{"id":317,"date":"2022-04-13T10:39:35","date_gmt":"2022-04-13T10:39:35","guid":{"rendered":"http:\/\/10thclass.deltapublications.in\/?page_id=317"},"modified":"2024-10-29T09:00:05","modified_gmt":"2024-10-29T09:00:05","slug":"p-9-areas-of-similar-figures","status":"publish","type":"page","link":"https:\/\/10thclass.deltapublications.in\/index.php\/p-9-areas-of-similar-figures\/","title":{"rendered":"P.9 Areas of similar figures"},"content":{"rendered":"\n

Areas 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
\n

The figures below are similar. The labelled sides are corresponding.<\/strong><\/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 centimetres<\/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 \/ 2)2<\/sup> = (2 \/ 1)2<\/sup> = 4\/1<\/p>\n\n\n\n

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

A1<\/sub> \/ A2<\/sub> = 16 \/ A2<\/sub><\/p>\n\n\n\n

Use these two ratios to set up a proportion and solve for A2<\/sub>.<\/p>\n\n\n\n

4 \/ 1 = 16 \/ A2<\/sub><\/p>\n\n\n\n

4 \/ 1 (A2<\/sub>) = 16 \/ A2<\/sub> (A2<\/sub>) Multiply both sides by<\/em> A2<\/sub> <\/p>\n\n\n\n

4 A2<\/sub> = 16 .1 Simplify<\/em><\/p>\n\n\n\n

4 A2<\/sub> = 16 Simplify<\/em><\/p>\n\n\n\n

4 A2<\/sub> \u00f7 4 = 16 \u00f7 4 Divide both sides by 4<\/em><\/p>\n\n\n\n

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

The area of the smaller square is 4 square centimetres.<\/p>\n<\/div><\/div>\n\n\n\n

\n
\n

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

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

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

A1<\/sub> = _______ square centimetres<\/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 \/ 8)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> = A1<\/sub> \/ 64<\/p>\n\n\n\n

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

1 \/ 4 = A1<\/sub> \/ 64<\/p>\n\n\n\n

1 \/ 4 (4 . 64 ) = A1<\/sub> \/ 64 (4 . 64 ) Multiply both sides by 4 \u00b7 64<\/em><\/p>\n\n\n\n

1 . 64 = 4A1<\/sub> Simplify<\/em><\/p>\n\n\n\n

64 = 4A1<\/sub> Simplify<\/em><\/p>\n\n\n\n

64 \u00f7  4 = 4A1<\/sub> \u00f7  4 Divide both sides by 4<\/em><\/p>\n\n\n\n

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

The area of the smaller triangle is 16 square millimetres.<\/p>\n<\/div><\/div>\n\n\n\n

\n
\n

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

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

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

A1<\/sub> = _______ square centimetres<\/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> = (2 \/ 5)2<\/sup> = (4 \/ 25) <\/p>\n\n\n\n

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

A1<\/sub> \/ A2<\/sub> = A1<\/sub> \/ 100<\/p>\n\n\n\n

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

4 \/ 25 = A1<\/sub> \/ 100<\/p>\n\n\n\n

4 \/ 25 (25 . 100 ) = A1<\/sub> \/ 100 (25 . 100 ) Multiply both sides by 25 . 100<\/em><\/p>\n\n\n\n

4 . 100 = 25A1<\/sub> Simplify<\/em><\/p>\n\n\n\n

400 = 25A1<\/sub> Simplify<\/em><\/p>\n\n\n\n

400 \u00f7  25 = 25A1<\/sub> \u00f7  25 Divide both sides by <\/em>25<\/p>\n\n\n\n

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

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

Let’s practice!<\/p>\n\n\n\n

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

Areas 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 labelledContinue reading “P.9 Areas 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-317","page","type-page","status-publish","hentry","entry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages\/317","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=317"}],"version-history":[{"count":9,"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages\/317\/revisions"}],"predecessor-version":[{"id":14336,"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages\/317\/revisions\/14336"}],"wp:attachment":[{"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/media?parent=317"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}