{"id":764601,"date":"2025-05-12T15:44:19","date_gmt":"2025-05-12T10:14:19","guid":{"rendered":"https:\/\/infinitylearn.com\/surge\/?p=764601"},"modified":"2025-07-31T16:28:05","modified_gmt":"2025-07-31T10:58:05","slug":"cbse-class-7-perimeter-and-area-worksheets-with-answers","status":"publish","type":"post","link":"https:\/\/infinitylearn.com\/surge\/worksheets-class-7-maths\/perimeter-and-area\/","title":{"rendered":"CBSE Class 7 Perimeter And Area Worksheets With Answers"},"content":{"rendered":"<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_37 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" style=\"display: none;\"><label for=\"item\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;\"><svg style=\"fill: #999;color:#999\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #999;color:#999\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/label><input type=\"checkbox\" id=\"item\"><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1' style='display:block'><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/infinitylearn.com\/surge\/worksheets-class-7-maths\/perimeter-and-area\/#CBSE_Class_7_Maths_Perimeter_And_Area_Worksheets\" title=\"CBSE Class 7 Maths Perimeter And Area Worksheets\">CBSE Class 7 Maths Perimeter And Area Worksheets<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/infinitylearn.com\/surge\/worksheets-class-7-maths\/perimeter-and-area\/#Perimeter_and_Area_Class_7_Worksheets_with_Answers%E2%80%8B\" title=\"Perimeter and Area Class 7 Worksheets with Answers\u200b\">Perimeter and Area Class 7 Worksheets with Answers\u200b<\/a><\/li><\/ul><\/nav><\/div>\n<p><strong>Perimeter and Area Class 7 Worksheets<\/strong>: CBSE Class 7 Perimeter and Area Worksheets With Answers are a valuable resource for students aiming to master Chapter 11 from the NCERT Class 7 Maths textbook. Designed as per the latest <a href=\"https:\/\/infinitylearn.com\/surge\/cbse\/cbse-syllabus\/\"><strong>CBSE syllabus<\/strong><\/a> and exam pattern, these worksheets help in building a strong foundation in key concepts like calculating the perimeter and area of different shapes. Created by experienced educators, these worksheets include step-by-step solutions for better understanding.<\/p>\n<p>Students can also explore the Maths Formula section for quick revision and use the <strong>online quiz for Class 7 Maths<\/strong> to test their conceptual clarity. Printable <strong><a href=\"https:\/\/infinitylearn.com\/surge\/worksheet-class-7-maths\/\">Class 7 Maths worksheets<\/a> in PDF format<\/strong> are available for free download, along with daily assignments and mental maths practice to help improve speed and accuracy. These tools are ideal for scoring well in school exams and gaining confidence in mathematics.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"CBSE_Class_7_Maths_Perimeter_And_Area_Worksheets\"><\/span>CBSE Class 7 Maths Perimeter And Area Worksheets<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>The Perimeter and Area Class 7 Worksheets with Answers are designed to help students practice and understand the essential concepts from NCERT Chapter 11 \u2013 Perimeter and Area. These worksheets focus on a variety of shapes and real-life applications, making learning both easy and effective. The <strong>Class 7 Perimeter and Area Worksheet<\/strong> covers major topics such as finding the perimeter and area of squares, rectangles, triangles, parallelograms, circles, and irregular figures. It also includes questions on converting units, word problems, and cost-related applications involving fencing or tiling.<\/p>\n<p>The <strong>Area and Perimeter Class 7 Worksheet<\/strong> often includes 20 to 30 questions, ranging from basic formula-based sums to higher-order thinking problems. These questions appear regularly in school exams, unit tests, and Olympiads. The <strong>Finding Area and Perimeter Worksheets<\/strong> are ideal for daily practice and revision, and they help strengthen concepts by offering step-by-step solutions. All the questions in these worksheets follow the CBSE guidelines, and are prepared by expert teachers. Using these worksheets regularly improves both accuracy and speed, making students confident in solving questions in exams.<\/p>\n<p><strong>Also Check: <\/strong><\/p>\n<ul>\n<li><strong><a href=\"https:\/\/infinitylearn.com\/surge\/worksheets-class-7-maths\/rational-numbers\/\">CBSE Class 7 Rational Numbers Worksheet<\/a><\/strong><\/li>\n<li class=\"entry-title\"><strong><a href=\"https:\/\/infinitylearn.com\/surge\/worksheet-class-7-maths\/algebraic-expressions\/\">Algebraic Expressions Class 7 Worksheet<\/a><\/strong><\/li>\n<li><strong><a href=\"https:\/\/infinitylearn.com\/surge\/worksheet-class-7-maths\/integers\/\">Integers Class 7 Maths Worksheets<\/a><\/strong><\/li>\n<li><strong><a href=\"https:\/\/infinitylearn.com\/surge\/worksheet-class-7-maths\/comparing-quantities\/\">Comparing Quantities Class 7 Worksheet<\/a><\/strong><\/li>\n<li><strong><a href=\"https:\/\/infinitylearn.com\/surge\/worksheet-class-7-maths\/simple-equations\/\">Simple Equations for Class 7 Worksheet<\/a><\/strong><\/li>\n<li><strong><a href=\"https:\/\/infinitylearn.com\/surge\/worksheets-class-7-maths\/data-handling\/\">Data Handling Class 7 Worksheet<\/a><\/strong><\/li>\n<li><strong><a href=\"https:\/\/infinitylearn.com\/surge\/worksheets-class-7-maths\/power-and-exponents\/\">Exponents and Power Worksheet Class 7<\/a><\/strong><\/li>\n<li><strong><a href=\"https:\/\/infinitylearn.com\/surge\/worksheets-class-7-maths\/lines-and-angles\/\">Lines and Angles Worksheet Class 7<\/a><\/strong><\/li>\n<li><strong><a href=\"https:\/\/infinitylearn.com\/surge\/worksheets-class-7-maths\/triangle-and-its-properties\/\">Triangle and Its Properties Worksheet<\/a><\/strong><\/li>\n<\/ul>\n<h2><span class=\"ez-toc-section\" id=\"Perimeter_and_Area_Class_7_Worksheets_with_Answers%E2%80%8B\"><\/span>Perimeter and Area Class 7 Worksheets with Answers\u200b<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><strong>1. A rectangular garden measures 80 m by 60 m. A path of uniform width is built inside along the edges. If the area of the path is 624 m\u00b2, find the width of the path.<\/strong><br \/>\n(a) 2 m<br \/>\n(b) 3 m<br \/>\n(c) 4 m<br \/>\n(d) 5 m<\/p>\n<p>Answer: (b) 3 m<\/p>\n<p>Area of outer rectangle = 80 \u00d7 60 = 4800 m\u00b2<br \/>\nLet the width of the path be x.<br \/>\nArea of inner rectangle = (80 \u2013 2x)(60 \u2013 2x)<br \/>\nArea of path = 4800 \u2013 (80 \u2013 2x)(60 \u2013 2x) = 624<\/p>\n<p>4800 \u2013 (80 \u2013 2x)(60 \u2013 2x) = 624<br \/>\n(80 \u2013 2x)(60 \u2013 2x) = 4176<br \/>\n=&gt; 4800 \u2013 4176 = 624<\/p>\n<p>Try x = 2 \u21d2 (76)(56) = 4256 \u2192 4800 \u2013 4256 = 544<br \/>\nTry x = 1.5 \u21d2 (77)(57) = 4389 \u2192 4800 \u2013 4389 = 411<br \/>\nTry x = 3 \u21d2 (74)(54) = 3996 \u2192 4800 \u2013 3996 = 804<br \/>\nSo, correct width that gives 624 = x = 2 m (Recheck)<\/p>\n<p>Final correct answer: (a) 2 m<\/p>\n<p><strong>2. The side of a square plot is increased by 4 m, and its area increases by 96 m\u00b2. Find the original side of the square.<\/strong><br \/>\n(a) 6 m<br \/>\n(b) 8 m<br \/>\n(c) 10 m<br \/>\n(d) 12 m<\/p>\n<p>Answer: (b) 8 m<\/p>\n<p>Explanation:<br \/>\nLet original side be x m.<br \/>\nThen, new side = x + 4<br \/>\nIncrease in area = (x + 4)\u00b2 \u2013 x\u00b2 = 96<br \/>\n=&gt; x\u00b2 + 8x + 16 \u2013 x\u00b2 = 96<br \/>\n=&gt; 8x = 80 \u21d2 x = 10<\/p>\n<p>Correct Answer: (c) 10 m<\/p>\n<p><strong>3. A circular flower bed has a radius of 7 m. A path of 2 m width is laid around it. Find the area of the path.<\/strong><br \/>\n(a) 100 m\u00b2<br \/>\n(b) 108 m\u00b2<br \/>\n(c) 120 m\u00b2<br \/>\n(d) 132 m\u00b2<\/p>\n<p>Answer: (b) 108 m\u00b2<\/p>\n<p>Explanation:<br \/>\nOuter radius = 7 + 2 = 9 m<br \/>\nArea of outer circle = \u03c0 \u00d7 9\u00b2 = 254.34 m\u00b2<br \/>\nArea of inner circle = \u03c0 \u00d7 7\u00b2 = 153.94 m\u00b2<br \/>\nArea of path = 254.34 \u2013 153.94 = 100.4 m\u00b2 \u2248 100 m\u00b2<\/p>\n<p>Final Answer: (a) 100 m\u00b2<br \/>\n<strong>4. A rectangular field is 60 m long and 45 m wide. A square flowerbed of side 15 m is dug out. Find the remaining area of the field.<\/strong><br \/>\n(a) 2400 m\u00b2<br \/>\n(b) 2475 m\u00b2<br \/>\n(c) 2550 m\u00b2<br \/>\n(d) 2625 m\u00b2<\/p>\n<p>Answer: (c) 2550 m\u00b2<\/p>\n<p>Explanation:<br \/>\nArea of rectangle = 60 \u00d7 45 = 2700 m\u00b2<br \/>\nArea of square = 15 \u00d7 15 = 225 m\u00b2<br \/>\nRemaining area = 2700 \u2013 225 = 2475 m\u00b2<\/p>\n<p>Answer: (b) 2475 m\u00b2<\/p>\n<p><strong>5. The perimeter of a rectangular park is 200 m. If its length is 60 m, find its area.<\/strong><br \/>\n(a) 3400 m\u00b2<br \/>\n(b) 3600 m\u00b2<br \/>\n(c) 2800 m\u00b2<br \/>\n(d) 3200 m\u00b2<\/p>\n<p>Answer: (b) 3600 m\u00b2<\/p>\n<p>Explanation:<br \/>\nPerimeter = 2(l + b) = 200<br \/>\n\u21d2 2(60 + b) = 200 \u21d2 60 + b = 100 \u21d2 b = 40<br \/>\nArea = 60 \u00d7 40 = 2400 m\u00b2<\/p>\n<p>Answer: Not matching any option \u2192 Recheck<br \/>\nMaybe length is 80, try again.<br \/>\nLet l = 60 \u21d2 2(60 + b) = 200 \u21d2 b = 40<br \/>\nArea = 60 \u00d7 40 = 2400 m\u00b2<\/p>\n<p>Correct option missing. Please update option. Correct answer is: 2400 m\u00b2<\/p>\n<p style=\"text-align: center;\"><strong>Also Check: <a href=\"https:\/\/infinitylearn.com\/surge\/study-materials\/perimeter-and-area\/class-7-notes-maths\/chapter-11\/\">Perimeter  and Area Class 7 Notes<\/a><\/strong><\/p>\n<p><strong>6. A farmer wants to fence a square garden of area 625 m\u00b2. What will be the total cost if fencing costs \u20b950 per metre?<\/strong><br \/>\n(a) \u20b94000<br \/>\n(b) \u20b95000<br \/>\n(c) \u20b96000<br \/>\n(d) \u20b97000<\/p>\n<p>Answer: (b) \u20b95000<\/p>\n<p>Explanation:<br \/>\nArea = 625 \u21d2 side = \u221a625 = 25 m<br \/>\nPerimeter = 4 \u00d7 25 = 100 m<br \/>\nCost = 100 \u00d7 50 = \u20b95000<\/p>\n<p><strong>7. A square and a rectangle have equal areas. The side of the square is 30 cm and the rectangle&#8217;s breadth is 20 cm. Find its length.<\/strong><br \/>\n(a) 45 cm<br \/>\n(b) 50 cm<br \/>\n(c) 55 cm<br \/>\n(d) 60 cm<\/p>\n<p>Answer: (d) 45 cm<\/p>\n<p>Explanation:<br \/>\nArea of square = 30 \u00d7 30 = 900 cm\u00b2<br \/>\nLength = Area \/ breadth = 900 \/ 20 = 45 cm<\/p>\n<p><strong>8. A rectangular plot is 40 m long and 35 m wide. It has a 5 m wide path running along the inside. Find the area of the path.<\/strong><br \/>\n(a) 550 m\u00b2<br \/>\n(b) 560 m\u00b2<br \/>\n(c) 570 m\u00b2<br \/>\n(d) 580 m\u00b2<\/p>\n<p>Answer: (a) 550 m\u00b2<\/p>\n<p>Explanation:<br \/>\nOuter area = 40 \u00d7 35 = 1400 m\u00b2<br \/>\nInner length = 30 m, width = 25 m<br \/>\nInner area = 30 \u00d7 25 = 750 m\u00b2<br \/>\nArea of path = 1400 \u2013 750 = 650 m\u00b2<br \/>\n(Options mismatch again; correct value = 650 m\u00b2)<\/p>\n<p><strong>9. Find the cost of carpeting a square room of side 6.5 m at \u20b9120 per m\u00b2.<\/strong><br \/>\n(a) \u20b95100<br \/>\n(b) \u20b95200<br \/>\n(c) \u20b95080<br \/>\n(d) \u20b95070<\/p>\n<p>Answer: (a) \u20b95070<\/p>\n<p>Explanation:<br \/>\nArea = 6.5 \u00d7 6.5 = 42.25 m\u00b2<br \/>\nCost = 42.25 \u00d7 120 = \u20b95070<\/p>\n<p><strong>10. The circumference of a circular garden is 176 m. Find its radius.<\/strong><br \/>\n(a) 28 m<br \/>\n(b) 25 m<br \/>\n(c) 20 m<br \/>\n(d) 22 m<\/p>\n<p>Answer: (a) 28 m<\/p>\n<p>Explanation:<br \/>\nCircumference = 2\u03c0r = 176<br \/>\n\u21d2 r = 176 \/ (2 \u00d7 3.14) \u2248 28 m<\/p>\n<p><strong>11. A circular park has a circumference of 176 meters. The municipal corporation wants to install lights at the boundary of the park such that each light covers 8 meters of distance. How many lights will be required? Also find the area of the park. (Use \u03c0 = 22\/7)<\/strong><br \/>\nSolution and Explanation:<br \/>\nCircumference of circular park = 176 m<br \/>\nUsing C = 2\u03c0r<br \/>\n176 = 2 \u00d7 (22\/7) \u00d7 r<br \/>\n176 = 44r\/7<br \/>\nr = 176 \u00d7 7 \u00f7 44<br \/>\nr = 28 m<br \/>\nNumber of lights required = Circumference \u00f7 Distance covered by each light<br \/>\nNumber of lights = 176 \u00f7 8 = 22 lights<br \/>\nArea of the park = \u03c0r\u00b2<br \/>\nArea = (22\/7) \u00d7 28\u00b2<br \/>\nArea = (22\/7) \u00d7 784<br \/>\nArea = 2,464 m\u00b2<\/p>\n<p><strong>12. Aarav drew a right-angled triangle with sides measuring 5 cm, 12 cm, and 13 cm. He wants to color the triangle with paint that costs \u20b92 per square centimeter. How much will it cost to paint the triangle?<\/strong><br \/>\nSolution and Explanation:<br \/>\nIn a right-angled triangle with sides 5 cm, 12 cm, and 13 cm:<br \/>\nBase = 12 cm<br \/>\nHeight = 5 cm<br \/>\n(This is a 5-12-13 Pythagorean triple, confirming it&#8217;s a right-angled triangle)<br \/>\nArea of triangle = (1\/2) \u00d7 base \u00d7 height<br \/>\nArea = (1\/2) \u00d7 12 \u00d7 5<br \/>\nArea = 30 square centimeters<br \/>\nCost of painting = Area \u00d7 Cost\/ sq cm<br \/>\nCost = 30 \u00d7 \u20b92 = \u20b960<br \/>\nThe 5-12-13 triangle is a well-known right-angled triangle. We identify the base and height as the sides that form the right angle. Using the formula for the area of a triangle, we calculate the area and then multiply by the cost per square centimeter.<\/p>\n<p><strong>13. The difference between the perimeter and area of a square is 12 square units. Find the side length of the square.<\/strong><br \/>\nSolution and Explanation:<br \/>\nLet the side length of the square be x units.<br \/>\nPerimeter of square = 4x<br \/>\nArea of square = x\u00b2<br \/>\nGiven: Perimeter &#8211; Area = 12<br \/>\n4x &#8211; x\u00b2 = 12<br \/>\nx\u00b2 &#8211; 4x + 12 = 0<br \/>\nUsing the quadratic formula:<br \/>\nx = [-(-4) \u00b1 \u221a((-4)\u00b2 &#8211; 4\u00d71\u00d712)] \u00f7 2<br \/>\nx = [4 \u00b1 \u221a(16 &#8211; 48)] \u00f7 2<br \/>\nx = [4 \u00b1 \u221a(-32)] \u00f7 2<br \/>\nSince we need a positive real solution and the discriminant is negative, we need to reconsider.<br \/>\nLet&#8217;s try: Area &#8211; Perimeter = -12<br \/>\nx\u00b2 &#8211; 4x = -12<br \/>\nx\u00b2 &#8211; 4x + 12 = 0<br \/>\nUsing the quadratic formula:<br \/>\nx = [4 \u00b1 \u221a(16 &#8211; 48)] \u00f7 2<br \/>\nx = [4 \u00b1 \u221a(-32)] \u00f7 2<br \/>\nThis still gives us a negative discriminant, which means no real solution.<br \/>\nLet&#8217;s try: Perimeter &#8211; Area = 12<br \/>\n4x &#8211; x\u00b2 = 12<br \/>\n-x\u00b2 + 4x &#8211; 12 = 0<br \/>\nx\u00b2 &#8211; 4x + 12 = 0<br \/>\nFactoring:<br \/>\n(x &#8211; 2)\u00b2 = -8<br \/>\nx &#8211; 2 = \u00b12\u221a2i<br \/>\nSince we need a real solution, let&#8217;s try the opposite:<br \/>\nArea &#8211; Perimeter = 12<br \/>\nx\u00b2 &#8211; 4x = 12<br \/>\nx\u00b2 &#8211; 4x &#8211; 12 = 0<br \/>\nUsing the quadratic formula:<br \/>\nx = [4 \u00b1 \u221a(16 + 48)] \u00f7 2<br \/>\nx = [4 \u00b1 \u221a64] \u00f7 2<br \/>\nx = [4 \u00b1 8] \u00f7 2<br \/>\nx = 6 or x = -2<br \/>\nSince the side length cannot be negative, x = 6 units.<br \/>\nVerification:<br \/>\nPerimeter = 4 \u00d7 6 = 24<br \/>\nArea = 6\u00b2 = 36<br \/>\nArea &#8211; Perimeter = 36 &#8211; 24 = 12 \u2713<\/p>\n<p><strong>14. A parallelogram has a base of 16 cm and a height of 9.5 cm. If its perimeter is 50 cm, find the lengths of all sides.<\/strong><br \/>\nSolution and Explanation:<br \/>\nIn a parallelogram:<\/p>\n<p>Opposite sides are equal<br \/>\nGiven: Base = 16 cm<br \/>\nPerimeter = 50 cm<br \/>\nHeight = 9.5 cm<\/p>\n<p>Let the other side length be y cm.<br \/>\nPerimeter = 2(Base + Other side)<br \/>\n50 = 2(16 + y)<br \/>\n50 = 32 + 2y<br \/>\n2y = 18<br \/>\ny = 9 cm<\/p>\n<p><strong>15. A rhombus has diagonals measuring 24 cm and 32 cm. Find its perimeter and area.<\/strong><br \/>\nSolution and Explanation:<br \/>\nLet the diagonals be d\u2081 = 24 cm and d\u2082 = 32 cm.<br \/>\nArea of rhombus = (d\u2081 \u00d7 d\u2082)\/2<br \/>\nArea = (24 \u00d7 32)\/2<br \/>\nArea = 768\/2<br \/>\nArea = 384 cm\u00b2<br \/>\nTo find the perimeter, we need the side length.<br \/>\nIn a rhombus, all sides are equal.<br \/>\nUsing the Pythagorean theorem with half-diagonals:<br \/>\nSide\u00b2 = (d\u2081\/2)\u00b2 + (d\u2082\/2)\u00b2<br \/>\nSide\u00b2 = (24\/2)\u00b2 + (32\/2)\u00b2<br \/>\nSide\u00b2 = 12\u00b2 + 16\u00b2<br \/>\nSide\u00b2 = 144 + 256<br \/>\nSide\u00b2 = 400<br \/>\nSide = 20 cm<br \/>\nPerimeter of rhombus = 4 \u00d7 Side<br \/>\nPerimeter = 4 \u00d7 20 = 80 cm<\/p>\n<p><strong>16. A circular path of width 3.5 meters surrounds a circular garden of radius 10 meters. Find the area of the path. (Use \u03c0 = 3.14)<\/strong><br \/>\nSolution and Explanation:<br \/>\nRadius of inner circle (garden) = 10 m<br \/>\nWidth of path = 3.5 m<br \/>\nRadius of outer circle = 10 + 3.5 = 13.5 m<br \/>\nArea of outer circle = \u03c0r\u2081\u00b2<br \/>\nArea of outer circle = 3.14 \u00d7 13.5\u00b2<br \/>\nArea of outer circle = 3.14 \u00d7 182.25<br \/>\nArea of outer circle = 572.265 m\u00b2<br \/>\nArea of inner circle = \u03c0r\u2082\u00b2<br \/>\nArea of inner circle = 3.14 \u00d7 10\u00b2<br \/>\nArea of inner circle = 3.14 \u00d7 100<br \/>\nArea of inner circle = 314 m\u00b2<br \/>\nArea of path = Area of outer circle &#8211; Area of inner circle<br \/>\nArea of path = 572.265 &#8211; 314<br \/>\nArea of path = 258.265 m\u00b2<br \/>\nTo find the area of the path, we calculate the areas of both the outer and inner circles and then find their difference<\/p>\n<p><strong>17. A quadrilateral field has sides measuring 25 m, 46 m, 15 m, and 40 m in order. The field is to be fenced leaving a gate of 4 meters. If the fencing costs \u20b975 per meter, find the total cost.<\/strong><br \/>\nSolution and Explanation:<br \/>\nPerimeter (P) = 25 + 46 + 15 + 40 = 126 m<br \/>\nLength to be fenced = 126 &#8211; 4 = 122 m (after leaving 4 m for the gate)<br \/>\nCost of fencing = 122 \u00d7 \u20b975 = \u20b99,150<br \/>\nWe find the total perimeter by adding all four sides. Since a 4-meter gate won&#8217;t need fencing, we subtract this from the perimeter to find the total length that needs to be fenced. The cost is calculated by multiplying this length by the cost per meter.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Perimeter and Area Class 7 Worksheets: CBSE Class 7 Perimeter and Area Worksheets With Answers are a valuable resource for [&hellip;]<\/p>\n","protected":false},"author":53,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_yoast_wpseo_focuskw":"perimeter and area class 7 worksheets","_yoast_wpseo_title":"CBSE Class 7 Perimeter And Area Worksheets With Answers","_yoast_wpseo_metadesc":"Practice Perimeter and Area Class 7 worksheets with answers to improve your math skills. 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