{"id":2091,"date":"2022-11-09T05:19:07","date_gmt":"2022-11-09T05:19:07","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=2091"},"modified":"2022-11-09T05:19:07","modified_gmt":"2022-11-09T05:19:07","slug":"ncert-solutions-class-7-maths-chapter-11-perimeter-and-area-ex-11-1","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-7-maths-chapter-11-perimeter-and-area-ex-11-1\/","title":{"rendered":"NCERT Solutions Class 7 Maths Chapter 11 Perimeter and Area Ex 11.1"},"content":{"rendered":"<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 7 Mathematics\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Chapter &#8211; 11 (Perimeter and Area)<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">The NCERT Solutions in English Language for Class 7 Mathematics <strong>Chapter &#8211; 11 Perimeter and Area <\/strong>Exercise 11.1 has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n<p><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Chapter : 11 Perimeter and Area<\/strong><\/span><\/p>\n<ul>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-7-maths-chapter-11-perimeter-and-area-ex-11-2\" target=\"_blank\" rel=\"noopener\"><span style=\"font-family: georgia, palatino, serif;\">NCERT Solution Class 7 Maths Exercise &#8211; 11.2<\/span><\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-7-maths-chapter-11-perimeter-and-area-ex-11-3\" target=\"_blank\" rel=\"noopener\"><span style=\"font-family: georgia, palatino, serif;\">NCERT Solution Class 7 Maths Exercise &#8211; 11.3<\/span><\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-7-maths-chapter-11-perimeter-and-area-ex-11-1\" target=\"_blank\" rel=\"noopener\"><span style=\"font-family: georgia, palatino, serif;\">NCERT Solution Class 7 Maths Exercise &#8211; 11.4<\/span><\/a><\/span><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 11.1\u00a0<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. The Length and the breadth of a rectangular piece of land are 500 m and 300 m, respectively. Find<br \/>\n<\/strong><strong>(i) Its area<br \/>\n(ii) the cost of the land, if 1 m<sup>2<\/sup>\u00a0of the land costs \u20b9 10,000.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;<br \/>\n<\/strong>Given that,<br \/>\nLength of the rectangular piece of land = 500 m<br \/>\nBreadth of the rectangular piece of land = 300 m<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i)<\/strong> Area of rectangle = Length \u00d7 Breadth<br \/>\n= 500 \u00d7 300<br \/>\n= 150000 m<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii)<\/strong> Cost of the land for 1 m<sup>2<\/sup>\u00a0= \u20b9 10000<br \/>\nCost of the land for 150000 m<sup>2<\/sup>\u00a0= 10000 \u00d7 150000<br \/>\n= \u20b9 1,50,00,00,000<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. Find the area of a square park whose perimeter is 320m.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;<br \/>\n<\/strong>Given that,<br \/>\nPerimeter of the square park = 320 m<br \/>\n4 \u00d7 Length of the side of park = 320 m<br \/>\nLength of the side of park = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{320}{4}\" alt=\"\\frac{320}{4}\" align=\"absmiddle\" \/> = 80 m<br \/>\nSo, Area of the square park = (length of the side of park)<sup>2<br \/>\n<\/sup>= (80m)<sup>2<br \/>\n<\/sup>= 6,400 m<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3. Find the breadth of a rectangular plot of land, if its area is 440 m<sup>2<\/sup>\u00a0and the length is 22 m. Also find its perimeter.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;<br \/>\n<\/strong>Given that,<br \/>\nArea of the rectangular plot = 440 m<sup>2<br \/>\n<\/sup>Length of the rectangular plot = 22 m<br \/>\nArea of the rectangle = Length \u00d7 Breadth<br \/>\n440 = 22 \u00d7 Breadth<br \/>\nBreadth = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{440}{22}\" alt=\"\\frac{440}{22}\" align=\"absmiddle\" \/><br \/>\nBreadth = 20 m<br \/>\nThen,<br \/>\nPerimeter of the rectangle = 2(Length + Breadth)<br \/>\n= 2 (22 + 20)<br \/>\n= 2(42)<br \/>\n= 84 m<br \/>\n\u2234 Perimeter of the rectangular plot is 84 m.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4. The perimeter of a rectangular sheet is 100 cm. If the length is 35 cm, find its breadth. <\/strong><strong>Also find the area.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;<br \/>\n<\/strong>Given that,<br \/>\nPerimeter of the a rectangular sheet = 100 cm<br \/>\nLength of the rectangular sheet = 35 cm<br \/>\nPerimeter of the rectangle = 2 (Length + Breadth)<br \/>\n100 = 2 (35 + Breadth)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{100}{2}\" alt=\"\\frac{100}{2}\" align=\"absmiddle\" \/> = 35 + Breadth<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">50 \u2013 35 = Breadth<br \/>\nBreadth = 15 cm<br \/>\nThen,<br \/>\nArea of the rectangle = Length \u00d7 Breadth<br \/>\n= 35 m\u00d7 15 m<br \/>\n= 525 cm<sup>2<br \/>\n<\/sup>\u2234 Area of the rectangular sheet is 525 cm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>5. The area of a square park is the same as of a rectangular park. If the side of the square park is 60 m and the length of the rectangular park is 90 m, find the breadth of the rectangular park.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;<br \/>\n<\/strong>Given that,<br \/>\nSide of the square park = 60 m<br \/>\nLength of the rectangular park = 90 m<br \/>\nArea of the square park = (one of the side of square)<sup>2<br \/>\n<\/sup>= 60<sup>2<br \/>\n<\/sup>= 3,600 m<sup>2<br \/>\n<\/sup>Area of the rectangular park = 3600 m<sup>2<\/sup><br \/>\nLength \u00d7 Breadth = 3600<br \/>\n90 \u00d7 Breadth = 3600<br \/>\nBreadth = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{3600}{90}\" alt=\"\\frac{3600}{90}\" align=\"absmiddle\" \/> <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">Breadth = 40 m<br \/>\nHence, the required breadth = 40 m.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>6. A wire is in the shape of a rectangle. Its length is 40 cm and breadth is 22 cm. If the same wire is rebent in the shape of a square, what will be the measure of each side? <\/strong><strong>Also find which shape encloses more area?<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;<br \/>\n<\/strong>Given that,<br \/>\nLength of the rectangle = 40 cm<br \/>\nBreadth of the square = 22 cm<br \/>\nPerimeter of the rectangle = Perimeter of the Square<br \/>\n2 (Length + Breadth) = 4 \u00d7 side<br \/>\n2 (40 + 22) = 4 \u00d7 side<br \/>\n2 (62) = 4 \u00d7 side<br \/>\n124 = 4 \u00d7 side<br \/>\nSide = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{124}{4}\" alt=\"\\frac{124}{4}\" align=\"absmiddle\" \/> <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">Side = 31 cm <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">So, Area of the rectangle = (Length \u00d7 Breadth)<br \/>\n= 40 m \u00d7 22 m<br \/>\n= 880 cm<sup>2<br \/>\n<\/sup>Area of square = side<sup>2<br \/>\n<\/sup>= 31<sup>2<br \/>\n<\/sup>= 31 \u00d7 31<br \/>\n= 961 cm<sup>2 <\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">Since 961 cm<sup>2<\/sup>\u00a0&gt; 880 cm<sup>2<\/sup><\/span><br \/>\n<span style=\"color: #000000;\">Hence, the square encloses more area.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>7. The perimeter of a rectangle is 130 cm. If the breadth of the rectangle is <\/strong><strong>30 cm, find its length. Also find the area of the rectangle.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;<br \/>\n<\/strong>Given that,<br \/>\nPerimeter of the rectangle = 130 cm<br \/>\nBreadth of the rectangle = 30<br \/>\nPerimeter of rectangle = 2 (Length + Breadth)<br \/>\n130 = 2 (length + 30)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{130}{2}\" alt=\"\\frac{130}{2}\" align=\"absmiddle\" \/> = length + 30 <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">Length + 30 = 65<br \/>\nLength = 65 \u2013 30<br \/>\nLength = 35 cm<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">Then,<br \/>\nArea of the rectangle = Length \u00d7 Breadth<br \/>\n= 35 m \u00d7 30 m<br \/>\n= 1050 cm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>8. A door of length 2 m and breadth 1 m is fitted in a wall. The length of the wall is 4.5 m and the breadth is 3.6 m (Fig). Find the cost of white washing the wall, if the rate of white washing the wall is \u20b9 20 per m<sup>2<\/sup>.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2100\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/10\/NCERT-Solutions-Class-7-Maths-Ex-11.1-Q8.png\" alt=\"NCERT Class 7 Maths Solution\" width=\"251\" height=\"156\" \/><br \/>\n<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;<br \/>\n<\/strong>Given that,<br \/>\nLength of the door = 2 m<br \/>\nBreadth of the door = 1 m<br \/>\nLength of the wall = 4.5 m<br \/>\nBreadth of the wall = 3.6 m<br \/>\nArea of the door = Length \u00d7 Breadth<br \/>\n= 2 m \u00d7 1 m<br \/>\n= 2 m<sup>2 <\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">Area of the wall = Length \u00d7 Breadth<br \/>\n= 4.5 \u00d7 3.6<br \/>\n= 16.2 m<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">So, Area to be white washed = 16.2 m<sup>2 <\/sup>\u2013 2 m<sup>2 <\/sup>= 14.2 m<sup>2<br \/>\n<\/sup>Cost of white washing 1 m<sup>2<\/sup> area = \u20b9 20<br \/>\nHence cost of whit washing 14.2 m<sup>2<\/sup> area = 14.2 \u00d7 20<br \/>\n= \u20b9 284<\/span><\/p>\n<p>&nbsp;<\/p>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 7 Mathematics\u00a0 Chapter &#8211; 11 (Perimeter and Area) The NCERT Solutions in English Language for Class 7 Mathematics Chapter &#8211; 11 Perimeter and Area Exercise 11.1 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter : 11 Perimeter and Area NCERT Solution Class 7 Maths [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[219],"tags":[341,342,283,5],"class_list":["post-2091","post","type-post","status-publish","format-standard","hentry","category-class-7-maths","tag-ncert-class-7-maths-chapter-11-perimeter-and-area-in-english","tag-ncert-solutions-class-7-maths-chapter-11-in-english","tag-ncert-solutions-class-7-maths-in-english","tag-ncert-solutions-in-english"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.9 (Yoast SEO v27.4) - 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