{"id":259,"date":"2022-04-13T10:29:52","date_gmt":"2022-04-13T10:29:52","guid":{"rendered":"http:\/\/10thclass.deltapublications.in\/?page_id=259"},"modified":"2025-02-15T06:20:14","modified_gmt":"2025-02-15T06:20:14","slug":"m-9-area-and-perimeter-of-similar-figures","status":"publish","type":"page","link":"https:\/\/10thclass.deltapublications.in\/index.php\/m-9-area-and-perimeter-of-similar-figures\/","title":{"rendered":"M.9 Area and perimeter of similar figures"},"content":{"rendered":"\n

Area and perimeter of similar figures<\/strong><\/h2>\n\n\n\n

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

The following proportion applies to similar shapes:<\/p>\n\n\n\n

( a\/b)2<\/sup> = A1<\/sub>\/A2<\/sub><\/p>\n\n\n\n

where a\/b is the ratio of the corresponding side lengths, and A1<\/sub>\/A2<\/sub> is the ratio of the areas.<\/p>\n\n\n\n

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

\n
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The figures below are similar. The labelled sides are corresponding.<\/p>\n\n\n

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

What is the area of the larger rectangle?<\/strong><\/p>\n\n\n\n

A2<\/sub> = ____________ square millimeters<\/p>\n<\/div><\/div>\n\n\n\n

Find the square of the ratio of the corresponding side lengths.<\/p>\n\n\n\n

(a\/b)2<\/sup> = (3\/6)2<\/sup> = (1\/2)2<\/sup> = 1\/4<\/p>\n\n\n\n

Find the ratio of the areas.<\/p>\n\n\n\n

A1<\/sub> \/ A2<\/sub> = 27 \/ A2<\/sub><\/p>\n\n\n\n

Use these two ratios to set up a proportion and solve for A<\/em>2.<\/p>\n\n\n\n

1\/4 = 27\/A2<\/sub><\/p>\n\n\n\n

1\/4 (4A2<\/sub>) = 27 \/ A2<\/sub> (4A2<\/sub>) Multiply both sides by 4A2<\/sub><\/p>\n\n\n\n

A2<\/sub> = 27 .4 Simplify<\/p>\n\n\n\n

A2<\/sub> = 108 Simplify<\/p>\n\n\n\n

The area of the larger rectangle is 108 square millimeters.<\/p>\n<\/div><\/div>\n\n\n\n

\n
\n

The figures below are similar. The labelled sides are corresponding.<\/p>\n\n\n

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

What is the area of the larger square?<\/strong><\/p>\n\n\n\n

A1<\/sub> = ____________ square millimeters<\/p>\n<\/div><\/div>\n\n\n\n

Find the square of the ratio of the corresponding side lengths.<\/p>\n\n\n\n

(a\/b)2<\/sup> = (4\/1)2<\/sup> = 16\/1<\/p>\n\n\n\n

Find the ratio of the areas.<\/p>\n\n\n\n

A1<\/sub> \/ A2<\/sub> = A1<\/sub> \/ 1 <\/p>\n\n\n\n

Use these two ratios to set up a proportion and solve for  A1<\/sub><\/p>\n\n\n\n

16\/1 = A1<\/sub>\/1<\/p>\n\n\n\n

16\/1 (1 .1) = A1<\/sub> \/ 1 (1 .1) Multiply both sides by 1 .1<\/p>\n\n\n\n

16 .1 = A1<\/sub> Simplify<\/p>\n\n\n\n

16 = A1<\/sub> Simplify<\/p>\n\n\n\n

The area of the larger rectangle is 16 square millimeters.<\/p>\n<\/div><\/div>\n\n\n\n

\n
\n

The figures below are similar. The labelled sides are corresponding.<\/p>\n\n\n

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

What is the area of the smaller square?<\/strong><\/p>\n\n\n\n

A2<\/sub> = ____________ square millimeters<\/p>\n<\/div><\/div>\n\n\n\n

Find the square of the ratio of the corresponding side lengths.<\/p>\n\n\n\n

(a\/b)2<\/sup> = (6\/4)2<\/sup> = (3\/2)2<\/sup> = 9\/4<\/p>\n\n\n\n

Find the ratio of the areas.<\/p>\n\n\n\n

A1<\/sub> \/ A2<\/sub> = 36 \/ A2<\/sub><\/p>\n\n\n\n

Use these two ratios to set up a proportion and solve for A<\/em>2.<\/p>\n\n\n\n

9\/4 = 36\/A2<\/sub><\/p>\n\n\n\n

9\/4 (4A2<\/sub>) = 36 \/ A2<\/sub> (4A2<\/sub>) Multiply both sides by 4A2<\/sub><\/p>\n\n\n\n

9 A2<\/sub> = 36 .4 Simplify<\/p>\n\n\n\n

9 A2<\/sub> = 144 Simplify<\/p>\n\n\n\n

9 A2<\/sub> \u00f7 9 = 144 \u00f7 9 Divide both sides by 9<\/p>\n\n\n\n

A2<\/sub> = 16<\/p>\n\n\n\n

The area of the larger rectangle is 16 square millimeters.<\/p>\n<\/div><\/div>\n\n\n\n

Let’s practice!<\/p>\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":"

Area and perimeter of similar figures Key Notes : The following proportion applies to similar shapes: ( a\/b)2 = A1\/A2 where a\/b is the ratio of the corresponding side lengths, and A1\/A2 is the ratio of the areas. Learn with an example The figures below are similar. The labelled sides are corresponding. What is theContinue reading “M.9 Area and perimeter of similar figures”<\/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-259","page","type-page","status-publish","hentry","entry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages\/259","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/comments?post=259"}],"version-history":[{"count":12,"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages\/259\/revisions"}],"predecessor-version":[{"id":17523,"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages\/259\/revisions\/17523"}],"wp:attachment":[{"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/media?parent=259"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}