{"id":27221,"date":"2021-06-29T19:01:53","date_gmt":"2021-06-29T13:31:53","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=27221"},"modified":"2021-06-29T19:02:28","modified_gmt":"2021-06-29T13:32:28","slug":"ncert-solutions-for-class-7-maths-chapter-11-ex-11-3","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-7-maths-chapter-11-ex-11-3\/","title":{"rendered":"NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3"},"content":{"rendered":"

These NCERT Solutions for Class 7 Maths<\/a> Chapter 11 Perimeter and Area Ex 11.3 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Exercise 11.3<\/h2>\n

Question 1.
\nFind the circumference of the circles with the following radius: (Take \u03c0 = \\(\\frac{22}{7}\\))
\n(a) 14 cm
\n(b) 28 mm
\n(c) 21 cm
\nAnswer:
\n(a) Radius of the circle (r) = 14 cm
\nCircumference of the circle = 2\u03c0r units
\n= 2 \u00d7 \\(\\frac{22}{7}\\) x 14 cm
\n= 2 \u00d7 22 \u00d7 2 cm
\n= 88 cm<\/p>\n

\"NCERT<\/p>\n

(b) Radius of the circle = 28 mm.
\n.’. Circumference of the circle = 2\u03c0r units
\n= 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 28 mm
\n= 2 \u00d7 22 \u00d7 4 mm
\n= 176 mm<\/p>\n

(c) Radius of the circle = 21 cm
\nCircumference of the circle = 2\u03c0r 22
\n= 2 \u00d7 \\(\\frac{22}{7}\\) x 21
\n= 2 \u00d7 22 \u00d7 3 cm
\n= 132 cm<\/p>\n

Question 2.
\nFind the area of the following circles, given that:
\n(a) Radius = 14 mm (\u03c0 = \\(\\frac{22}{7}\\))
\n(b) Diameter = 49 m
\n(c) Radius = 5 cm
\nAnswer:
\n(a) Radius of the circle = 14 mm
\nArea of the circle = \u03c0r2<\/sup> sq.m
\n= \\(\\frac{22}{7}\\) \u00d7 142<\/sup>
\n= \\(\\frac{22}{7}\\) \u00d7 196
\n= 22 \u00d7 28
\n= 616 mm2<\/sup><\/p>\n

(b) Diameter of the circle = 49 m.
\nRadius of the circle = \\(\\frac{49}{2}\\) m
\n= 24.5 m
\nArea of the circle = \u03c0r2<\/sup>
\n= \\(\\frac{22}{7}\\) \u00d7 (24.5)2<\/sup>
\n= \\(\\frac{22}{7}\\) \u00d7 600.25
\n= 22 \u00d7 85.75
\n= 1886.5 m2<\/sup>
\nArea of the circle = \u03c0r2<\/sup> sq units<\/p>\n

(c) Radius of the circle = 5 cm
\nArea of the circle = \u03c0r2<\/sup> sq.unit
\n= \\(\\frac{22}{7}\\) \u00d7 5 \u00d7 5 cm2<\/sup>
\n= \\(\\frac{22 \\times 25}{7}\\) cm2<\/sup>
\n= \\(\\frac{550}{7}\\) cm2<\/sup> or 78.57 cm2<\/sup>.<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nIf the circumference of a circular sheet is 154 m, find its radius. Also find the area of the sheet. (Take \u03c0 = \\(\\frac{22}{7}\\))
\nAnswer:
\nCircumference of a circle = 154 m
\n2\u03c0r = 154
\n2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 r = 154
\n\"NCERT<\/p>\n

Area of the circle = \u03c0r2<\/sup> sq.m
\n= \\(\\frac{22}{7}\\) \u00d7 (24.5)2<\/sup>
\n= \\(\\frac{22}{7}\\) \u00d7 600.25
\n= 22 \u00d7 85.75 m2<\/sup>
\n= 1886.5 m2<\/sup><\/p>\n

Question 4.
\nA gardener wants to fence a circular garden of diameter 21 m. 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 cost \u20b9 4 per meter. (Take \u03c0 = \\(\\frac{22}{7}\\))
\n\"NCERT
\nAnswer:
\nDiameter of the circular garden = 21 m
\nradius of the circular garden = \\(\\frac{21}{2}\\)
\nCircumference of the garden = 2\u03c0r units
\n= 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 \\(\\frac{21}{2}\\) m = 22 x 3m = 66m
\nLength of the rope required for one round fence = 66 m
\nLength of the rope required for two round fence = 66 m \u00d7 2 m = 132 m
\nCost of rope per meter = \u20b9 4
\nTotal cost of the rope = \u20b9132 \u00d7 4 = \u20b9 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. (Take \u03c0 = 3.14)
\n\"NCERT
\nAnswer:
\nRadius of outer circle (R) = 4 cm
\nRadius of inner circle (r) = 3 cm
\n= Area of the outer circle – Area of the inner circle
\nArea of the remaining sheet = n (R2<\/sup> – r2<\/sup>)
\n= 3.14 (42<\/sup> – 32<\/sup>)
\n= 3.14 \u00d7 (16-9)
\n= 3.14 x 7 = 21.98 cm2<\/sup>
\nSo, the area of the remaining sheet is 21.98 cm2<\/sup>.<\/p>\n

\"NCERT<\/p>\n

Question 6.
\nSaima wants to put a lace on the edge of a circular table cover of diameter 1.5 m. Find the length of the lace required and also find its cost if one meter of the lace costs \u20b9 15. (Take \u03c0 = 3.14)
\nAnswer:
\nDiameter of the table cover = 1.5 m
\nRadius of the table cover = \\(\\frac { 1.5 }{ 2 }\\) m
\nCircumference of the table cover = 2\u03c0r units
\n= 2 x 3.14 x \\(\\frac { 1.5 }{ 2 }\\) = 3.14 x 1.5 m = 4.71 m
\nCost of lace for one metre = \u20b915
\nCost of lace for table cover = \u20b9 4.71 x 15
\n= \u20b9 70.71<\/p>\n

Question 7.
\nFind the perimeter of the given figure, which is a semicircle including its diameter.
\n\"NCERT
\nAnswer:
\nDiameter of the semicircle = 10 cm
\nRadius of the semicircle = \\(\\frac { 10 }{ 2 }\\) = 5 cm
\nCircumference of the semicircle = \\(\\frac { 1 }{ 2 }\\) x 2\u03c0r unit
\n= \u03c0r = \\(\\frac { 22 }{ 7 }\\) x 5
\n= 15.71 cm
\n\u2234 Perimeter of the semicircle
\n= 15.71 + 10 cm
\n= 25.71 cm<\/p>\n

Question 8.
\nFind the cost of polishing a circular table-top of diameter 1.6 m, if the rate of polishing is \u20b9 15\/m2<\/sup>. (Take \u03c0 = 3.14)
\nAnswer:
\nDiameter of the table-top = 1.6 m
\nRadius of the table-top = \\(\\frac { 1.6 }{ 2 }\\) = 0.8m
\nArea of the table-top = \u03c0r2<\/sup> sq. m
\n= 3.14 \u00d7 0.8 \u00d7 0.8 m2<\/sup> = 2.0096 m2<\/sup>
\nRate of Polishing = \u20b9 15 per m2<\/sup>
\nCost of polishing the table-top = \u20b9 2.0096 \u00d7 15 =\u20b9 30.14 (approx.)<\/p>\n

\"NCERT<\/p>\n

Question 9.
\nShazli 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?
\n(Take \u03c0 = \\(\\frac { 22 }{ 7 }\\))
\nAnswer:
\nLength of the wire = 44 cm
\nLet the radius of the circle be r.
\nCircumference of the circle = 44 cm
\n2\u03c0r = 44
\n2 \u00d7 \\(\\frac { 22 }{ 7 }\\) \u00d7 r = 44
\nr = \\(\\frac{44 \\times 7}{2 \\times 22}\\) cm =7 cm
\nArea of the circle = \u03c0r2<\/sup> sq.m
\n= \\(\\frac { 22 }{ 7 }\\) \u00d7 7 \u00d7 7 cm2<\/sup>
\n= 154 cm<\/p>\n

Since, the wire is rebent to form a square. Perimeter of the square = Length of the wire
\n4a = 44
\n[Perimeter of a square = 4a] 44
\na = \\(\\frac { 44 }{ 6 }\\) = 11 cm 6
\nSide of a square =11 cm
\nArea of the square = side \u00d7 side
\n= 11 \u00d7 11 cm2<\/sup> = 121 cm2<\/sup>
\n154 cm2<\/sup> > 121 cm2<\/sup>
\nArea of the circle > Area of the square
\n\u2234 The circle encloses greater 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 given figure). Find the area of the remaining 22
\nsheet. (Take \u03c0 = \\(\\frac { 22 }{ 7 }\\))
\n\"NCERT
\nAnswer:
\nRadius of the circular card sheet =14 cm
\nArea of the sheet = \u03c0r2<\/sup> sq m
\n= \\(\\frac { 22 }{ 7 }\\) \u00d7 14 \u00d7 14 cm2<\/sup>
\n= 22 \u00d7 2 \u00d7 14 cm2 = 616 cm2<\/sup>
\nRadius of a small circle = 3.5 cm
\nArea of 2 small circles = 2 \u00d7 \u03c0r2<\/sup> sq m
\n= 2 \u00d7 \\(\\frac { 22 }{ 7 }\\) \u00d7 3.5 \u00d7 3.5 cm2<\/sup>
\n= 2 \u00d7 22 \u00d7 0.5 \u00d7 3.5 cm = 77 cm2<\/sup>
\nLength of a small rectangle = 3 cm
\nBreadth of a small rectangle = 1 cm.
\nArea of small rectangle = l \u00d7 b sq.m
\n= 3 \u00d7 1 cm2<\/sup> = 3 cm2<\/sup>
\nArea of the remaining sheet = Area of the circular sheet – (Area of two small circles + Area of rectangle)
\n= 616 cm2<\/sup>– (77 + 3)cm2<\/sup>
\n= 616 cm2<\/sup> – 80 cm2<\/sup> = 536 cm2<\/sup>
\n\u2234 Required area of the remaining sheet = 536 cm2<\/sup><\/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 left over aluminium sheet? (Take \u03c0 = 3.14)
\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 left over aluminium sheet? (Take \u03c0 = 3.14)
\nAnswer:
\nSide of the square = 6 cm
\nArea of the square = side \u00d7 side
\n= 6 cm \u00d7 6 cm = 36 cm2<\/sup>
\n\"NCERT
\nRadius of the circle cut out from the sheet = 2 cm
\nArea of the circle = \u03c0r2<\/sup> sq.m
\n= 3.14 \u00d7 2 \u00d7 2 cm2<\/sup> = 12.56 cm2<\/sup>
\nArea of the remaining sheet
\n= 36 cm2<\/sup> – 12.56 cm2<\/sup> = 23.44 cm2<\/sup><\/p>\n

\"NCERT<\/p>\n

Question 12.
\nThe circumference of a circle is 31.4 cm. Find the radius and the area of the circle? (Take n = 3.14)
\nAnswer:
\nLet the radius of the circle be \u2018r\u2019
\nCircumference of a circle = 31.4 cm
\n2\u03c0r2<\/sup> = 31.4
\n2 \u00d7 3.14 \u00d7 r = 31.4
\nr = \\(\\frac{31.4}{2 \\times 3.14}\\)
\n= \\(\\frac{314 \\times 10}{2 \\times 314}\\)
\n= \\(\\frac{10}{2}\\)
\n= 5 cm
\nArea of the circle = 2\u03c0r2<\/sup> sq. units = 3.14 \u00d7 5 \u00d7 5 cm2<\/sup> = 78.5 cm2<\/sup><\/p>\n

Question 13.
\nA circular flower bed is surrounded by a path 4 m wide. The diameter of the flower
\nbed is 66 m. What is the area of this path? (\u03c0 = 3.14)
\nAnswer:
\nDiameter of the flower bed = 66 m
\nRadius of the flower bed = \\(\\frac { 66 }{ 2 }\\) m
\n(r) = 33 m
\n\"NCERT<\/p>\n

width of the surrounding path = 4 m
\nRadius of the outer circle (R)
\n= 33 m + 4 m
\n(R) = 37 m
\nArea of the pathway = Area of the outer circle – Area of the inner circle
\n= \u03c0R2<\/sup> – \u03c0r2<\/sup>
\n= \u03c0(R2<\/sup> – r2<\/sup>) sq. units
\n= 3.14 (372<\/sup> – 332<\/sup> ) m2<\/sup>
\n= 3.14 (37 + 33) (37 – 33) m2<\/sup>
\n= 3.14 \u00d7 70 \u00d7 4 m2<\/sup>
\n= 3.14 \u00d7 280 m2<\/sup>
\n= 879.2 m2<\/sup>
\nThus, the area of the path = 879.2 m2<\/sup><\/p>\n

\"NCERT<\/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 the sprinkler water
\nthe entire garden? (Take \u03c0 = 3.14)
\nAnswer:
\nLet the radius of the garden be ‘r’
\nArea of the circular garden = 314 m2<\/sup>
\n\u03c0r2<\/sup> =314
\n3.14 \u00d7 r2<\/sup> = 314
\nr2<\/sup> = \\(\\frac{314}{3.14}=\\frac{314 \\times 100}{314}\\)
\nr2<\/sup> = 100
\nr2<\/sup>= 102<\/sup>
\nr = 10 m
\nRadius of the area covered by the sprinkler = 12 m
\nSince 12 m > 10 m
\nThe sprinkler covers an area beyond the garden.
\nYes, the entire garden is covered by the sprinkler.<\/p>\n

Question 15.
\nFind the circumference of the inner and the outer circles, shown in the adjoining figure? (Take \u03c0 = 3.14)
\n\"NCERT
\nAnswer:
\nRadius of the outer circle (R) = 19 m.
\nCircumference of the outer circle
\n= 2\u03c0r unit
\n= 2 \u00d7 3.14 \u00d7 19 m
\n= 119.32 m
\nRadius of the inner circle (r)
\n= 19- 10 = 9m
\nCircumference of the inner circle
\n= 2\u03c0 = 2 \u00d7 3.14 \u00d7 9 m = 56.52 m<\/p>\n

Question 16.
\nHow many times a wheel of radius 28 cm must rotate to go 352 m? (Take \u03c0 = \\(\\frac { 22 }{ 7 }\\))
\nAnswer:
\nRadius of a wheel = 28 cm
\n\u2234 Circumference of the wheel = 2\u03c0r units
\n= 2 \u00d7 \\(\\frac { 22 }{ 7 }\\) \u00d7 28
\n= 2 \u00d7 22 \u00d7 4 cm = 176 cm
\nTotal distance covered
\n= 352 m
\n= 352 \u00d7 100 cm
\n= 35200 cm
\nNumber of rotations
\n= \\(\\begin{aligned}
\n&\\frac{\\text { Total distance }}{\\text { umference of the wheel }}\\\\&\\text { Circum }
\n\\end{aligned}\\)
\n= \\(\\frac{35200}{176}\\) = 200
\nThus, the distance of 352 m will be covered in 200 rotations.<\/p>\n

\"NCERT<\/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)
\nAnswer:
\nLength of the minute hand =15 cm
\nradius of the circle (r) = 15 cm
\n(made by the tip of the minute hand)
\nPerimeter of the circle = 2\u03c0r units
\n= 2 \u00d7 3.14 \u00d7 15 cm = 94.2 cm
\nDistance covered by the minute hand = 94.2 cm<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Exercise 11.3 Question 1. Find the circumference of the circles with the following radius: (Take \u03c0 = ) (a) …<\/p>\n

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