Worksheet on Area and Perimeter of Rectangle Problems will help the students to explore their knowledge of Rectangle word Problems. Solve all the Problems to learn the formula of Area of Rectangle and Perimeter of a Rectangle. To know the definition, properties, derivation, Problems with Solutions, Formulas of Rectangle you can visit our website. We have given the complete Rectangle concept along with examples. Check out the Area and Perimeter of Rectangle Problems Worksheet and know the various strategies to solve problems in an easy and understandable way.<\/p>\n
Also Read :<\/p>\n
A Rectangle is a quadrilateral with two equal sides and two parallel lines and four right angles. Four right angles vertices are equal to 90 degrees, it is also called an equiangular quadrilateral.<\/p>\n
The perimeter of the rectangle is defined as the sum of all the sides of the rectangle.\u00a0 Rectangle has two lengths and breadths, it is denoted by P, it is measured in units. For finding the perimeter of the rectangle we have to add the length and breadth.<\/p>\n
Perimeter of the Rectangle, P = 2(l + b)<\/p>\n
The area of the rectangle is defined as to calculate the length and breadth of the two- dimensional closed figure. For finding the area of the rectangle we have to multiply the length and breadth, it is denoted by A, measured in square units.<\/p>\n
Area of the rectangle , A = l x b<\/p>\n
1. Find the Area and Perimeter of the following rectangles whose dimensions are :<\/p>\n
(i) length = 15 cm\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0breadth = 12 cm<\/p>\n
(ii) length = 7.9 m\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 breadth = 6.2 m<\/p>\n
(iii) length = 4 m\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 breadth = 36 cm<\/p>\n
(iv) length = 2 m\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 breadth = 6 dm<\/p>\n (i) Given, length = 15 cm, breadth = 12 cm<\/p>\n we know that, Perimeter of rectangle = 2 (length + breadth)<\/p>\n substitute the given values in above formula, we get<\/p>\n Perimeter of rectangle = 2 (15 + 12) cm<\/p>\n = 2 \u00d7 27 cm<\/p>\n = 54 cm<\/p>\n We know that, area of rectangle = length \u00d7 breadth<\/p>\n Therefore, substituting the\u00a0 values in above formula, we get<\/p>\n Area of rectangle = 15 cm x 12 cm<\/p>\n = (15 \u00d7 12) cm\u00b2<\/p>\n = 195 cm\u00b2<\/p>\n Therefore, Area of rectangle is 195 cm\u00b2<\/p>\n (ii) Given, length = 7 m, breadth = 6.2 m<\/p>\n we know that, Perimeter of rectangle = 2 (length + breadth)<\/p>\n substitute the values in the formula , we get<\/p>\n Perimeter of rectangle = 2 (7.9 + 6.2) m<\/p>\n = 2 \u00d7 14. 1 m<\/p>\n = 28. 2 m<\/p>\n We know that, area of rectangle = length \u00d7 breadth<\/p>\n Therefore, substituting the value we get,<\/p>\n Area of rectangle = 7.9 m x 6.2 m<\/p>\n = (7.9 \u00d7 6.2) m\u00b2<\/p>\n = 48. 98m\u00b2<\/p>\n Therefore, Area of rectangle = 48. 98 m\u00b2<\/p>\n (iii) Given, length = 4 m<\/p>\n breadth = 36 cm = 36\/ 100 = 0. 36 m ( cm is converted to m)<\/p>\n we know that, Perimeter of rectangle = 2 (length + breadth)<\/p>\n substitute the values in the formula, we get<\/p>\n Perimeter of\u00a0 a rectangle\u00a0 = 2 (4 + 36) m<\/p>\n = 2 \u00d7 40 m<\/p>\n The perimeter of a rectangle is 80 m<\/p>\n We know that, area of rectangle = length \u00d7 breadth<\/p>\n substituting the value we get,<\/p>\n Area of rectangle\u00a0 = 4 m\u00d7 36 m<\/p>\n = (4 x 36) m\u00b2<\/p>\n = 144 m\u00b2<\/p>\n Therefore, Area of a rectangle is 144 m\u00b2<\/p>\n (iv) Given, length = 2 m<\/p>\n breadth = 60 dm<\/p>\n 1m = 10 dm<\/p>\n so we get, 60 dm =\u00a0 6 m ( dm is converted to m)<\/p>\n we know that, Perimeter of rectangle = 2 (length + breadth)<\/p>\n substitute the values in the formula, we get<\/p>\n Perimeter of rectangle\u00a0 = 2 (2 + 6) m<\/p>\n = 2 \u00d7 8 m<\/p>\n = 16 m<\/p>\n We know that, area of rectangle = length \u00d7 breadth<\/p>\n substituting the value we get,<\/p>\n Area of rectangle\u00a0 = 2 m\u00d7 6 m<\/p>\n = (2x 6) m\u00b2 = 12 m\u00b2<\/p>\n Therefore, the Area of a rectangle is\u00a0 12m\u00b2.<\/p>\n<\/details>\n 2. The perimeter of the rectangle is\u00a0 140 cm. If the length of the rectangle is 30 cm, find its breadth and area of the rectangle?<\/p>\n Given, Perimeter of the rectangle is, 140 cm<\/p>\n The length of the rectangle is, 30 cm<\/p>\n we know that, Perimeter of the rectangle = 2(l + b)<\/p>\n substitute the value in the above formula, we get<\/p>\n 140 = 2( 30 + b)<\/p>\n 70 = 30<\/p>\n 40 = b<\/p>\n Therefore, breadth = 40 cm<\/p>\n Now, Area of Rectangle = length x breadth<\/p>\n = 30 x 40 = 120\u00a0 cm\u00b2<\/p>\n Therefore, the Area of a rectangle is 120 cm\u00b2<\/p>\n<\/details>\n 3. The area of a rectangle is 78 cm\u00b2. If the breadth of the rectangle is 6 cm, find its length and perimeter?<\/p>\n Given, Area of a rectangle is 78 cm\u00b2<\/p>\n The breadth of the rectangle is 6 cm<\/p>\n we know that, Area of rectangle = length x breadth<\/p>\n substitute the given value, we get<\/p>\n 78 cm = length x 6 cm<\/p>\n 78\/ 6 = length<\/p>\n Length of the rectangle =12 cm<\/p>\n Now, perimeter of rectangle = 2 (l + b)<\/p>\n substitute the value, we get<\/p>\n Perimeter of rectangle = 2(12 + 6)<\/p>\n = 2 x 18<\/p>\n = 36 cm<\/p>\n Therefore, the perimeter of the rectangle = 36 cm<\/p>\n<\/details>\n 4. How many boxes whose length and breadth are 9 cm and 5 cm respectively are needed to cover a rectangular region whose length and breadth are 420 cm and 90 cm?<\/p>\n Given,\u00a0 Length of the box is 9 cm<\/p>\n The breadth of the box is 5 cm<\/p>\n Region length is 420 cm<\/p>\n Region breadth is 90 cm<\/p>\n we know the formula,<\/p>\n The area of a rectangle is l x b<\/p>\n Therefore, Area of region = l x b<\/p>\n substitute the value, we get<\/p>\n Area of region = 420 cm x 90 cm<\/p>\n = 37800 cm\u00b2<\/p>\n Again use the\u00a0 area of a rectangle formula,<\/p>\n Area of one box is = 9 cm x 5 cm<\/p>\n = 45 cm\u00b2<\/p>\n Number of boxes = Area of region \/Area of one box = 37800\/45 = 840<\/p>\n Thus, 840 boxes are required.<\/p>\n<\/details>\n 5. If it costs $500 to fence a rectangular park of length 40 m at the rate of $25 per m\u00b2, find the breadth of the park and its perimeter. Also, find the area of the field?<\/p>\n Given, Cost of Rectangular park fencing is $500<\/p>\n Length of the\u00a0 rectangular park = 20 m<\/p>\n Rate of fencing 1 m\u00b2 = $25<\/p>\n Area of a rectangle = l x b<\/p>\n Now we find\u00a0 area , therefore\u00a0 Area = 500\/ 25 = 20<\/p>\n substitute the\u00a0 value\u00a0 in formula, we get<\/p>\n 20 = 20 x breadth<\/p>\n breadth = area \/ length<\/p>\n b = 20 \/ 20 = 1 m<\/p>\n Now finding the Perimeter,<\/p>\n Perimeter of a rectangle = 2 (l + b)<\/p>\n substituting the values ,<\/p>\n Perimeter\u00a0 of a rectangle=\u00a0 2 (20 + 1)<\/p>\n =\u00a0 2 (21)<\/p>\n Therefore, the Perimeter of a rectangle =\u00a0 41 m<\/p>\n<\/details>\n 6. A rectangular tile has a length equals to 20 cm and a perimeter equals 70 cm. Find its width?<\/p>\n Given, Perimeter of the tile = 80 cm<\/p>\n Length of the tile = 20 cm<\/p>\n Let W be the width of the tile<\/p>\n we know that,<\/p>\n Perimeter of a rectangle = 2(length + width)<\/p>\n Substituting the values, we get,<\/p>\n The perimeter of a tile = 80 cm<\/p>\n Therefore, 80 = 2 (20 + Width)<\/p>\n 80\/ 2 = 20 + Width<\/p>\n 40 = 20 + Width<\/p>\n 40 – 20 = Width<\/p>\n Therefore, Width = W = 20.<\/p>\n<\/details>\n 7. Find the area of a rectangle, Perimeter of a rectangle, and diagonal of a rectangle whose length and breadth 12 cm and 16 cm respectively.<\/p>\n Given, length of the rectangle = 12 cm<\/p>\n Breadth of the rectangle = 16 cm<\/p>\n we know the formulae,<\/p>\n Area of a rectangle = l x b<\/p>\n substitute the values in the above formula, we get<\/p>\n Area of a rectangle = 12 x 16 = 192 cm\u00b2<\/p>\n we know, Perimeter of a rectangle = 2 (l + b)<\/p>\n substitute the values, we get<\/p>\n Perimeter of a rectangle = 2 (12 + 16)<\/p>\n = 2 (192) = 384 m<\/p>\n Now, we finding the diagonal of a rectangle<\/p>\n The diagonal of a rectangle is d\u00b2 = l\u00b2 + b\u00b2<\/p>\n substitute the values, we get<\/p>\n d\u00b2 = (12)\u00b2 + (16)\u00b2<\/p>\n d\u00b2 = (12 + 16)\u00b2<\/p>\n d = \u221a(12 + 16)\u00b2<\/p>\n square and root both will be cancelled,<\/p>\n d = 12 + 16 = 28<\/p>\n Therefore, the Diagonal of a rectangle = 28 cm.<\/p>\n<\/details>\n 8. Find the cost of tiling a rectangular plot of land 200 m long and 120 m wide at the rate of $6 per hundred square m?<\/p>\n Given,<\/p>\n Cost of tiling rectangular plot of land 200 m long and 120 m wide<\/p>\n The cost of tiling per 100 sq.m is $6<\/p>\n we know the area of a rectangle formula,<\/p>\n Area of a rectangle = length\u00a0 x breadth<\/p>\n substituting the values in the above formula, we get<\/p>\n Area of a rectangle = 200 m x 120 m<\/p>\n = 24000 m\u00b2<\/p>\n Therefore, the Area of a rectangle is 24000 m\u00b2<\/p>\n Now, we finding the total cost of tiling<\/p>\n Total cost of tiling =\u00a0 (6 x 24000) \/ 100<\/p>\n = 144000\/100<\/p>\n =\u00a0 $1440<\/p>\n Therefore, the Total cost of tiling is $1440<\/p>\n<\/details>\n 9. The length of a rectangular board is thrice its width. If the width of the board is 140 cm, find the cost of framing it at the rate of $5 for 30 cm.<\/p>\n Given, the width of the board = 140 cm<\/p>\n length of the board is thrice<\/p>\n so ,length = 3 x width<\/p>\n length =\u00a0 3 x 140 = 420 cm<\/p>\n 30 cm rate is $5<\/p>\n Circumference of rectangle = 2 ( l+ b)<\/p>\n substitute the values in the above formula, we get<\/p>\n Circumference of rectangle = 2 ( 420 + 140)<\/p>\n = 2 x 560 = 1120 cm Now, 30 cm cost is equal to rs. 5<\/p>\n So, 1 cm = 5\/ 30<\/p>\n But, we want the cost of framing<\/p>\n So, 1120 = (5 x 1120)\/ 20 = \u00a0rs. 280<\/p>\n Therefore, the cost of framing is Rs. 280<\/p>\n<\/details>\n 10. The Perimeter of a rectangular pool is 46 meters. If the length of the pool is 16 meters, then find its width. Here the perimeter and length of the rectangular pool are given. we have to find the width of the pool.<\/p>\n Given,<\/p>\n The perimeter of a rectangular pool is 46 meters<\/p>\n The length of the pool is 16 meters<\/p>\n Now we find the width of the pool.<\/p>\n we know the formula,<\/p>\n Perimeter of a rectangle = 2(l + b)<\/p>\n substituting the values, we get<\/p>\n 46 = 2( 16) +2( w)<\/p>\n 46= 32 + 2w<\/p>\n 46 – 32 = 2W<\/p>\n 14 = 2W<\/p>\n W= 14\/2 = 7 meters<\/p>\n Therefore, the width of the Pool is 7 meters.<\/p>\n<\/details>\n 11. The sides of a rectangle are in the ratio of 4: 5 and its perimeter is 90 cm. Find the dimensions of the rectangle and hence its area.<\/p>\n Given, Perimeter of a rectangle is 90 cm<\/p>\n Length of the sides = 4 : 5<\/p>\n Let the common ratio be X<\/p>\n So the sides will be 4X and 5X<\/p>\n we know that,<\/p>\n The Sum of all sides of the rectangle is equal to the perimeter.<\/p>\n so, Perimeter of a rectangle = 2 ( length + breadth)<\/p>\n substituting the values, we get<\/p>\n 90 = 2(l) + 2(b)<\/p>\n 90 = 2(4X) + 2 (5X)<\/p>\n 90 = 8X + 10 X<\/p>\n 18 X = 90<\/p>\n Therefore, X = 90\/18 = 5<\/p>\n Hence , length = 4X\u00a0 and breadth = 5X<\/p>\n substitute the ‘X’ value, we get<\/p>\n length = 4(5) = 20 , Breadth = 5(5) = 25<\/p>\n Now we find the area of a rectangle,<\/p>\n Area of a rectangle = length x breadth<\/p>\n substitute the values in the formula, we get<\/p>\n Area of a rectangle = 20\u00a0 cm x 25 cm<\/p>\n = 500 cm\u00b2<\/p>\n Therefore, the Area of a rectangle is 500 cm\u00b2.<\/p>\n<\/details>\n Worksheet on Area and Perimeter of Rectangle Problems will help the students to explore their knowledge of Rectangle word Problems. Solve all the Problems to learn the formula of Area of Rectangle and Perimeter of a Rectangle. To know the definition, properties, derivation, Problems with Solutions, Formulas of Rectangle you can visit our website. We … Read more<\/a><\/p>\n","protected":false},"author":11,"featured_media":72522,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[5],"tags":[],"jetpack_featured_media_url":"https:\/\/bigideasmathanswers.com\/wp-content\/uploads\/2021\/03\/Worksheet-on-Area-and-Perimeter-of-Rectangles.png","_links":{"self":[{"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts\/70589"}],"collection":[{"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/users\/11"}],"replies":[{"embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/comments?post=70589"}],"version-history":[{"count":1,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts\/70589\/revisions"}],"predecessor-version":[{"id":92282,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts\/70589\/revisions\/92282"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/media\/72522"}],"wp:attachment":[{"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/media?parent=70589"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/categories?post=70589"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/tags?post=70589"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Solution:<\/summary>\n
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\nTherefore, the circumference of rectangle = 1120 cm<\/p>\n
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