{"id":3080,"date":"2023-11-29T11:36:21","date_gmt":"2023-11-29T06:06:21","guid":{"rendered":"https:\/\/wordpress-505192-1602719.cloudwaysapps.com\/?p=3080"},"modified":"2023-11-30T12:23:29","modified_gmt":"2023-11-30T06:53:29","slug":"samacheer-kalvi-7th-maths-term-2-chapter-2-ex-2-4","status":"publish","type":"post","link":"https:\/\/samacheerkalviguru.com\/samacheer-kalvi-7th-maths-term-2-chapter-2-ex-2-4\/","title":{"rendered":"Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 2 Measurements Ex 2.4"},"content":{"rendered":"

Students can Download Maths Chapter 2 Measurements Ex 2.4 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 Ex 2.4<\/h2>\n

Miscellaneous Practice Problems<\/span><\/p>\n

Question 1.
\nA wheel of a car covers a distance of 3520 cm in 20 rotations. Find the radius of the wheel?
\nSolutions:
\nDistance covered by circular wheel in 20 rotation = 3520 cm
\n\u2234 Distance covered ini rotation = \\(\\frac { 3520 }{ 20 } \\) cm = 176 cm
\n\u2234 Circumference of the wheel = 176 cm
\n\u2234 2\u03c0r = 176
\n2 \u00d7 \\(\\frac { -2 }{ 6 } \\) \u00d7 r = 176
\nr = \\(\\frac{176 \\times 7}{2 \\times 22}\\)
\nr = 28 cm
\nRadius of the wheel = 28 cm<\/p>\n

Question 2.
\nThe cost of fencing a circular race course at the rate of \u20b9 8 per metre is \u20b92112. Find the diameter of the race course.
\nSolution:
\nCost of fencing the circumference = \u20b9 2112
\nCost of fencing one meter = \u20b9 8
\n\u2234 Circumference of the circle = \\(\\frac { 2112 }{ 8 } \\) = 264 m
\n\u03c0d = 264 m
\n\\(\\frac { 22 }{ 7 } \\) \u00d7 d = 264
\nd = \\(\\frac{264 \\times 7}{22}\\) = 12 \u00d7 7 m = 84 m
\n\u2234 Diameter of the race cource = 84 m<\/p>\n

Question 3.
\nA path 2 m long and 1 m broad is constructed around a rectangular ground of dimensions 120 m and 90 m respectively. Find the area of the path.
\nSolution:
\nLength of the rectangular ground l = 120 m
\nBreadth b = 90 m
\nLength of the path W1<\/sub> = 2m
\nLength of the path W2<\/sub> = 1m
\nLength of the ground with path L = 1 + 2 (W2<\/sub>) = 120 + 2(1) m
\n= 120 + 2 = 122 m
\nBreadth of the ground with path B = l + 2(W1<\/sub>) units
\n= 90 + 2(2) m = 90 + 4 m = 94 m
\n\u2234 Area of the path = (L \u00d7 B) – (1 \u00d7 b) sq. units
\n= (122 \u00d7 94) – (122 \u00d7 94) m2<\/sup> = 668 m2<\/sup>
\n\u2234 Area of the path = 668 m2<\/sup><\/p>\n

\"Samacheer<\/p>\n

Question 4.
\nThe cost of decorating the circumference of a circular lawn of a house at the rate of \u20b955 per metre is \u20b916940. What is the radius of the lawn?
\nSolution:
\nCost of decorating the circumference = \u20b9 16,940
\nCost of decorating per meter = \u20b9 55
\n\u2234 Length of the circumference = \\(\\frac { 16940 }{ 55 } \\) m = 308 m
\nCircumference of the circular lawn = 308 m
\n2 \u00d7 \u03c0r = 308 m
\n2 \u00d7 \\(\\frac { 22 }{ 7 } \\) \u00d7 r = 308 m
\nr = \\(\\frac{308 \\times 7}{2 \\times 22}\\)
\nr = 49 m
\nRadius of the lawn = 49 m<\/p>\n

Question 5.
\nFour circles are drawn side by side in a line and enclosed by a rectangle as shown below.
\nIf the radius of each of the circles is 3 cm, then calculate:
\n(i) The area of the rectangle.
\n(ii) The area of each circle.
\n(iii) The shaded area inside the rectangle.
\n\"Samacheer
\nSolution:
\nGiven radius of a circle r = 3 cm
\nDiameter of the circle = 2r = 2 \u00d7 3 = 6 cm
\nBreadth of the rectangle = Diamter of the circle
\nB = 6cm
\nLength of the rectangle L = 4 \u00d7 diameter of a circle
\nL = 4 \u00d7 6
\nL = 24cm<\/p>\n

(i) Area of the rectangle = L \u00d7 B sq. units
\n= 24 \u00d7 6 cm2<\/sup>
\nArea of the rectangle = 144 cm2<\/sup><\/p>\n

(ii) Area of the circle = \u03c0r2<\/sup> sq. units
\n= \\(\\frac { 22 }{ 7 } \\) \u00d7 3 \u00d7 3 cm2<\/sup>
\n= \\(\\frac { 198 }{ 7 } \\) cm2<\/sup>
\n= 28.28 cm2<\/sup><\/p>\n

(iii) Area of the shaded area = Area of the rectangle – Area of the 4 circles
\n= 144 – (4 \u00d7 \\(\\frac { 198 }{ 7 } \\)) cm2<\/sup> = 144 – \\(\\frac { 792 }{ 7 } \\) cm2<\/sup>
\n= 144 – 113.14 cm2<\/sup> = 30.85 cm2<\/sup><\/p>\n

\"Samacheer<\/p>\n

Challenge Problems<\/span><\/p>\n

Question 6.
\nA circular path has to be constructed around a circular lawn. If the outer and inner circumferences of the path are 88 cm and 44 cm respectively, find the width and area of the path.
\nSolution:
\nOuter circumference of the circular lawn = 88 cm
\n2\u03c0R = 88 cm
\nInner circumference of the lawn 2\u03c0r = 44 cm
\n2\u03c0R – 2\u03c0r = 88 – 44
\n2 \u00d7 \\(\\frac { 22 }{ 7 } \\) (R – r) = 44
\n(R – r) = \\(\\frac{44 \\times 7}{2 \\times 22}\\)
\nOuter radius – Inner radius = 7 cm
\n\u2234 Width of the lawn = 7 cm
\nAlso 2\u03c0R + 2\u03c0r = 88 + 44
\n2\u03c0 (R + r) = 132
\n\u03c0 (R + r) = \\(\\frac { 132 }{ 2 } \\) = 66 cm
\nArea of the path = \u03c0R2<\/sup> – \u03c0r2<\/sup> sq. units
\n= \u03c0 (R + r) (R – r) = 66 \u00d7 7
\nArea of the path = 462cm2<\/sup><\/p>\n

Question 7.
\nA cow is tethered with a rope of length 35 m at the centre of the rectangular field of length 76 m and breadth 60 m. Find the area of the land that the cow cannot graze?
\nSolution:
\nLength of the field l = 76 m
\nBreadth of the field b = 60m
\nArea of the field A = l \u00d7 b sq. units = 76 \u00d7 60 m2<\/sup>
\nArea of the field A = 4560 m2<\/sup>
\nLength of the rope = 35m
\nRadius of the land that the cow can graze = 35m
\nArea of the land tha the cow can graze = circle of radius 35 m = \u03c0r2<\/sup> sq.units
\n\u03c0 \u00d7 35 \u00d7 35 m2<\/sup> = \\(\\frac { 22 }{ 7 } \\) \u00d7 35 \u00d7 35 m2<\/sup>
\n= 3850 m2<\/sup>
\nArea of the land the cow cannot graze = Area of the field – Area that the cow can graze
\n= 4560 – 3860 m2<\/sup> = 710 m2<\/sup>
\nArea of the land that the cow cannot graze = 710 m2<\/sup><\/p>\n

\"Samacheer<\/p>\n

Question 8.
\nA path 5 m wide runs along the inside of the rectangular field. The length of the rectangular field is three times the breadth.of the field. If the area of the path is 500 m2<\/sup> then find the length and breadth of the field.
\nSolution:
\nLet the length of the rectangular field = ‘L’ m
\nBreadth of the rectangular field = = \u2018B\u2019 m
\nArea of the rectangular field = (L \u00d7 B) m2<\/sup>
\nAlso given length = 3 \u00d7 Breadth
\nL = 3B
\nWidth of the path (W) = 5m
\nLenth of the inner rectangle = L – 2W = l – 2(5)
\n= 3B – 10m
\nBreadth of the inner rectangle = B – 2W
\n= B – 2(5)
\n= B – 10 m
\nArea of the inner rectangle = (3B – 10) (B – 10)
\n= 3B2<\/sup> – 10B – 30B + 100
\nArea of the path = Area of outer rectangle
\n– Area of inner rectangle
\n= (L \u00d7 B) – (3B2<\/sup> – 10B – 30B + 100)
\n3B \u00d7 B – (3B2<\/sup> – 40B + 100)
\n= 3B2<\/sup> – 3B2<\/sup> + 40B – 100
\nArea of the path = 40B – 100
\nGiven area of the path = 500 m2<\/sup>
\n40B – 100 = 500
\n40B = 500 + 100 = 600
\nB = \\(\\frac { 600 }{ 40 } \\)
\nB = 15m
\nLength of the field = 45 m; Breadth of the field = 15 m<\/p>\n

Question 9.
\nA circular path has to be constructed around a circular ground. 1f the areas of the outer and inner circles are 1386 m2 and 616 m2 respectively, find the width and area of the path.
\nSolution:
\nArea of the outer circle = 1386 m2<\/sup>
\n\u03c0R2<\/sup> = 1386m2<\/sup>
\nArea of the inner circle = 616 m2<\/sup>
\n\u03c0r2<\/sup> = 616m2<\/sup>
\nArea of the path = Area of outer circle – Area of the inner circle
\n1386 m2<\/sup> – 616 m2<\/sup>
\nArea of the path = 770m2<\/sup>
\nAlso \u03c0R2<\/sup> = 1386
\nR2<\/sup> = \\(\\frac{1386 \\times 7}{22}\\)
\nR2<\/sup> = 63 \u00d7 7
\nR2<\/sup> = 9 \u00d7 7 \u00d7 7
\nR2<\/sup> = 32 \u00d7 72
\nR = 3 \u00d7 7
\nOuter Radius R = 21 m
\nAgain \u03c0r2<\/sup> = 616
\n\\(\\frac { 22 }{ 7 } \\) \u00d7 r2<\/sup> = 616
\nr2<\/sup> = 28 \u00d7 7
\nr2<\/sup> = 4 \u00d7 7 \u00d7 7
\nr2<\/sup> = 22 \u00d7 72
\nr = 2 \u00d7 7
\nInner radius r = 14m
\nWidth of the path = Outer radius – Inner radius = 21 – 14
\nWidth of the path = 7m<\/p>\n

Question 10.
\nA goat is tethered with a rope of length 45 m at the centre of the circular grass land whose radius is 52 m. Find the area of the grass land that the goat cannot graze.
\nSolution:
\nLength of the rope = 45 m = Radius of the inner circle
\n\u2234 Area of the circular area that the goat graze = \u03c0r2<\/sup> sq. units
\n= \\(\\frac { 22 }{ 7 } \\) \u00d7 45 \u00d7 45 m2<\/sup> = 6364.28 m2<\/sup>
\nRadius of the gross land = 52 m
\nArea of the grass land = \\(\\frac { 22 }{ 7 } \\) \u00d7 52 \u00d7 52 = 8,498.28 m2<\/sup>
\nArea that the goat cannot graze
\n= Area of the outer circle – Area of the inner circle
\n= 8498.28 – 6364.28 = 2134 m2<\/sup>
\nArea of the goat cannot grass = 2134 m2<\/sup><\/p>\n

\"Samacheer<\/p>\n

Question 11.
\nA strip of 4 cm wide is cut and removed from all the sides of the rectangular cardboard with dimensions 30 cm \u00d7 20 cm. Find the area of the removed portion and area of the remaining cardboard.
\nSolution:
\nArea of the outer rectangular cardboard
\n= L \u00d7 B sq.units = 30 \u00d7 20 cm2<\/sup> = 600 cm2<\/sup>
\nWidth of the stip = 4 cm
\nLength of the inner rectangle = L – 2W
\nl = 30 – 2(4) = 30 – 8
\nl = 22cm
\nBreadth of the inner rectangle B = 2W = 20 – 2(4) = 20 – 8
\nb = 12cm
\nArea of the inner rectangle = l \u00d7 b sq.units = 22 \u00d7 12 cm2<\/sup> = 264 cm2<\/sup>
\nArea of the remaining cardboard = 264 cm2<\/sup>
\nArea of the removed portion = Area of outer rectangle
\n– Area of the inner rectangle
\n= 600 – 264 cm2<\/sup>
\nArea of the removed portion = 336 cm2<\/sup><\/p>\n

Question 12.
\nA rectangular field is of dimension 20 m \u00d7 15 m. Two paths run parallel to the sides of the rectangle through the centre of the field. The width of the longer path is 2m and that of the shorter path is 1 m. Find (i) the area of the paths (ii) the area of the remaining portion of the field (iii) the cost of constructing the roads at the rate of \u20b9 10 per sq.m.
\nSolution:
\nLength of the rectangular field L = 20 m
\nBreadth B = 15m
\nArea = L \u00d7 B
\n20 \u00d7 15 m2<\/sup>
\nArea of outer rectangle = 300 m2<\/sup>
\n\"Samacheer
\nArea of inner small rectangle = \\(\\frac { 19 }{ 2 } \\) \u00d7 \\(\\frac { 13 }{ 2 } \\) = 61.75 cm2<\/sup><\/p>\n

(i) Area of the path = Area of the outer rectangle
\n– Area of 4 inner small rectangles
\n= 300 – 4(61.75) = 300 – 247 = 53 m2<\/sup>
\nArea of the paths = 53 m2<\/sup><\/p>\n

(ii) Area of the remaining portion of the field
\n= Area of the outer rectangle – Area of the paths
\n= 300 – 53 m2<\/sup> = 247 m2<\/sup>
\nArea of the remaining portion = 247 m2<\/sup><\/p>\n

(iii) Cost of constructing 1 m2<\/sup> road = \u20b910
\n\u2234 Cost of constructing 53 m2<\/sup> road = \u20b910 \u00d7 53 = \u20b9530
\n\u2234 Cost of constructing road = \u20b9530<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 2 Measurements Ex 2.4 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 Ex 2.4 Miscellaneous Practice …<\/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\/3080"}],"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=3080"}],"version-history":[{"count":1,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/3080\/revisions"}],"predecessor-version":[{"id":60033,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/3080\/revisions\/60033"}],"wp:attachment":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/media?parent=3080"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/categories?post=3080"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/tags?post=3080"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}