cm, 4 cm and 5 cm.<\/strong><\/p>\n\n\n\n<p><strong>(b) An equilateral triangle of side 9 cm.<\/strong><\/p>\n\n\n\n<p><strong>(c) An isosceles triangle with equal sides 8 cm each and third side 6 cm.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>(a) Perimeter of \u0394ABC\u0394ABC = AB + BC + CA<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/06_math_ncert_ch10_03.jpg\" alt=\"NCERT Solutions for 6th Class Maths: Chapter 10-Mensuration\"\/><\/figure>\n\n\n\n<p>= 3 cm + 5 cm + 4 cm = 12 cm<\/p>\n\n\n\n<p>(b) Perimeter of equilateral \u0394ABC\u0394ABC = 3\u00d7side<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/06_math_ncert_ch10_04.jpg\" alt=\"NCERT Solutions for 6th Class Maths: Chapter 10-Mensuration\"\/><\/figure>\n\n\n\n<p>= 3\u00d79 cm = 27 cm<\/p>\n\n\n\n<p>(c) Perimeter of \u0394ABC\u0394ABC = AB + BC + CA<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/06_math_ncert_ch10_05.png\" alt=\"NCERT Solutions for 6th Class Maths: Chapter 10-Mensuration\"\/><\/figure>\n\n\n\n<p>= 8 cm + 6 cm + 8 cm = 22 cm<\/p>\n\n\n\n<p><br><strong>7.&nbsp;Find the perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Perimeter of triangle = Sum of all three sides<\/p>\n\n\n\n<p>= 10 cm + 14 cm + 15 cm = 39 cm<\/p>\n\n\n\n<p>Thus, perimeter of triangle is 39 cm.<\/p>\n\n\n\n<p><strong>8. Find the perimeter of a regular hexagon with each side measuring 8 cm.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Perimeter of Hexagon = 6 \u00d7&nbsp;length of one side<\/p>\n\n\n\n<p>= 6 x 8 m = 48 m<\/p>\n\n\n\n<p>Thus, the perimeter of hexagon is 48 m.<\/p>\n\n\n\n<p><strong>9. Find the side of the square whose perimeter is 20 m.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Perimeter of square = 4 \u00d7&nbsp;side<\/p>\n\n\n\n<p>\u21d2 20 = 4 \u00d7&nbsp;side<\/p>\n\n\n\n<p>\u21d2 side = 20\/4 = 5 cm<\/p>\n\n\n\n<p>Thus, the side of square is 5 cm.<\/p>\n\n\n\n<p><strong>10. The perimeter of a regular pentagon is 100 cm. How long is its each side?<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Perimeter of regular pentagon = 100 cm<\/p>\n\n\n\n<p>\u21d2 5 \u00d7&nbsp;side = 100 cm<\/p>\n\n\n\n<p>\u21d2 side = 100\/5 = 20 cm<\/p>\n\n\n\n<p>Thus, the side of regular pentagon is 20 cm.<\/p>\n\n\n\n<p><strong>11. A piece of string is 30 cm long. What will be the length of each side if the string is used to form:<\/strong><\/p>\n\n\n\n<p><strong>(a) a square<\/strong><\/p>\n\n\n\n<p><strong>(b) an equilateral triangle<\/strong><\/p>\n\n\n\n<p><strong>(c) a regular hexagon?<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Length of string = Perimeter of each figure<\/p>\n\n\n\n<p>(a) Perimeter of square = 30 cm<\/p>\n\n\n\n<p>\u21d2 4 \u00d7 side = 30 cm<\/p>\n\n\n\n<p>\u21d2 side = 30\/4 = 7.5 cm<\/p>\n\n\n\n<p>Thus, the length of each side of square is 7.5 cm.<\/p>\n\n\n\n<p>(b) Perimeter of equilateral triangle = 30 cm<\/p>\n\n\n\n<p>\u21d2 3 \u00d7&nbsp;side = 30 cm<\/p>\n\n\n\n<p>\u21d2 side = 30\/3 = 10 cm<\/p>\n\n\n\n<p>Thus, the length of each side of equilateral triangle is 10 cm.<\/p>\n\n\n\n<p>(c) Perimeter of hexagon = 30 cm<\/p>\n\n\n\n<p>\u21d2 6 \u00d7&nbsp;side = 30 cm<\/p>\n\n\n\n<p>\u21d2 side = 30\/6 = 5 cm<\/p>\n\n\n\n<p>Thus, the length of each side of hexagon is 5 cm.<\/p>\n\n\n\n<p><strong>12. Two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is the third side?<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Let the length of third side be xx cm.<\/p>\n\n\n\n<p>Length of other two side are 12 cm and 14 cm.<\/p>\n\n\n\n<p>Now, Perimeter of triangle = 36 cm<\/p>\n\n\n\n<p>\u21d2 12+14+x = 36<\/p>\n\n\n\n<p>\u21d2 26+x = 6<\/p>\n\n\n\n<p>\u21d2 x = 36\u221226<\/p>\n\n\n\n<p>\u21d2 x=10<\/p>\n\n\n\n<p>Thus, the length of third side is 10 cm.<\/p>\n\n\n\n<p><strong>13. Find the cost of fencing a square park of side 250 m at the rate of Rs 20 per meter.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Side of square = 250 m<\/p>\n\n\n\n<p>Perimeter of square = 4 \u00d7&nbsp;side<\/p>\n\n\n\n<p>= 4 \u00d7&nbsp;250 = 1000 m<\/p>\n\n\n\n<p>Since, cost of fencing per meter = Rs. 20<\/p>\n\n\n\n<p>Therefore, cost of fencing of 1000 meters = 20\u00d71000 = Rs. 20,000<\/p>\n\n\n\n<p><strong>14. Find the cost of fencing a rectangular park of length 175 m and breadth 125 m at the rate of Rs. 12 per meter.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Length of rectangular park = 175 m<\/p>\n\n\n\n<p>Breadth of rectangular park = 125 m<\/p>\n\n\n\n<p>Perimeter of park = 2 x (length + breadth)<\/p>\n\n\n\n<p>= 2 \u00d7&nbsp;(175 + 125)<\/p>\n\n\n\n<p>= 2\u00d7300 = 600 m<\/p>\n\n\n\n<p>Since, cost of fencing park per meter = Rs. 12<\/p>\n\n\n\n<p>Therefore, cost of fencing park of 600 m = 12\u00d7600 = Rs. 7,200<\/p>\n\n\n\n<p><strong>15. Sweety runs around a square park of side 75 m. Bulbul runs around a rectangular park with length of 60 m and breadth 45 m. Who covers less distance?<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Distance covered by Sweety = Perimeter of square park<\/p>\n\n\n\n<p>Perimeter of square = 4\u00d7side<\/p>\n\n\n\n<p>= 4\u00d775 = 300 m<\/p>\n\n\n\n<p>Thus, distance covered by Sweety is 300 m.<\/p>\n\n\n\n<p>Now, distance covered by Bulbul = Perimeter of rectangular park<\/p>\n\n\n\n<p>Perimeter of rectangular park = 2\u00d7(length + breadth)<\/p>\n\n\n\n<p>= 2\u00d7(60 + 45)<\/p>\n\n\n\n<p>= 2\u00d7105 = 210 m<\/p>\n\n\n\n<p>Thus, Bulbul covers the distance of 210 m.<\/p>\n\n\n\n<p>So, Bulbul covers less distance.<\/p>\n\n\n\n<p><strong>16. What is the perimeter of each of the following figures? What do you infer from the answer?<br><\/strong><img loading=\"lazy\" decoding=\"async\" width=\"320\" height=\"261\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/06_math_ncert_ch10_06.jpg\"><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>(a) Perimeter of square = 4\u00d7side<\/p>\n\n\n\n<p>= 4\u00d725 = 100 cm<\/p>\n\n\n\n<p>(b) Perimeter of rectangle = 2\u00d7(length + breadth)<\/p>\n\n\n\n<p>= 2\u00d7(40 + 10)<\/p>\n\n\n\n<p>= 2\u00d750 = 100 cm<\/p>\n\n\n\n<p>(c) Perimeter of rectangle = 2\u00d7(length + breadth)<\/p>\n\n\n\n<p>= 2\u00d7(30 + 20)<\/p>\n\n\n\n<p>= 2\u00d750 = 100 cm<\/p>\n\n\n\n<p>(d) Perimeter of triangle = Sum of all sides<\/p>\n\n\n\n<p>= 30 cm + 30 cm + 40 cm = 100 cm<\/p>\n\n\n\n<p>Thus, all the figures have same perimeter.<\/p>\n\n\n\n<p><strong>17.&nbsp;Avneet buys 9 square paving slabs, each with a side 1212 m. He lays them in the form of a square<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/06_math_ncert_ch10_05.jpg\" alt=\"NCERT Solutions for 6th Class Maths: Chapter 10-Mensuration\"\/><\/figure>\n\n\n\n<p><strong>(a) What is the perimeter of his arrangement?<\/strong><\/p>\n\n\n\n<p><strong>(b) Shari does not like his arrangement. She gets him to lay them out like a cross. What is the<\/strong> <strong>perimeter of her arrangement?<\/strong><\/p>\n\n\n\n<p><strong>(c) Which has greater perimeter?<\/strong><\/p>\n\n\n\n<p><strong>(d) Avneet wonders, if there is a way of getting an even greater perimeter. Can you find a way of doing this? (The paving slabs must meet along complete edges, i.e., they cannot be broken.)<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>(a) 6 m<\/p>\n\n\n\n<p>(b) 10 m<\/p>\n\n\n\n<p>(c) Second arrangement has greater perimeter.<\/p>\n\n\n\n<p>(d) Yes, if all the squares are arranged in row, the perimeter be 10 cm.<\/p>\n\n\n\n<p>NCERT 6th Maths Chapter 10, class 6 Maths Chapter 10 solutions<\/p>\n\n\n\n<h4 class=\"wp-block-heading\" id=\"h-exercise-10-2\">Exercise 10.2<\/h4>\n\n\n\n<p><strong>1. Find the areas of the following figures by counting squares:<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/06_math_ncert_ch10_07.jpg\" alt=\"NCERT Solutions for 6th Class Maths: Chapter 10-Mensuration\"\/><\/figure>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>(a) Number of filled square = 9<\/p>\n\n\n\n<p>Area covered by squares = 9 \u00d7&nbsp;1 = 9 sq. units<\/p>\n\n\n\n<p>(b) Number of filled squares = 5<\/p>\n\n\n\n<p>Area covered by filled squares = 5 \u00d7&nbsp;1 = 5 sq. units<\/p>\n\n\n\n<p>(c) Number of full filled squares = 2<\/p>\n\n\n\n<p>Number of half filled squares = 4<\/p>\n\n\n\n<p>Area covered by full filled squares = 2 \u00d7&nbsp;1 = 2 sq. units<\/p>\n\n\n\n<p>And, Area covered by half filled squares = 1\/2 \u00d74 = 2 sq. units<\/p>\n\n\n\n<p>Total area = 2 + 2 = 4 sq. units<\/p>\n\n\n\n<p>(d) Number of filled squares = 8<\/p>\n\n\n\n<p>Area covered by filled squares = 8 \u00d7&nbsp;1 = 8 sq. units<\/p>\n\n\n\n<p>(e) Number of filled squares = 10<\/p>\n\n\n\n<p>Area covered by filled squares = 10 \u00d7&nbsp;1 = 10 sq. units<\/p>\n\n\n\n<p>(f) Number of full filled squares = 2<\/p>\n\n\n\n<p>Number of half filled squares = 4<\/p>\n\n\n\n<p>Area covered by full filled squares = 2 \u00d7&nbsp;1 = 2 sq. units<\/p>\n\n\n\n<p>And Area covered by half filled squares = 1\/2 \u00d74 = 2 sq. units<\/p>\n\n\n\n<p>Total area = 2 + 2 = 4 sq. units<\/p>\n\n\n\n<p>(g) Number of full filled squares = 4<\/p>\n\n\n\n<p>Number of half filled squares = 4<\/p>\n\n\n\n<p>Area covered by full filled squares = 4 \u00d7&nbsp;1 = 4 sq. units<\/p>\n\n\n\n<p>And, Area covered by half filled squares = 1\/2 \u00d74 = 2 sq. units<\/p>\n\n\n\n<p>Total area = 4 + 2 = 6 sq. units<\/p>\n\n\n\n<p>(h) Number of filled squares = 5<\/p>\n\n\n\n<p>Area covered by filled squares = 5 \u00d7 1 = 5 sq. units<\/p>\n\n\n\n<p>(i) Number of filled squares = 9<\/p>\n\n\n\n<p>Area covered by filled squares = 9 \u00d7&nbsp;1 = 9 sq. units<\/p>\n\n\n\n<p>(j) Number of full filled squares = 2<\/p>\n\n\n\n<p>Number of half filled squares = 4<\/p>\n\n\n\n<p>Area covered by full filled squares = 2 \u00d7&nbsp;1 = 2 sq. units<\/p>\n\n\n\n<p>And Area covered by half filled squares = 1\/2 \u00d74 = 2 sq. units<\/p>\n\n\n\n<p>Total area = 2 + 2 = 4 sq. units<\/p>\n\n\n\n<p>(k) Number of full filled squares = 4<\/p>\n\n\n\n<p>Number of half filled squares = 2<\/p>\n\n\n\n<p>Area covered by full filled squares = 4 \u00d7&nbsp;1 = 4 sq. units<\/p>\n\n\n\n<p>And Area covered by half filled squares = 1\/2 \u00d72 = 1 sq. units<\/p>\n\n\n\n<p>Total area = 4 + 1 = 5 sq. units<\/p>\n\n\n\n<p>(l) Number of full filled squares = 3<\/p>\n\n\n\n<p>Number of half filled squares = 10<\/p>\n\n\n\n<p>Area covered by full filled squares = 3 \u00d7&nbsp;1 = 3 sq. units<\/p>\n\n\n\n<p>And Area covered by half filled squares = 1\/2 \u00d710 = 5 sq. units<\/p>\n\n\n\n<p>Total area = 3 + 5 = 8 sq. units<\/p>\n\n\n\n<p>(m) Number of full filled squares = 7<\/p>\n\n\n\n<p>Number of half filled squares = 14<\/p>\n\n\n\n<p>Area covered by full filled squares = 7 \u00d7&nbsp;1 = 7 sq. units<\/p>\n\n\n\n<p>And Area covered by half filled squares = 1\/2 \u00d714 = 7 sq. units<\/p>\n\n\n\n<p>Total area = 7 + 7 = 14 sq. units<\/p>\n\n\n\n<p>(n) Number of full filled squares = 10<\/p>\n\n\n\n<p>Number of half filled squares = 16<\/p>\n\n\n\n<p>Area covered by full filled squares = 10 \u00d7&nbsp;1 = 10 sq. units<\/p>\n\n\n\n<p>And Area covered by half filled squares = 1\/2 \u00d716 = 8 sq. units<\/p>\n\n\n\n<p>Total area = 10 + 8 = 18 sq. units<\/p>\n\n\n\n<p>NCERT 6th Maths Chapter 10, class 6 Maths Chapter 10 solutions<\/p>\n\n\n\n<h4 class=\"wp-block-heading\" id=\"h-exercise-10-3\">Exercise 10.3<\/h4>\n\n\n\n<p><strong>1. Find the areas of the rectangles whose sides are:<\/strong><\/p>\n\n\n\n<p><strong>(a) 3 cm and 4 cm<\/strong><\/p>\n\n\n\n<p><strong>(b) 12 m and 21 m<\/strong><\/p>\n\n\n\n<p><strong>(c) 2 km and 3 km<\/strong><\/p>\n\n\n\n<p><strong>(d) 2 m and 70 cm<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>(a) Area of rectangle = length \u00d7&nbsp;breadth<\/p>\n\n\n\n<p>= 3 cm \u00d7&nbsp;4 cm = 12 cm<sup>2<\/sup><\/p>\n\n\n\n<p>(b) Area of rectangle = length \u00d7&nbsp;breadth<\/p>\n\n\n\n<p>= 12 m \u00d7&nbsp;21 m = 252 m<sup>2<\/sup><\/p>\n\n\n\n<p>(c) Area of rectangle = length \u00d7&nbsp;breadth<\/p>\n\n\n\n<p>= 2 km \u00d7&nbsp;3 km = 6 km<sup>2<\/sup><\/p>\n\n\n\n<p>(d) Area of rectangle = length \u00d7&nbsp;breadth<\/p>\n\n\n\n<p>= 2 m \u00d7&nbsp;70 cm = 2 m \u00d7&nbsp;0.7 m = 1.4 m<sup>2<\/sup><\/p>\n\n\n\n<p><strong>2. Find the areas of the squares whose sides are:<\/strong><\/p>\n\n\n\n<p><strong>(a) 10 cm (b) 14 cm (c) 5 m<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>(a) Area of square = side \u00d7&nbsp;side&nbsp;<\/p>\n\n\n\n<p>= 10 cm \u00d7&nbsp;10 cm = 100 cm<sup>2<\/sup><\/p>\n\n\n\n<p>(b) Area of square = side x side&nbsp;<\/p>\n\n\n\n<p>= 14 cm \u00d7&nbsp;14 cm = 196 cm<sup>2<\/sup><\/p>\n\n\n\n<p>(c) Area of square = side x side&nbsp;<\/p>\n\n\n\n<p>= 5 m \u00d7&nbsp;5 m = 25 m<sup>2<\/sup><\/p>\n\n\n\n<p><strong>3. The length and the breadth of three rectangles are as given below:<\/strong><\/p>\n\n\n\n<p><strong>(a) 9 m and 6 m<\/strong><\/p>\n\n\n\n<p><strong>(b) 17 m and 3 m<\/strong><\/p>\n\n\n\n<p><strong>(c) 4 m and 14 m<\/strong><\/p>\n\n\n\n<p><strong>Which one has the largest area and which one has the smallest?<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>(a) Area of rectangle = length \u00d7&nbsp;breadth<\/p>\n\n\n\n<p>= 9 m \u00d7&nbsp;6 m = 54 m<sup>2<\/sup><\/p>\n\n\n\n<p>(b) Area of rectangle = length \u00d7&nbsp;breadth<\/p>\n\n\n\n<p>= 3 m \u00d7&nbsp;17 m = 51 m<sup>2<\/sup><\/p>\n\n\n\n<p>(c) Area of rectangle = length x breadth<\/p>\n\n\n\n<p>= 4 m \u00d7&nbsp;14 m = 56 m<sup>2<\/sup><\/p>\n\n\n\n<p>Thus, the rectangle (c) has largest area, i.e. 56 m<sup>2<\/sup>&nbsp;and rectangle (b) has smallest area, i.e., 51 m<sup>2<\/sup>.<\/p>\n\n\n\n<p><strong>4. The area of a rectangular garden 50 m long is 300 m<sup>2<\/sup>, find the width of the garden.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Length of rectangle = 50 m and Area of rectangle = 300 m<sup>2<\/sup><\/p>\n\n\n\n<p>Since, Area of rectangle = length x breadth<\/p>\n\n\n\n<p>Therefore, Breadth =&nbsp;<img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/fgh.png\">=300\/50 = 6 m<\/p>\n\n\n\n<p>Thus, the breadth of the garden is 6 m.<\/p>\n\n\n\n<p><strong>5. What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the rate of Rs. 8 per hundred sq. m?<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Length of land = 500 m and Breadth of land = 200 m<\/p>\n\n\n\n<p>Area of land = length x breadth = 500 m \u00d7&nbsp;200 m = 1,00,000 m<sup>2<\/sup><\/p>\n\n\n\n<p>\u2235 Cost of tiling 100 sq. m of land = Rs. 8<\/p>\n\n\n\n<p>\u2234 Cost of tilling 1,00,000 sq. m of land = 8\/100 \u00d7100000 = Rs. 8000<\/p>\n\n\n\n<p><strong>6. A table-top measures 2 m by 1 m 50 cm. What is its area in square meters?<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Length of table = 2 m and breadth of table = 1 m 50 cm = 1.50 m<\/p>\n\n\n\n<p>Area of table = length \u00d7&nbsp;breadth<\/p>\n\n\n\n<p>= 2 m \u00d7&nbsp;1.50 m = 3 m<sup>2<\/sup><\/p>\n\n\n\n<p><strong>7. A room is 4 m long and 3 m 50 cm wide. How many square meters of carpet is needed to cover the floor of the room?<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Length of room = 4 m and breadth of room = 3 m 50 cm = 3.50 m<\/p>\n\n\n\n<p>Area of carpet = length \u00d7&nbsp;breadth<\/p>\n\n\n\n<p>= 4 \u00d7&nbsp;3.50 = 14m<sup>2<\/sup><\/p>\n\n\n\n<p>Therefore, 14m<sup>2<\/sup> of carpet required to cover the floor.<\/p>\n\n\n\n<p><strong>8. A floor is 5 m long and 4 m wide. A square carpet of sides 3 m is laid on the floor. Find the area of the floor that is not carpeted.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Length of floor = 5 m and breadth of floor = 4 m<\/p>\n\n\n\n<p>Area of floor = length \u00d7&nbsp;breadth<\/p>\n\n\n\n<p>= 5 m \u00d7&nbsp;4 m = 20 m<sup>2<\/sup><\/p>\n\n\n\n<p>Now, Side of square carpet = 3 m<\/p>\n\n\n\n<p>Area of square carpet = side x side = 3 \u00d7&nbsp;3 = 9 m<sup>2<\/sup><\/p>\n\n\n\n<p>Area of floor that is not carpeted = 20 m<sup>2<\/sup>&nbsp;\u2013 9 m<sup>2<\/sup>&nbsp;= 11 m<sup>2<\/sup><\/p>\n\n\n\n<p><strong>9. Five square flower beds each of sides 1 m are dug on a piece of land 5 m long and 4 m wide. What is the area of the remaining part of the land?<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Side of square bed = 1 m<\/p>\n\n\n\n<p>Area of square bed = side \u00d7&nbsp;side = 1 m \u00d7&nbsp;1 m = 1 m<sup>2<\/sup><\/p>\n\n\n\n<p>\u2234 Area of 5 square beds = 1 \u00d7&nbsp;5 = 5 m<sup>2<\/sup><\/p>\n\n\n\n<p>Now, Length of land = 5 m and breadth of land = 4 m<\/p>\n\n\n\n<p>\u2234 Area of land = length \u00d7&nbsp;breadth = 5 m \u00d7&nbsp;4 m = 20 m<sup>2<\/sup><\/p>\n\n\n\n<p>Area of remaining part = Area of land \u2013 Area of 5 flower beds<\/p>\n\n\n\n<p>= 20 m<sup>2<\/sup> \u2013 5 m<sup>2<\/sup> = 15 m<sup>2<\/sup><\/p>\n\n\n\n<p><strong>10. By splitting the following figures into rectangles, find their areas. (The measures are given in centimeters)<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/06_math_ncert_ch10_08.jpg\" alt=\"NCERT Solutions for 6th Class Maths: Chapter 10-Mensuration\"\/><\/figure>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>(a) The given figure can be broken into rectangles as<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/06_math_ncert_ch10_09.jpg\" alt=\"NCERT Solutions for 6th Class Maths: Chapter 10-Mensuration\"\/><\/figure>\n\n\n\n<p>Area of 1st rectangle = 12 \u00d7&nbsp;2 = 24 cm<sup>2<\/sup><\/p>\n\n\n\n<p>Area of 2nd rectangle = 8 \u00d7&nbsp;2 = 16 cm<sup>2<\/sup><\/p>\n\n\n\n<p>Total area of the figure = 24 + 16 = 40 cm<sup>2<\/sup><\/p>\n\n\n\n<p>(b) The given figure can be broken into rectangles as follow<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/06_math_ncert_ch10_10.jpg\" alt=\"NCERT Solutions for 6th Class Maths: Chapter 10-Mensuration\"\/><\/figure>\n\n\n\n<p>Area of 1st rectangle = 21 \u00d7&nbsp;7&nbsp; = 147 cm<sup>2<\/sup>&nbsp;<\/p>\n\n\n\n<p>Area of 1st square = 7 \u00d7&nbsp;7 = 49 cm<sup>2<\/sup><\/p>\n\n\n\n<p>Area of 2nd square = 7 \u00d7&nbsp;7 = 49 cm<sup>2<\/sup><\/p>\n\n\n\n<p>Total area of the figure = 147 + 49 + 49 = 245 cm<sup>2<\/sup><\/p>\n\n\n\n<p>(c) The given figure can be broken into rectangles as follow<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/06_math_ncert_ch10_11.jpg\" alt=\"NCERT Solutions for 6th Class Maths: Chapter 10-Mensuration\"\/><\/figure>\n\n\n\n<p>Area of 1st rectangle = 5 \u00d7&nbsp;1 = 5 cm<sup>2<\/sup><\/p>\n\n\n\n<p>Area of 2nd rectangle = 4 \u00d7&nbsp;1 = 4 cm<sup>2<\/sup><\/p>\n\n\n\n<p>Total area of the figure = 5 + 4 = 9 cm<sup>2<\/sup><\/p>\n\n\n\n<p><strong>11. Split the following shapes into rectangles and find their areas. (The measures are given in centimetres)<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/ncert-solutions-for-class-6-maths-chapter-10-exerc-8.png\" alt=\"NCERT Solutions for Class 6 Maths Chapter 10 Exercise 10.3 - 4\" title=\"NCERT Solutions for Class 6 Maths Chapter 10 Exercise 10.3 - 4\"\/><\/figure>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p><strong>(a)<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/ncert-solutions-for-class-6-maths-chapter-10-exerc-9.png\" alt=\"NCERT Solutions for Class 6 Maths Chapter 10 Exercise 10.3 - 5\" title=\"NCERT Solutions for Class 6 Maths Chapter 10 Exercise 10.3 - 5\"\/><\/figure>\n\n\n\n<p>Total area of the figure = 12 \u00d7 2 + 8 \u00d7 2<\/p>\n\n\n\n<p>= 40 cm<sup>2<\/sup><\/p>\n\n\n\n<p>(b)<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/ncert-solutions-for-class-6-maths-chapter-10-exerc-10.png\" alt=\"NCERT Solutions for Class 6 Maths Chapter 10 Exercise 10.3 - 6\" title=\"NCERT Solutions for Class 6 Maths Chapter 10 Exercise 10.3 - 6\"\/><\/figure>\n\n\n\n<p>There are 5 squares. Each side is 7 cm<\/p>\n\n\n\n<p>Area of 5 squares = 5 \u00d7 7<sup>2<\/sup><\/p>\n\n\n\n<p>= 245 cm<sup>2<\/sup><\/p>\n\n\n\n<p>(c)<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/ncert-solutions-for-class-6-maths-chapter-10-exerc-11.png\" alt=\"NCERT Solutions for Class 6 Maths Chapter 10 Exercise 10.3 - 7\" title=\"NCERT Solutions for Class 6 Maths Chapter 10 Exercise 10.3 - 7\"\/><\/figure>\n\n\n\n<p>Area of grey rectangle = 2 \u00d7 1<\/p>\n\n\n\n<p>= 2 cm<sup>2<\/sup><\/p>\n\n\n\n<p>Area of brown rectangle = 2 \u00d7 1<\/p>\n\n\n\n<p>= 2 cm<sup>2<\/sup><\/p>\n\n\n\n<p>Area of orange rectangle = 5 \u00d7 1<\/p>\n\n\n\n<p>= 5 cm<sup>2<\/sup><\/p>\n\n\n\n<p>Total area = 2 + 2 + 5<\/p>\n\n\n\n<p>= 9 cm<sup>2<\/sup><\/p>\n\n\n\n<p><strong>12. 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:<br>(a)100 cm and 144 cm<br>(b)70 cm and 36 cm<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/ncert-solutions-for-class-6-maths-chapter-10-exerc-11.png\" alt=\"NCERT Solutions for 6th Class Maths: Chapter 10-Mensuration\"\/><\/figure>\n\n\n\n<p>NCERT 6th Maths Chapter 10, class 6 Maths Chapter 10 solutions<\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-ncert-solutions-for-6th-class-maths-chapter-10-nbsp-download-pdf\">NCERT Solutions for 6th Class Maths: Chapter 10:&nbsp;<strong>Download PDF<\/strong><\/h2>\n\n\n\n<p>NCERT Solutions for 6th Class Maths: Chapter 10-Mensuration<\/p>\n\n\n\n<p><a href=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/NCERT-Solutions-for-6th-Class-Maths_-Chapter-10-Mensuration.pdf\"><strong>Download PDF<\/strong>: NCERT Solutions for 6th Class Maths: Chapter 10-Mensuration PDF<\/a><\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-chapterwise-ncert-solutions-for-class-6-maths\"><strong>Chapterwise NCERT Solutions for Class 6 Maths :<\/strong><\/h2>\n\n\n\n<ul class=\"wp-block-list\">\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-6th-class-maths-chapter-1-knowing-our-numbers\/\">Chapter 1 Knowing Our Numbers<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-6th-class-maths-chapter-2-whole-numbers\/\">Chapter 2 Whole Numbers<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-6th-class-maths-chapter-3-playing-with-numbers\/\">Chapter 3 Playing with Numbers<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-6th-class-maths-chapter-4-basic-geometrical-ideas\/\">Chapter 4 Basic Geometrical Ideas<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-6th-class-maths-chapter-5-understanding-elementary-shapes\/\">Chapter 5 Understanding Elementary Shapes<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-6th-class-maths-chapter-6-integers\/\">Chapter 6 Integers<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-6th-class-maths-chapter-7-fractions\/\">Chapter 7 Fractions<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-6th-class-maths-chapter-8-decimals\/\">Chapter 8 Decimals<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-6th-class-maths-chapter-9-data-handling\/\">Chapter 9 Data Handling<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-6th-class-maths-chapter-10-mensuration\/\">Chapter 10 Mensuration<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-6th-class-maths-chapter-11-algebra\/\">Chapter 11 Algebra<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-6th-class-maths-chapter-12-ratio-and-proportion\/\">Chapter 12 Ratio and Proportion<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-6th-class-maths-chapter-13-symmetry\/\">Chapter 13 Symmetry<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-6th-class-maths-chapter-14-practical-geometry\/\">Chapter 14 Practical Geometry<\/a><\/li>\n<\/ul>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-about-ncert\"><strong><br><\/strong>About NCERT<\/h2>\n\n\n\n<p>The National Council of Educational Research and Training is an autonomous organization of the Government of India which was established in 1961 as a literary, scientific, and charitable Society under the Societies Registration Act. Its headquarters are located at Sri Aurbindo Marg in New Delhi. <a href=\"https:\/\/ncert.nic.in\/\" target=\"_blank\" rel=\"noreferrer noopener\">Visit the Official NCERT website<\/a> to learn more. <\/p>\n\n\n\n<div class=\"wp-block-buttons is-layout-flex wp-block-buttons-is-layout-flex\">\n<div class=\"wp-block-button\"><a class=\"wp-block-button__link has-primary-background-color has-background wp-element-button\" href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions\/\">NCERT Solutions<\/a><\/div>\n\n\n\n<div class=\"wp-block-button\"><a class=\"wp-block-button__link has-primary-background-color has-background wp-element-button\" href=\"https:\/\/www.indcareer.com\/schools\/ncert-class-vi\/\">NCERT Solutions for Class 6<\/a><\/div>\n\n\n\n<div class=\"wp-block-button\"><a class=\"wp-block-button__link has-primary-background-color has-background wp-element-button\" href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-class-6-maths\/\">NCERT Solutions for Class 6 Maths<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Class 6: Maths Chapter 10 solutions. Complete Class 6 Maths Chapter 10 Notes. NCERT Solutions for 6th Class Maths: Chapter 10-Mensuration NCERT 6th Maths Chapter 10, class 6 Maths Chapter 10 solutions Exercise 10.1 1.&nbsp;Find the perimeter of each of the following figures: Answer (a) Perimeter = Sum of all the sides = 4 cm [&hellip;]<\/p>\n","protected":false},"author":302,"featured_media":627101,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"newspack_featured_image_position":"","newspack_post_subtitle":"","newspack_article_summary_title":"Overview:","newspack_article_summary":"","newspack_hide_updated_date":false,"newspack_show_updated_date":false,"footnotes":""},"categories":[1411,876],"tags":[1556],"boards":[1180],"class_list":["post-164055","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-book-solutions","category-class-6","tag-ncert-maths-class-6","boards-ncert","entry"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v27.0 (Yoast SEO v27.1.1) - https:\/\/yoast.com\/product\/yoast-seo-premium-wordpress\/ -->\n<title>NCERT Solutions Class 6, Maths Chapter 10 - IndCareer Schools<\/title>\n<meta name=\"description\" content=\"NCERT Solutions for 6th Class Maths: Chapter 10-Mensuration | Browse Class 6 Maths Chapters NCERT books- IndCareer Schools\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-6th-class-maths-chapter-10-mensuration\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for 6th Class Maths: Chapter 10-Mensuration\" \/>\n<meta property=\"og:description\" content=\"Class 6: Maths Chapter 10 solutions. Complete Class 6 Maths Chapter 10 Notes. NCERT Solutions for 6th Class Maths: Chapter 10-Mensuration NCERT 6th Maths\" \/>\n<meta property=\"og:url\" content=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-6th-class-maths-chapter-10-mensuration\/\" \/>\n<meta property=\"og:site_name\" content=\"IndCareer Schools\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/indcareer\" \/>\n<meta property=\"article:published_time\" content=\"2021-02-24T09:06:54+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2023-09-14T05:44:21+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/NCERT-Solutions-25-scaled.jpg\" \/>\n\t<meta property=\"og:image:width\" content=\"1600\" \/>\n\t<meta property=\"og:image:height\" content=\"900\" \/>\n\t<meta property=\"og:image:type\" content=\"image\/jpeg\" \/>\n<meta name=\"author\" content=\"Pooja\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:creator\" content=\"@indcareer\" \/>\n<meta name=\"twitter:site\" content=\"@indcareer\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"Pooja\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"18 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-6th-class-maths-chapter-10-mensuration\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-6th-class-maths-chapter-10-mensuration\/\"},\"author\":{\"name\":\"Pooja\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/#\/schema\/person\/d6945cf059726f162259ba738092301e\"},\"headline\":\"NCERT Solutions for 6th Class Maths: Chapter 10-Mensuration\",\"datePublished\":\"2021-02-24T09:06:54+00:00\",\"dateModified\":\"2023-09-14T05:44:21+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-6th-class-maths-chapter-10-mensuration\/\"},\"wordCount\":2365,\"publisher\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/#organization\"},\"image\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-6th-class-maths-chapter-10-mensuration\/#primaryimage\"},\"thumbnailUrl\":\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/NCERT-Solutions-25-scaled.jpg\",\"keywords\":[\"NCERT Maths ( Class 6)\"],\"articleSection\":[\"Book Solutions\",\"class 6\"],\"inLanguage\":\"en-US\"},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-6th-class-maths-chapter-10-mensuration\/\",\"url\":\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-6th-class-maths-chapter-10-mensuration\/\",\"name\":\"NCERT Solutions Class 6, Maths Chapter 10 - 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