{"id":279,"date":"2022-04-12T05:30:17","date_gmt":"2022-04-12T05:30:17","guid":{"rendered":"http:\/\/4thclass.deltapublications.in\/?page_id=279"},"modified":"2025-01-24T09:55:50","modified_gmt":"2025-01-24T09:55:50","slug":"n-5-use-perimeter-to-determine-cost","status":"publish","type":"page","link":"https:\/\/4thclass.deltapublications.in\/index.php\/n-5-use-perimeter-to-determine-cost\/","title":{"rendered":"N.5 Use perimeter to determine cost"},"content":{"rendered":"\n

Use perimeter to determine cost<\/strong><\/h2>\n\n\n\n

Key Notes:<\/strong><\/p>\n\n\n\n

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Find the perimeter of a rectangle by adding the lengths of the sides.<\/p>\n\n\n\n

Learn with an example<\/strong><\/p>\n\n\n\n

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\ud83d\udd14 Each side of a square courtyard is 17 metres long. It costs \u20b91,370.00 per metre to replace the brick wall around the courtyard.<\/strong><\/p>\n\n\n\n

How much would it cost to replace the brick wall?<\/strong><\/p>\n\n\n\n

\u20b9_____<\/p>\n<\/div><\/div>\n\n\n\n

Picture the square:<\/p>\n\n\n

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\"\"<\/figure><\/div>\n\n\n

Add the lengths:<\/p>\n\n\n\n

17 + 17 + 17 + 17 = 68<\/p>\n\n\n\n

The perimeter is 68 metres. Now multiply the price per metre by the perimeter.<\/p>\n\n\n

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\"\"<\/figure><\/div>\n\n\n

It would cost \u20b993,160.00 to replace the brick wall.<\/p>\n<\/div><\/div>\n\n\n\n

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\ud83d\udd14 A rectangular portrait is 1 metre wide and 2 metres high. It costs \u20b91,922.00 per metre to put a gold frame around the portrait. <\/strong><\/p>\n\n\n\n

How much will the frame cost?<\/strong><\/p>\n\n\n\n

\u20b9_______<\/p>\n<\/div><\/div>\n\n\n\n

Picture the rectangle:<\/p>\n\n\n

\n
\"\"<\/figure><\/div>\n\n\n

Add the lengths:<\/p>\n\n\n\n

1 + 2 + 1 + 2 = 6<\/p>\n\n\n\n

The perimeter is 6 metres. Now multiply the price per metre by the perimeter.<\/p>\n\n\n

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\"\"<\/figure><\/div>\n\n\n

The frame will cost \u20b911,532.00.<\/p>\n<\/div><\/div>\n\n\n\n

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\ud83d\udd14 A beach volleyball court is 8 metres wide and 17 metres long. The rope used for the boundary line costs \u20b9166.00 per metre.<\/strong><\/p>\n\n\n\n

How much would it cost to buy a new boundary line for the court?<\/strong><\/p>\n\n\n\n

\u20b9_______<\/p>\n<\/div><\/div>\n\n\n\n

Picture the rectangle:<\/p>\n\n\n

\n
\"\"<\/figure><\/div>\n\n\n

Add the lengths:<\/p>\n\n\n\n

8 + 17 + 8 + 17 = 50<\/p>\n\n\n\n

The perimeter is 50 metres. Now multiply the price per metre by the perimeter.<\/p>\n\n\n

\n
\"\"<\/figure><\/div>\n\n\n

It would cost \u20b98,300.00 to buy a new boundary line for the court.<\/p>\n<\/div><\/div>\n\n\n\n

let’s practice!<\/p>\n<\/div><\/div>\n\n\n\n

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\"\"<\/a><\/figure>\n<\/div>\n\n\n\n
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\"\"<\/a><\/figure>\n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Use perimeter to determine cost Key Notes: Find the perimeter of a rectangle by adding the lengths of the sides. Learn with an example \ud83d\udd14 Each side of a square courtyard is 17 metres long. It costs \u20b91,370.00 per metre to replace the brick wall around the courtyard. How much would it cost to replace the brickContinue reading “N.5 Use perimeter to determine cost”<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"om_disable_all_campaigns":false,"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"footnotes":""},"class_list":["post-279","page","type-page","status-publish","hentry","entry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/4thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages\/279","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/4thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/4thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/4thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/4thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/comments?post=279"}],"version-history":[{"count":11,"href":"https:\/\/4thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages\/279\/revisions"}],"predecessor-version":[{"id":14621,"href":"https:\/\/4thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages\/279\/revisions\/14621"}],"wp:attachment":[{"href":"https:\/\/4thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/media?parent=279"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}