{"id":3204,"date":"2023-11-30T06:35:57","date_gmt":"2023-11-30T01:05:57","guid":{"rendered":"https:\/\/wordpress-505192-1602719.cloudwaysapps.com\/?p=3204"},"modified":"2023-12-01T11:58:26","modified_gmt":"2023-12-01T06:28:26","slug":"samacheer-kalvi-7th-maths-term-2-chapter-2-additional-questions","status":"publish","type":"post","link":"https:\/\/samacheerkalviguru.com\/samacheer-kalvi-7th-maths-term-2-chapter-2-additional-questions\/","title":{"rendered":"Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 2 Measurements Additional Questions"},"content":{"rendered":"

Students can Download Maths Chapter 2 Measurements Additional Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions<\/a> Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.<\/p>\n

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 2 Measurements Additional Questions<\/h2>\n

Additional Questions and Answers<\/span><\/p>\n

Exercise 2.1<\/span><\/p>\n

Question 1.
\nA circular disc of radius 28 cm is divided into two equal parts. What is the perimeter of each semicircular shape disc? Also find the perimeter of the circular disc.
\nAnswer:
\nRadius of the circular disc = 28 cm
\n\u2234 circumference of the circle = 2 \u03c0 r units
\n\"Samacheer
\n= 2 \u00d7 \\(\\frac { 22 }{ 7 } \\) \u00d7 28 cm
\n= 2 \u00d7 22 \u00d7 4 = 44 \u00d7 4 = 176 cm
\nPerimeter of the circular disc = 176 cm
\nNow circumference of the semicircular disc = \\(\\frac { 1 }{ 2 } \\) \u00d7 (2 \u03c0 r) = \u03c0 r units
\nPerimeter of the semicircular disk = \u03c0 r + r + r = \\(\\frac { 22 }{ 7 } \\) \u00d7 28 + 28 + 28
\n= 22 \u00d7 4 + 28 + 28 = 88 + 28 + 28 = 144
\nPerimeter of each semicircular disc = 144 cm<\/p>\n

\"Samacheer<\/p>\n

Question 2.
\nA gardener wants to fence a circular garden of diameter 14m. Find the length of the rope he needs to purchase if he makes 3 rounds of fence. Also find the cost of the rope, if it costs \u20b9 4 per metre.
\nSolution:
\nDiameter of the circular garden (d) = 14 m
\nCircumference = \u03c0 d = \\(\\frac { 22 }{ 7 } \\) \u00d7 14 = 44 m
\nSince the rope makes 3 rounds of fence,
\nlength of the rope needed = 3 \u00d7 circumference of the garden
\n= 3 \u00d7 44 m = 132 m
\nCost of rope per meter = \u20b9 4 \u00d7 132 = \u20b9 528<\/p>\n

Question 3.
\nLatha wants to put a lace on the edge of a circular table cover of diameter 3 m. Find the length of the lace required and also find the cost if one metre of the lace costs \u20b9 30 (Take \u03c0 = 3.15 )
\nSolution:
\nDiameter of the circular table cover = 3m
\nCircumference C = \u03c0 d units = 3.15 \u00d7 3 m = 9.45 m
\nLength of the lace required = 9.45 m
\nCost of lace per meter = \u20b9 30
\n\u2234 Cost of 9.45m lace = \u20b9 30 \u00d7 9.45 = \u20b9 283.50<\/p>\n

\"Samacheer<\/p>\n

Exercise 2.2<\/span><\/p>\n

Question 1.
\nThe circumference of two circles are on the ratio 5 : 6. Find the ratio of their areas.
\nSolution:
\nLet the radii of the given circles be r1<\/sub> and r2<\/sub>
\nLet their circumference be C1<\/sub> and C2<\/sub> respectively
\nC1<\/sub> = 2\u03c0r1<\/sub> and C2<\/sub> = 2\u03c0r2<\/sub>
\n\"Samacheer
\nNow let the area of the given circles the A1<\/sub> and A2
\n\"Samacheer
\n<\/sub>\u2234 The areas of two circles are in the ratio 25 : 36.<\/p>\n

Question 2.
\nFind the area of the circle whose circumference is 88 cm.
\nSolution:
\nThe circumference of the circle = 88 cm
\n2\u03c0r = 88 cm
\n2 \u00d7 \\(\\frac { 22 }{ 7 } \\) \u00d7 r = 88 cm
\nr = \\(\\frac{88 \\times 7}{2 \\times 22}\\) = 14 cm
\nr = 14 cm
\nArea of the circle A = \u03c0r2<\/sup>
\n= \\(\\frac { 22 }{ 7 } \\) \u00d7 14 \u00d7 14 cm2<\/sup>
\n= 22 \u00d7 2 \u00d7 14 cm2<\/sup>
\n= 616 cm2<\/sup><\/p>\n

\"Samacheer<\/p>\n

Question 3.
\nFind the cost of polishing a circular table top of diameter 16 dm if the rate of polishing is \u20b9 15 per dm2<\/sup>.
\nSolution:
\nDiameter of the circular table top = 16 dm
\nRadius (r) = \\(\\frac { 16 }{ 2 } \\) = 8cm
\nArea of the circular table top
\n= \u03c0r2<\/sup>\u00a0sq.units
\n= \\(\\frac { 22 }{ 7 } \\) \u00d7 8 \u00d7 8 dm2<\/sup> = \\(\\frac { 1408 }{ 7 } \\) dm2<\/sup>
\n= 201.14 dm2<\/sup>
\nRate of the polishing per dm2<\/sup> = \u20b9 15
\n\u2234 Rate of the polishing 201.14 dm2<\/sup>
\n= \u20b9 15 \u00d7 201.14
\n= \u20b9 3,017<\/p>\n

\"Samacheer<\/p>\n

Exercise 2.3<\/span><\/p>\n

Question 1.
\nFrom a circular sheet of radius 5 cm a circle of radius 3 cm is removed. Find the area of the remaining sheet.
\nSolution:
\nRadius of the outer circle R = 5 cm
\nRadius of the inner circle r = 3 cm
\nArea of the remaining sheet = Area of the outer circle
\n– Area of the inner circle
\n= \u03c0R2<\/sup> – \u03c0r2<\/sup> sq. units = \u03c0(R2<\/sup> – r2<\/sup>) sq. units
\n= \\(\\frac { 22 }{ 7 } \\) (52<\/sup> – 32<\/sup>) cm2<\/sup> = \\(\\frac { 22 }{ 7 } \\) (52<\/sup> – 32<\/sup>) cm2<\/sup>
\n= \\(\\frac { 22 }{ 7 } \\) \u00d7 (5 + 3) (5 – 3) = \\(\\frac { 22 }{ 7 } \\) \u00d7 (8) (2)
\n= \\(\\frac { 352 }{ 7 } \\) = 50.28 cm2<\/sup><\/p>\n

\"Samacheer<\/p>\n

Question 2.
\nA picture is painted on a cardborad 8 cm long and 5 cm wide such that there is a margin of 1.5 cm along each of its sides. Find the total area of the margin.
\nSolution:
\nLength of the outer rectangle L = 8 cm
\nBreadth of the outer rectangle B = 5 cm
\nArea of the outer rectangle = L \u00d7 B sq. units = 8 \u00d7 5 cm2<\/sup> = 40 cm2<\/sup>
\nLength of the inner rectangle l = L – 2W = 8 – 2(1.5)cm = 8 – 3cm
\nl = 5cm
\nBreadth of thqnner rectangle b = B – 2W = 5 – 2(1.5) cm
\n= 5 – 3cm = 2cm
\n\u2234 Area of the margin = Area of the outer rectangle
\n– Area of the inner rectangle
\n= (40 – 10) cm2<\/sup> = 30cm2<\/sup><\/p>\n

\"Samacheer<\/p>\n

Question 3.
\nA circular piece of radius 2 cm is cut from a rectangle sheet of length 5 cm and breadth 3 cm. Find the area left in the sheet.
\nSolution:
\nRadius of the portion removed r = 2 cm
\nArea of the circular sheet = \u03c0r2<\/sup> sq.units
\n= \\(\\frac { 22 }{ 7 } \\) \u00d7 2 \u00d7 2 cm2<\/sup> = \\(\\frac { 88 }{ 7 } \\) cm2<\/sup> = 12.57 cm2<\/sup>
\nLength of the rectangular sheet L = 5 cm
\nBreadth of the rectangular sheet B = 3 cm
\nArea of the rectangle = L \u00d7 B sq. units = 5 \u00d7 3 cm2<\/sup> = 15 cm2<\/sup>
\nArea of the sheet left over = Area of the rectangle – Area of the circle
\n= 15 cm2<\/sup> – 12.57 cm2<\/sup> = 2.43 cm2<\/sup><\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 2 Measurements Additional Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations. Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 2 Measurements Additional Questions Additional Questions and …<\/p>\n","protected":false},"author":4,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":"","footnotes":""},"categories":[5],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/3204"}],"collection":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/users\/4"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/comments?post=3204"}],"version-history":[{"count":1,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/3204\/revisions"}],"predecessor-version":[{"id":60044,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/3204\/revisions\/60044"}],"wp:attachment":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/media?parent=3204"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/categories?post=3204"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/tags?post=3204"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}