{"id":43724,"date":"2022-05-29T21:00:01","date_gmt":"2022-05-29T15:30:01","guid":{"rendered":"https:\/\/cbselibrary.com\/?p=43724"},"modified":"2023-01-25T12:17:10","modified_gmt":"2023-01-25T06:47:10","slug":"ml-aggarwal-class-6-solutions-for-icse-maths-chapter-14-ex-14-1","status":"publish","type":"post","link":"https:\/\/cbselibrary.com\/ml-aggarwal-class-6-solutions-for-icse-maths-chapter-14-ex-14-1\/","title":{"rendered":"ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 14 Mensuration Ex 14.1"},"content":{"rendered":"
Question 1.
\nFind the perimeter of each of the following figures:
\n
\nSolution:
\nPerimeter = Sum of all the sides.
\n(i) Perimeter = 5 cm + 3 cm + 2 cm + 7 cm = 17 cm
\n(ii) Perimeters = 31 cm + 38 cm + 48 cm + 38 cm = 155 cm
\n(iii) Perimeter = 19 cm + 19 cm + 19 cm + 19 cm = 76 cm
\n(iv) Perimeter = 7 cm + 7 cm + 7 cm + 7 cm + 7 cm = 35 cm<\/p>\n
Question 2.
\nFind the perimeter of each of the following shapes:
\n(i) A triangle of sides 3 cm, 4 cm and 6 cm.
\n(ii) A triangle of sides 7 cm, 5.4 cm and 10.2 cm.
\n(iii) An equilateral triangle of side 11 cm.
\n(iv) An isosceles triangle with equal sides 10 cm each and third side 7 cm.
\nSolution:
\n(i) Perimeter of the triangle with sides 3 cm, 4 cm and 6 cm
\n= 3 cm + 4 cm + 6 cm
\n= 13 cm
\n(ii) Perimeter of the triangle with sides 7 cm, 5.4 cm, 10.2 cm
\n= 7 cm + 5.4 cm + 10.2 cm
\n= 22.6 cm
\n(iii) Perimeter of an equilateral triangle
\n= 3 \u00d7 length of a side
\n= 3 \u00d7 11 cm
\n= 33 cm
\n(iv) Perimeter of isosceles triangle
\n= 10 cm + 10 cm + 7 cm
\n= 27 cm<\/p>\n
Question 3.
\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?
\nSolution:
\nLength of the tape required
\n= Perimeter of the rectangular box
\n= 2 \u00d7 (Length + Breadth)
\n= 2 \u00d7 (40 cm + 10 cm)
\n= 2 \u00d7 (50 cm)
\n= 100 cm
\n= 1 m<\/p>\n
Question 4.
\nTable-Top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the table-top?
\nSolution:
\nPerimeter of the table-top
\n= 2 \u00d7 (Length + Breadth)
\n= 2 \u00d7 (2 m 25 cm + 1 m 50 cm)
\n= 2 \u00d7 (2.25 m + 1.50 m)
\n= 2 \u00d7 (3.75 m)
\n= 7.5 m<\/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?
\nSolution:
\nPerimeter of the rectangle
\n= 2 \u00d7 (Length + Breadth)
\n= 2 \u00d7 (0.7 km + 0.5 km)
\n= 2 \u00d7 (1.2 km)
\n= 2.4 km
\nLength of the wire needed
\n= 4 \u00d7 Perimeter of the rectangle
\n= 4 \u00d7 (2.4 km)
\n= 9.6 km<\/p>\n
Question 6.
\nFind the perimeter of a regular hexagon with each side measuring 7.5 m.
\nSolution:
\nThe perimeter of a regular hexagon
\n= 6 \u00d7 Length of a side
\n= 6 \u00d7 7.5 m
\n= 45 m<\/p>\n
Question 7.
\nThe lengths of two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is the length of its third side?
\nSolution:
\nPerimeter of a triangle = 12 cm + 14 cm + l
\n\u21d2 36 cm = 12 cm + 14 cm + l
\n\u21d2 36 cm = 26 cm + l
\n\u21d2 l = 36 cm – 26 cm
\n\u21d2 l = 10 cm<\/p>\n
Question 8.
\nThe perimeter of a regular pentagon is 100 cm. How long is its each side?
\nSolution:
\nPerimeter of the regular pentagon = 5 \u00d7 Length of a side
\n\u21d2 Length of one (each) side
\n\\(\\begin{array}{l}{=\\frac{\\text { Perimeter of the regular pentagon }}{5}} \\\\ {=\\frac{100}{5} \\mathrm{cm}=20 \\mathrm{cm}}\\end{array}\\)<\/p>\n
Question 9.
\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?
\nSolution:
\n(a) Perimeter of the square = 4 \u00d7 Length of aside
\n\u21d2 Length of a side
\n= \\(\\frac{\\text { Perimeter of the square }}{4}=\\frac{30}{4} \\mathrm{cm}\\)
\n= 7.5 cm<\/p>\n
(b) Perimeter of the equilateral triangle = 3 \u00d7 Length of a side
\n\u21d2 Length of a side
\n= \\(\\frac{\\text { Perimeter of the equilateral triangle }}{3}\\)
\n= \\(\\frac{30}{3} \\mathrm{cm}=10 \\mathrm{cm}\\)<\/p>\n
(c) Perimeter of the regular hexagon = 6 \u00d7 Length of a side
\n\u21d2 Length of a side
\n\\(\\begin{array}{l}{=\\frac{\\text { Perimeter of the regular hexagon }}{6}} \\\\ {=\\frac{30}{6} \\mathrm{cm}=5 \\mathrm{cm}}\\end{array}\\)<\/p>\n
Question 10.
\nFind the cost of fencing a rectangular park of length 225 m and breadth 115 m at the rate of \u20b913 per metre.
\nSolution:
\nPerimeter of the rectangular park
\n= 2 \u00d7 (Length + Breadth)
\n= 2 \u00d7 (225 m + 115 m)
\n= 2 \u00d7 (340 m)
\n= 680 m
\n\u2234 Cost of fencing the rectangular park
\nat the rate of \u20b913 per metre = \u20b913 \u00d7 680 m = \u20b98840<\/p>\n
Question 11.
\nMeera went to a rectangular park 140 m long and 90 m wide. She took 5 complete rounds on its boundary. What is the distance covered by her?
\nSolution:
\nLength (l) = 140 m
\nWidth (b) = 90 m
\n
\n\u2234 Perimeter of park = 2(l + b)
\n= 2 (140 + 90)
\n= 2(230)
\n= 460 m
\nShe takes 5 complete round,
\ntherefore distance covered by her = 5 \u00d7 460m = 2300 m<\/p>\n
Question 12.
\nPinky runs 8 times around a rectangular park with length 80 m and breadth 55 m while Pankaj runs 7 times around a square park of side 75 cm. Who covers more distance and by how much?
\nSolution:
\nPerimeter of rectangular park = 2 \u00d7 (Length + Breadth)
\n= 2 \u00d7 (80 + 55)
\n= 270 m
\nPinky runs 8 times = 8 \u00d7 270 = 2160 m
\nAnd, perimeter of square park = 4 \u00d7 Length of a side
\n= 4 \u00d7 75
\n= 300
\nPankaj runs 7 times = 7 \u00d7 300 = 2100 m
\n\u2234 Pinky covers more distance i.e., 2160 – 2100 = 60 m<\/p>\n
ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 14 Mensuration Ex 14.1 Question 1. Find the perimeter of each of the following figures: Solution: Perimeter = Sum of all the sides. (i) Perimeter = 5 cm + 3 cm + 2 cm + 7 cm = 17 cm (ii) Perimeters = 31 cm + … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[3034],"tags":[],"yoast_head":"\n