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>\nNCERT 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
<\/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 <\/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 <\/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>\nQuestion 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 \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 \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 <\/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 \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 <\/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>\nSince, 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>\nQuestion 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 \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>\nQuestion 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 \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 <\/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>\nQuestion 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 <\/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 <\/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>\nQuestion 15. \nFind the circumference of the inner and the outer circles, shown in the adjoining figure? (Take \u03c0 = 3.14) \n \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
<\/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
NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[9],"tags":[],"yoast_head":"\nNCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3 - MCQ Questions<\/title>\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n