{"id":25610,"date":"2021-06-23T15:41:15","date_gmt":"2021-06-23T10:11:15","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=25610"},"modified":"2021-06-25T15:28:29","modified_gmt":"2021-06-25T09:58:29","slug":"ncert-solutions-for-class-6-maths-chapter-10-ex-10-1","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-6-maths-chapter-10-ex-10-1\/","title":{"rendered":"NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.1"},"content":{"rendered":"

These NCERT Solutions for Class 6 Maths<\/a> Chapter 10 Mensuration Ex 10.1 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Exercise 10.1<\/h2>\n

Question 1.
\nFind the perimeter of each of the following figures:
\n\"NCERT
\nAnswer:
\n(a) Perimeter = Sum of all the sides = 4 cm + 2 cm + 1 cm + 5 cm = 12 cm
\n(b) Perimeter = Sum of all the sides = 23 cm + 35 cm + 40 cm + 35 cm = 133 cm
\n(c) Perimeter = Sum of all the sides =15 cm + 15 cm + 15 cm + 15 cm = 60 cm
\n(d) Perimeter = Sum of all the sides = 4 cm + 4 cm + 4 cm + 4 cm + 4 cm = 20 cm
\n(e) Perimeter = Sum of all the sides 1 cm + 4 cm + 0.5 cm + 2.5 cm + 2.5 cm + 0.5 cm + 4 cm = 15 cm
\n(f) Perimeter = Sum of all the sides = 4
\ncm + 1 cm + 3 cm + 2 cm + 3 cm + 4
\ncm + 1 cm + 3 cm + 2 cm + 3 cm + 4
\ncm + 1 cm + 3 cm + 2 cm + 3 cm + 4
\ncm + 1 cm + 3 cm + 2 cm + 3 cm = 52 cm<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nThe 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?
\nAnswer:
\nTotal length of tape required
\n= Perimeter of rectangle
\n= 2 (length + breadth) = 2 (40 + 10)
\n= 2 \u00d7 50 = 100 cm = 1 m
\nThus, the total length of tape required is 100 cm or 1 m.<\/p>\n

Question 3.
\nA table-top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the table-top?
\nAnswer:
\nLength of table top = 2 m 25 cm = 2.25 m
\nBreadth of table top = 1 m 50 cm = 1.50 m
\nPerimeter of table top = 2 \u00d7 (length + breadth) = 2 \u00d7 (2.25 + 1.50) = 2 \u00d7 3.75 = 7.5 m
\nThus, the perimeter of tabletop is 7.5 m.<\/p>\n

Question 4.
\nWhat is the length of the wooden strip required to frame a photograph of length and breadth 32 cm and 21 cm respectively?
\nAnswer:
\nLength of wooden strip = Perimeter of photograph
\nPerimeter of photograph
\n= 2 x (length + breadth)
\n= 2 (32 + 21) = 2 \u00d7 53cm =106 cm
\nThus, the length of the wooden strip required is equal to 106 cm.<\/p>\n

\"NCERT<\/p>\n

Question 5.
\nA 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?
\nAnswer:
\nSince the 4 rows of wires are needed.
\nTherefore the total length of wires is equal to 4 times the perimeter of rectangle.
\nPerimeter of field = 2 \u00d7 (length + breadth)
\n= 2 \u00d7 (0.7 + 0.5) = 2 \u00d7 1.2 = 2.4 km
\n= 2.4 \u00d7 1000 m = 2400 m
\nThus, the length of wire = 4 \u00d7 2400 = 9600 m = 9.6 km<\/p>\n

Question 6.
\nFind the perimeter of each of the following shapes:
\n(a) A triangle of sides 3 cm, 4 cm and 5 cm.
\n(b) An equilateral triangle of side 9 cm.
\n(c) An isosceles triangle with equal sides 8 cm each and third side 6 cm.
\nAnswer:
\n(a) Perimeter of \u0394ABC = AB + BC + CA = 3 cm + 5 cm + 4 cm = 12 cm
\n\"NCERT<\/p>\n

(b) Perimeter of equilateral ABC
\n= 3 \u00d7 side = 3 \u00d7 9 cm = 27 cm
\n\"NCERT<\/p>\n

(c) Perimeter of AABC = AB + BC + CA = 8 cm + 6 cm + 8 cm = 22 cm
\n\"NCERT<\/p>\n

Question 7.
\nFind the perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.
\nAnswer:
\nPerimeter of triangle
\n= Sum of all three sides
\n= 10 cm + 14 cm + 15 cm = 39 cm
\nThus, the perimeter of triangle is 39 cm<\/p>\n

\"NCERT<\/p>\n

Question 8.
\nFind the perimeter of a regular hexagon with each side measuring 8 m.
\nAnswer:
\nPerimeter of Hexagon
\n= 6 \u00d7 length of one side
\n= 6 \u00d7 8 m = 48 m
\nThus, the perimeter of hexagon is 48 m.<\/p>\n

Question 9.
\nFind the side of the square whose perimeter is 20 m.
\nAnswer:
\nPerimeter of square = 4 \u00d7 side
\n\u21d2 20 = 4 \u00d7 side
\n\u21d2 Side = \\(\\frac{20}{4}\\) = 5 cm
\nThus, the side of square is 5 cm.<\/p>\n

Question 10.
\nThe perimeter of a regular pentagon is 100 cm. How long is its each side?
\nAnswer:
\nPerimeter of regular pentagon = 100 cm
\n\u21d2 5 \u00d7 side = 100 cm
\n\u21d2 Side = \\(\\frac{100}{5}\\) = 20 cm
\nThus, the side of regular pentagon is 20 cm.<\/p>\n

\"NCERT<\/p>\n

Question 11.
\nA piece of string is 30 cm long. What will be the length of each side if the string is used to form:
\n(a) a square?
\n(b) an equilateral triangle?
\n(c) a regular hexagon?
\nAnswer:
\nLength of string = Perimeter of each figure
\n(a) Perimeter of square = 30 cm
\n\u21d2 4 \u00d7 side = 30 cm
\n\u21d2 Side = \\(\\frac{30}{4}\\) = 7.5 cm 4
\nThus, the length of each side of square is 7.5 cm.<\/p>\n

(b) Perimeter of equilateral triangle = 30 cm
\n\u21d2 3 \u00d7 side = 30 cm
\n\u21d2 Side = \\(\\frac{30}{3}\\) = 10 cm
\nThus, the length of each side of equilateral triangle is 10 cm.<\/p>\n

(c) Perimeter of hexagon = 30 cm
\n\u21d2 6 \u00d7 side = 30 cm
\n\u21d2 Side = \\(\\frac{30}{6}\\) = 5 cm
\nThus, the side of each side of hexagon is 5 cm.<\/p>\n

Question 12.
\nTwo sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is its third side?
\nAnswer:
\nLet the length of third side be X cm.
\nLength of other two side are 12 cm and 14 cm.
\nNow, Perimeter of triangle = 36 cm
\n\u21d2 12 +14 + X = 36
\n\u21d2 26 + X = 36
\n\u21d2 X = 36 – 26 = 10 cm
\nThus, the length of third side is 10 cm<\/p>\n

\"NCERT<\/p>\n

Question 13.
\nFind the cost of fencing a square park of side 250 m at the rate of \u20b9 20 per metre.
\nAnswer:
\nSide of square = 250 m
\nPerimeter of square
\n= 4 \u00d7 side = 4 \u00d7 250 = 1000 m
\nSince, cost of fencing of per meter = \u20b920
\nTherefore, the cost of fencing of 1000 meters = 20 \u00d7 1000 = \u20b9 20,000<\/p>\n

Question 14.
\nFind the cost of fencing a rectangular park of length 175 m and breadth 125 m at the rate of \u00d7 12 per metre.
\nAnswer:
\nLength of rectangular park = 175 m
\nBreadth of rectangular park = 125 m
\nPerimeter of park
\n= 2 \u00d7 (length + breadth)
\n= 2 \u00d7 (175 + 125) = 2 \u00d7 300 = 600 m
\nSince, the cost of fencing park per meter = \u20b9 12
\nTherefore, the cost of fencing park of 600 m = 12 \u00d7 600 = \u20b9 7,200<\/p>\n

Question 15.
\nSweety runs around a square park of side 75 m. Bulbul runs around a rectangular park with length 60 m and breadth 45 m. Who covers less distance?
\nAnswer:
\nDistance covered by Sweety
\n= Perimeter of square park
\nPerimeter of square = 4 \u00d7 side
\n= 4 \u00d7 75 = 300 m
\nThus, distance covered by Sweety is 300 m. Now, distance covered by Bulbul =
\nPerimeter of rectangular park
\nPerimeter of rectangular park
\n= 2 \u00d7 (length + breadth)
\n= 2 \u00d7 (60 + 45) = 2 \u00d7 105 = 210 m
\nThus, Bulbul covers the distance of 210 m and Bulbul covers less distance.<\/p>\n

\"NCERT<\/p>\n

Question 16.
\nWhat is the perimeter of each of the following figures? What do you infer from the answers?
\n\"NCERT
\nAnswer:
\n(a) Perimeter of square
\n= 4 \u00d7 side = 4 \u00d7 25 = 100 cm<\/p>\n

(b) Perimeter of rectangle
\n= 2 \u00d7 (length + breadth)
\n= 2 \u00d7 (40 + 10) = 2 \u00d7 50 = 100 cm<\/p>\n

(c) Perimeter of rectangle
\n= 2 \u00d7 (length + breadth)
\n= 2 \u00d7 (30 + 20) = 2 \u00d7 50 = 100 cm<\/p>\n

(d) Perimeter of triangle = Sum of all sides
\n= 30 cm + 30 cm + 40 cm = 100 cm
\nThus, all the figures have same perimeter.<\/p>\n

\"NCERT<\/p>\n

Question 17.
\nAvneet buys 9 square paving slabs, each with a side of 1\/2 m. He lays them in the form of a square.
\n(a) What is the perimeter of his arrangement [Fig (i)]?
\n(b) Shari does not like his arrangement. She gets him to lay them out like a cross. What is the perimeter of her arrangement [(Fig (ii)]?
\n(c) Which has greater perimeter?
\n(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.)
\n\"NCERT
\nAnswer:
\n(a) The arrangement so formed in Fig (i) is a square of side equal to 3 \u00d7 side of the square slab
\n\u2234 Perimeter of his arrangement in Fig (i)
\n= 4 \u00d7 ( 3 \u00d7 side of the square slab)
\n= 4 \u00d7 ( 3 \u00d7 0.5) m = 4 \u00d7 1.5 m = 6 m<\/p>\n

(b) The cross arrangement so formed in Fig (ii) by 9 square slabs, has perimeter, which consists of 20 length segments each equal to side of I a square slab i.e. 0.5 m
\n\u2234 Perimeter of her arrangement in Fig (ii)
\n= 20 \u00d7 ( side of the square slab)
\n= 20 \u00d7 0.5 m = 10 m<\/p>\n

(c) \u2235 10m > 6m
\n\u2234 Cross arrangement has greater perimeter<\/p>\n

(d) \u2235 Total number of tiles = 9
\n\u2234 The following arrangement will also have greater perimeter.
\n\"NCERT
\nSince, this arrangement is in the form of rectangle with length and breadth as \\(\\frac { 9 }{ 2 }\\) m and \\(\\frac { 1 }{ 2 }\\) respectively.
\n\u2234 Perimeter = 2 \\(\\left(\\frac{9}{2}+\\frac{1}{2}\\right)\\)
\n= 2 \u00d7 = 10m \\(\\left[\\frac{10}{2}\\right]\\) = 10m<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.1 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Exercise 10.1 Question 1. Find the perimeter of each of the following figures: Answer: (a) Perimeter = Sum of all the sides = …<\/p>\n

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.1<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[10],"tags":[],"yoast_head":"\nNCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.1 - NCERT Solutions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-6-maths-chapter-10-ex-10-1\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.1 - NCERT Solutions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.1 Questions and Answers are prepared by our highly skilled subject experts. 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NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Exercise 10.1 Question 1. 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