{"id":309,"date":"2022-04-13T10:38:27","date_gmt":"2022-04-13T10:38:27","guid":{"rendered":"http:\/\/10thclass.deltapublications.in\/?page_id=309"},"modified":"2025-03-03T09:16:47","modified_gmt":"2025-03-03T09:16:47","slug":"p-5-similar-triangles-and-indirect-measurement","status":"publish","type":"page","link":"https:\/\/10thclass.deltapublications.in\/index.php\/p-5-similar-triangles-and-indirect-measurement\/","title":{"rendered":"P.5 Similar triangles and indirect measurement"},"content":{"rendered":"\n

Similar triangles and indirect measurement<\/strong><\/h2>\n\n\n\n

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

The sides of similar triangles are proportional.<\/p>\n\n\n\n

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

\n
\n

Find r<\/em>.<\/strong><\/p>\n\n\n

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

Write your answer as a whole number or a decimal. Do not round.<\/p>\n\n\n\n

r = _______ metres<\/p>\n<\/div><\/div>\n\n\n\n

The original diagram included a smaller triangle inside a larger triangle. Redraw them as separate triangles.<\/p>\n\n\n

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

\u25b3KJI<\/em> is similar to  \u25b3GJH<\/em> because all three pairs of corresponding angles are congruent.<\/p>\n\n\n\n

One pair of corresponding side lengths is KJ<\/em> and GJ<\/em> .Together they form the fraction KJ \/ GJ <\/p>\n\n\n\n

An other pair of corresponding side lengths is IK<\/em> and HG<\/em>. Together they form the fraction IK \/ HG .<\/p>\n\n\n\n

In both fractions, the numerator comes from \u25b3KJI<\/em>  and the denominator comes from \u25b3GJH<\/em> Since the triangles are similar, you can use the fractions to set up a proportion and solve for r<\/em>.<\/p>\n\n\n\n

KJ \/ GJ = IK \/ HG<\/p>\n\n\n\n

3 \/ 6 = 4 \/ r Plug in the side lengths<\/em><\/p>\n\n\n\n

3 \/ 6 (6r) = 4 \/ r (6r) Multiply both sides by 6<\/em>r<\/p>\n\n\n\n

3 r = 4 . 6 Simplify<\/em><\/p>\n\n\n\n

3r = 24 Simplify<\/em><\/p>\n\n\n\n

3r \u00f7 3 = 24 \u00f7 3 Divide both sides by 3<\/em><\/p>\n\n\n\n

r = 8<\/p>\n\n\n\n

The missing length is 8 metres.<\/p>\n<\/div><\/div>\n\n\n\n

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\n

In the diagram below, \u25b3GHD<\/em> ~ \u25b3 EFD . Find w<\/em>.<\/strong><\/p>\n\n\n

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

Write your answer as a whole number or a decimal. Do not round.<\/p>\n\n\n\n

w = _______ metres<\/p>\n<\/div><\/div>\n\n\n\n

The original diagram included a smaller triangle and a larger triangle. Redraw them as separate triangles with corresponding sides in the same colour.<\/p>\n\n\n

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

\u25b3GHD<\/em> ~ <\/strong>\u25b3 EFD means that \u25b3GHD<\/em> is similar to \u25b3 EFD<\/p>\n\n\n\n

One pair of corresponding side lengths is DG<\/em> and DE<\/em>. Together they form the fraction DG \/ DE .<\/p>\n\n\n\n

Another pair of corresponding side lengths is GH<\/em> and EF<\/em>. Together they form the fraction GH \/ EF . <\/p>\n\n\n\n

In both fractions, the numerator comes from \u25b3GHD<\/em> and the denominator comes from \u25b3 EFD .Since the triangles are similar, you can use the fractions to set up a proportion and solve for w<\/em>.<\/p>\n\n\n\n

DG \/ DE = GH \/ EF<\/p>\n\n\n\n

4 \/ 8 = 2 \/ w Plug in the side lengths<\/em><\/p>\n\n\n\n

4 \/ 8 (8w) = 2 \/ w (8w) Multiply both sides by <\/em>8w<\/p>\n\n\n\n

4 w = 2 . 8 Simplify<\/em><\/p>\n\n\n\n

4w = 16 Simplify<\/em><\/p>\n\n\n\n

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

w = 4<\/p>\n\n\n\n

The missing length is 4 metres.<\/p>\n<\/div><\/div>\n\n\n\n

\n
\n

Find k<\/em>.<\/strong><\/p>\n\n\n

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

Write your answer as a whole number or a decimal. Do not round.<\/p>\n\n\n\n

k = _______ metres<\/p>\n<\/div><\/div>\n\n\n\n

The original diagram included a smaller triangle inside a larger triangle. Redraw them as separate triangles.<\/p>\n\n\n

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

\u25b3IJK is similar to  \u25b3MJL  because all three pairs of corresponding angles are congruent.<\/p>\n\n\n\n

One pair of corresponding side lengths is MJ and IJ . Together they form the fraction MJ \/ IJ .<\/p>\n\n\n\n

Another pair of corresponding side lengths is LM and KI. Together they form the fraction LM \/ KI .<\/p>\n\n\n\n

In both fractions, the numerator comes from \u25b3MJL and the denominator comes from \u25b3IJK .Since the triangles are similar, you can use the fractions to set up a proportion and solve for K.<\/p>\n\n\n\n

MJ \/ IJ = LM \/ KI <\/p>\n\n\n\n

6 \/ 3 = 8 \/ K Plug in the side lengths<\/em><\/p>\n\n\n\n

6 \/ 3 (3k) = 8 \/ k (3k) Multiply both sides by <\/em>3k <\/p>\n\n\n\n

6 k = 8 . 3 Simplify<\/em><\/p>\n\n\n\n

6k = 24 Simplify<\/em><\/p>\n\n\n\n

6k \u00f7 6 = 24 \u00f7 6 Divide both sides by <\/em>6<\/p>\n\n\n\n

k = 4<\/p>\n\n\n\n

The missing length is 4 metres.<\/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":"

Similar triangles and indirect measurement Key Notes : The sides of similar triangles are proportional. Learn with an example Find r. Write your answer as a whole number or a decimal. Do not round. r = _______ metres The original diagram included a smaller triangle inside a larger triangle. Redraw them as separate triangles. \u25b3KJI is similarContinue reading “P.5 Similar triangles and indirect measurement”<\/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-309","page","type-page","status-publish","hentry","entry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages\/309","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=309"}],"version-history":[{"count":16,"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages\/309\/revisions"}],"predecessor-version":[{"id":17557,"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages\/309\/revisions\/17557"}],"wp:attachment":[{"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/media?parent=309"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}