{"id":321,"date":"2022-04-13T10:41:09","date_gmt":"2022-04-13T10:41:09","guid":{"rendered":"http:\/\/10thclass.deltapublications.in\/?page_id=321"},"modified":"2024-05-24T05:42:57","modified_gmt":"2024-05-24T05:42:57","slug":"q-1-pythagoras-theorem","status":"publish","type":"page","link":"https:\/\/10thclass.deltapublications.in\/index.php\/q-1-pythagoras-theorem\/","title":{"rendered":"Q.1 Pythagoras’ Theorem"},"content":{"rendered":"\n

Pythagoras’ Theorem<\/strong><\/h2>\n\n\n\n

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

\ud83d\udcda In a right triangle, a<\/em>2<\/sup> + b<\/em>2<\/sup> = c<\/em>2<\/sup>, where a<\/em> and b<\/em> are the lengths of the legs and c<\/em> is the length of the hypotenuse. This is called Pythagoras’ theorem.<\/p>\n\n\n\n

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

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

What is the perimeter?<\/strong><\/p>\n\n\n\n

________ millimetres<\/p>\n<\/div><\/div>\n\n\n\n

Use Pythagoras’ theorem, with a<\/em> = 56 and b<\/em> = 33.<\/p>\n\n\n\n

a<\/em>2<\/sup> + b<\/em>2<\/sup>= c<\/em>2<\/sup> Pythagoras’ theorem<\/em><\/p>\n\n\n\n

562<\/sup> + 332<\/sup>= c<\/em>2<\/sup> Plug in a = 56 and b = 33<\/em><\/p>\n\n\n\n

3,136 + 1,089 =  c<\/em>2<\/sup> Square<\/em><\/p>\n\n\n\n

4,225 =  c<\/em>2<\/sup> Add<\/em><\/p>\n\n\n\n

\u221a4225 = \u221ac<\/em>2<\/sup> Take the square root of both sides<\/em><\/p>\n\n\n\n

65 =  c<\/em> Simplify<\/em><\/p>\n\n\n\n

The length of the hypotenuse is 65 millimetres.<\/p>\n\n\n\n

Find the perimeter. Add the lengths of the two sides and the hypotenuse.<\/p>\n\n\n\n

a + b +c = 56 + 33 +65<\/p>\n\n\n\n

= 154<\/p>\n\n\n\n

The perimeter is 154 millimetres.<\/p>\n<\/div><\/div>\n\n\n\n

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Marco’s school is due west of his house and due south of his friend Jonah’s house. The distance between the school and Jonah’s house is 15 kilometres and the straight-line distance between Marco’s house and Jonah’s house is 17 kilometres. How far is Marco’s house from school?<\/p>\n\n\n\n

_______ Kilometres <\/strong><\/p>\n<\/div><\/div>\n\n\n\n

Draw a diagram.<\/p>\n\n\n

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

Use Pythagoras’ theorem, with a<\/em> = 15 and c<\/em> = 17.<\/p>\n\n\n\n

a<\/em>2<\/sup> + b<\/em>2<\/sup>= c<\/em>2<\/sup> Pythagoras’ theorem<\/em><\/p>\n\n\n\n

152<\/sup> + b2<\/sup>= 172<\/sup> Plug in a = 15 and c = <\/em>17<\/p>\n\n\n\n

225+ b2<\/sup> =  289 Square<\/em><\/p>\n\n\n\n

b2<\/sup> =  64 Add<\/em><\/p>\n\n\n\n

\u221ab2<\/sup> = \u221a64 Take the square root of both sides<\/em><\/p>\n\n\n\n

b =  8 Simplify<\/em><\/p>\n\n\n\n

Marco’s house is 8 kilometres from school.<\/p>\n<\/div><\/div>\n\n\n\n

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

What is the length of the missing leg?<\/strong><\/p>\n\n\n\n

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

Use Pythagoras’ theorem, with a<\/em> = 55 and c<\/em> = 73.<\/p>\n\n\n\n

a<\/em>2<\/sup> + b<\/em>2<\/sup>= c<\/em>2<\/sup> Pythagoras’ theorem<\/em><\/p>\n\n\n\n

552<\/sup> + b2<\/sup>= 732<\/sup> Plug in a = 15 and c = <\/em>17<\/p>\n\n\n\n

3025+ b2<\/sup> =  5329 Square<\/em><\/p>\n\n\n\n

b2<\/sup> =  2304 Add<\/em><\/p>\n\n\n\n

\u221ab2<\/sup> = \u221a2304 Take the square root of both sides<\/em><\/p>\n\n\n\n

b =  48 Simplify<\/em><\/p>\n\n\n\n

The length of the missing leg is 48 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":"

Pythagoras’ Theorem Key Notes : \ud83d\udcda In a right triangle, a2 + b2 = c2, where a and b are the lengths of the legs and c is the length of the hypotenuse. This is called Pythagoras’ theorem. Learn with an example What is the perimeter? ________ millimetres Use Pythagoras’ theorem, with a = 56 and b = 33. a2 + b2= c2 Pythagoras’ theorem 562 + 332= c2 Plug in a = 56 and b = 33 3,136 +Continue reading “Q.1 Pythagoras’ Theorem”<\/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-321","page","type-page","status-publish","hentry","entry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages\/321","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=321"}],"version-history":[{"count":8,"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages\/321\/revisions"}],"predecessor-version":[{"id":12701,"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages\/321\/revisions\/12701"}],"wp:attachment":[{"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/media?parent=321"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}