{"id":15730,"date":"2020-09-10T11:39:50","date_gmt":"2020-09-10T06:09:50","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=15730"},"modified":"2021-07-02T15:01:53","modified_gmt":"2021-07-02T09:31:53","slug":"kseeb-solutions-for-class-7-maths-chapter-11-ex-11-3","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-for-class-7-maths-chapter-11-ex-11-3\/","title":{"rendered":"KSEEB Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3"},"content":{"rendered":"

Students can Download Chapter 11 Perimeter and Area Ex 11.3, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths<\/a>, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.<\/p>\n

Karnataka State Syllabus Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3<\/h2>\n

Question 1.
\nFind the circumference of the circles with the following radius :
\n(Take n = \\(\\frac{22}{7}\\))
\na) 14 cm
\nradius = r = 14 cm
\n\u2234 circumference of a circle = 2\u03c0r
\n\"KSEEB<\/p>\n

b) 28 mm
\nradius = r = 28 cm
\n\u2234 circumference of a circle = 2\u03c0r
\n\"KSEEB<\/p>\n

c) 21 cm
\nradius = r = 21 cm
\ncircumference of a circle = 2\u03c0r
\n\"KSEEB<\/p>\n

\"KSEEB<\/p>\n

Question 2.
\nFind the area of the following circles, given that:<\/p>\n

a) radius = 14 mm (Take \u03c0 = \\(\\frac{22}{7}\\))
\nradius = 14 mm
\n\u2234 Area of the circle = \u03c0r2<\/sup>
\n\"KSEEB
\n= 616 sq.cms<\/p>\n

b) diameter = 49 m
\n\"KSEEB
\nArea of the circle = \u03c0r2<\/sup>
\n\"KSEEB
\n= 1,886.5 sq.metrs<\/p>\n

c) radius = 5 cm
\nradius = r = 5 cm
\n\u2234 Area of the circle = \u03c0r2<\/sup>
\n\"KSEEB<\/p>\n

Question 3.
\nIf the circumference of a circular sheet is 154m, find its radius. Also find the area of the sheet. (Take \u03c0 = \\(\\frac{22}{7}\\))
\nSolution:
\nC = circumference of a circle = 154m = 2\u03c0r
\n\"KSEEB
\nArea of the sheet = \u03c0r2<\/sup>
\n\"KSEEB
\n= 1,886.5 sq.metrs<\/p>\n

\"KSEEB<\/p>\n

Question 4.
\nA gardener wants to fence a circular garden of diameter 21m. Find the length of the rope he needs to purchase, if he makes 2 rounds of fence. Also find the cost of the rope, if it costs \u20b9 4 per meter.
\n(Take \u03c0 = \\(\\frac{22}{7}\\))
\nd = diameter = 21 m
\n\"KSEEB
\n\u2234 radius of the circular garden
\n\"KSEEB
\ncircumference of the circular garden = 2\u03c0r
\n\"KSEEB
\nLength of the rope needed to make one round of fence } = 66 mts
\nLength of the rope needed to make 2 rounds =
\n66 \u00d7 2 = 132 mts
\ncost of rope per metre = Rs. 4
\nTotal cost of rope = 132 \u00d7 4 = Rs. 528<\/p>\n

Question 5.
\nFrom a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet.
\n(Take \u03c0 = 3.14)
\nSolution:
\nRadius of the outer circle = 4 cms.
\n\"KSEEB
\n\u2234 Area of the outer circle = \u03c0r2<\/sup>
\n= 3.14 \u00d7 4 \u00d7 4
\n= 28.26 sq.cms
\n= 50.24 sq.cms
\n\u03c0 = 3.14
\nRadius of the inner circle = 3 cms
\n\u2234 Area of the inner circle = \u03c0r2<\/sup>
\n= 3.14 \u00d7 3 \u00d7 3 = 28.26 sq. cms
\nArea of the remaining part = 50.24 – 28.26 = 21.98 sq.cms<\/p>\n

\"KSEEB<\/p>\n

Question 6.
\nSaima wants to put lace on the edge of a circular table cover of diameter 1.5m. Find the length of the lace required and also find its cost if one meter of the lace
\ncosts \u20b9 15. (Take \u03c0 = 3.14)
\nSolution:
\nDiameter of the table cover = 1.5m
\nCircumference of the table cover }
\n2\u03c0r \u2013 \u03c0d = 3.14 \u00d7 1.5 (\u2235 d = 2r) = 4.710 mts
\n\u2234 Length of the lace required = 4.71 mt
\nCost of lace per meter = Rs. 15
\n\u2234 Total cost of the lace =4.71 \u00d7 15 = Rs 70.65<\/p>\n

Question 7.
\nFind the perimeter of the adjoining figure, which is a semicircle including its diameter.
\n\"KSEEB
\nSolution:
\nDiameter of the semi-circle = 10 cm
\nCircumference of semi-circle =
\n\"KSEEB
\nPerimeter of the semi-circle(including its diameter) } = 15.7 + 10 = 25.7 cms<\/p>\n

\"KSEEB<\/p>\n

Question 8.
\nFind the cost of polishing a circular table-top of diameter 1.6m, if the rate of polishing is ? 15\/m2<\/sup>. (Take \u03c0 = 3.14)
\nSolution:
\nDiameter of the Table Top = 1.6 m
\nArea of the Table Top = \u03c0r2<\/sup>
\n\"KSEEB
\nRate of polishing = Rs. 15\/sq.m
\nTotal cost of polishing = 2.0096 \u00d7 15
\n= Rs.30.14<\/p>\n

\"KSEEB<\/p>\n

Question 9.
\nShazil took a wire of length 44 cm and bent it into the shape of a circle. Find the radius of that circle. Also find its area. If the same wire is bent into the shape of a square, what will be the length of each of its sides ? Which figure encloses more area, the circle or the square? (Take \u03c0 = \\(\\frac{22}{7}\\))
\nSolution:
\nLength of the wire = 44 cms
\nTotal length of the circle = perimeter of the circle
\n\"KSEEB
\n\u2234 radius = 7 cms.
\n\"KSEEB
\n\u2234 Length of wire = perimeter of the square = 154 sq.cms
\n44 = 44
\n\"KSEEB
\nArea of the square = a2<\/sup> = 11 \u00d7 11 = 121 sq.cms.
\n\u2234 The area of the circle is more means the circle encloses more area.<\/p>\n

Question 10.
\nFrom a circular card sheet of radius 14cm, two circles of radius 3.5 cm and a rectangle of length 3 cm and breadth 1 cm are removed, (as shown in the adjoining figure). Find the area of the remaining sheet. (Take \u03c0 = \\(\\frac{22}{7}\\))
\nSolution:
\nRadius of the circular Card = 14 cm
\nArea of the Card
\n\"KSEEB
\n\"KSEEB
\nArea of small 2 circles =
\n\"KSEEB
\n= 77 sq. cms.
\nArea of rectangle = l \u00d7 b = 3 \u00d7 1 = 3 sq. cms
\nArea of the remaining sheet
\n= 616 – (77 + 3)
\n= 616 = 80 = 536 sq.cm<\/p>\n

\"KSEEB<\/p>\n

Question 11.
\nA circle of radius 2 cm is cut out from a square piece of an aluminium sheet of side 6 cm. What is the area of the leftover aluminium sheet? (Take \u03c0 = 3.14)
\nSolution:
\nArea of square aluminium sheet = 6 \u00d7 6 = 36 sq. cms
\nSide = 6 cm radius = 2 cm
\nArea of circle =
\n\"KSEEB
\nArea of the left over Aluminium sheet = 36 – 12.56 = 23.44sq.cm<\/p>\n

Question 12.
\nThe circumference of a circle is 31.4. Find the radius and the area of the circle? (Take \u03c0 = 3.14)
\nSolution:
\nThe circumference of a circle = 2\u03c0r = 31.4 cm
\n\"KSEEB
\nArea of the cirlce = \u03c0r2<\/sup>
\n= 3.14 \u00d7 5 \u00d7 5
\n= 78.5 sq.cms<\/p>\n

Question 13.
\nA circular flower bed is surrounded by a path 4 m wide. The diameter of the flower bed is 66m. What is the area of this path? (\u03c0 = 3.14)
\nSolution:
\n\"KSEEB
\nDiameter of the flower bed = 66 m
\nRadius = 66\/2 = 33 m
\nArea of the flower bed = \u03c0r2<\/sup>
\n= 3.14 \u00d7 33 \u00d7 33
\n= 3419.46 sq.cm
\nRadius of the flower bed with path = 33 + 4 = 37 ms.
\nArea of the flower bed with path =
\n= 3.14 \u00d7 37 \u00d7 37
\n= 4298.66 sq.cm
\nArea of the path = Area of the flower bed with path \u2013 Area of the flower bed
\n= 4298.66 \u2013 3419.46
\n= 879.2 sq.mts<\/p>\n

\"KSEEB<\/p>\n

Question 14.
\nA circular flower garden has an area of 314 m2<\/sup>. A sprinkler at the centre of the garden can cover an area that has a radius of 12 m. Will sprinkler water the
\nentire garden? (Take \u03c0 = 3.14)
\nSolution:
\nArea of the circular flower garden = 314 sq.m
\nRadius of the sprinkler = 12m
\nArea of the sprinkler = \u03c0r2<\/sup>\u00a0= 3.14 \u00d7 12 \u00d7 12
\n= 452.16 sq.mt
\n\u2234 452.16 > 314
\n\u2234 Yes, the sprinkler water the entire garden.<\/p>\n

Question 15.
\nFind the circumference of the inner and the outer circles, shown in the adjoining figure? (Take \u03c0 = 3.14)
\nSolution:
\n\"KSEEB
\nCircumference of the outer circle = 2\u03c0r
\nradius = 19 m
\n= 2 \u00d7 3.14 \u00d7 19
\n= 119.32 mts
\nCircumference of the inner circle = 2\u03c0r
\nradius = 19 \u2013 10 = 9 cm.
\n= 2 \u00d7 3.14 \u00d7 9
\n= 56.52 cm<\/p>\n

Question 16.
\nHow many times a wheel of radius 25cm must to go 352 m ? (Take \u03c0 = \\(\\frac{22}{7}\\))
\nSolution:
\nradius of the wheel = r = 28 cm
\ncircumference of the wheel = 2\u03c0r
\n\"KSEEB
\n= 176 cms
\nTo cover the distance of 352 mtrs (\u2235 1 m = 100 cm)
\nNo. of rotation
\n\"KSEEB<\/p>\n

\"KSEEB<\/p>\n

Question 17.
\nThe minute hand of a circular clock is 15 cm long. How far does the tip of the minute hand move in 1 hour? (Take \u03c0 = 3.14)
\nSolution:
\nLength of the minute hand = 15 cm
\n1 hour = 1 round = circumference = 2\u03c0r
\n= 2 \u00d7 3.14 \u00d7 15
\n= 94.2 cms
\n\u2234 The tip of the minute hand in one hour moves 94.2 cms.<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Chapter 11 Perimeter and Area Ex 11.3, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations. Karnataka State Syllabus Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3 Question 1. Find the …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[84],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/15730"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=15730"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/15730\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=15730"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=15730"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=15730"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}