{"id":403,"date":"2022-04-13T10:53:57","date_gmt":"2022-04-13T10:53:57","guid":{"rendered":"http:\/\/10thclass.deltapublications.in\/?page_id=403"},"modified":"2025-02-22T09:20:36","modified_gmt":"2025-02-22T09:20:36","slug":"u-4-minimum-and-maximum-area-and-volume","status":"publish","type":"page","link":"https:\/\/10thclass.deltapublications.in\/index.php\/u-4-minimum-and-maximum-area-and-volume\/","title":{"rendered":"U.4 Minimum and maximum area and volume"},"content":{"rendered":"\n

Minimum and maximum area and volume<\/strong><\/h2>\n\n\n\n

key notes :<\/p>\n\n\n\n

\ud83c\udfaf To find the minimum possible volume, subtract the greatest possible error from each measurement before calculating.<\/p>\n\n\n\n

\ud83c\udfaf To find the maximum possible volume, add the greatest possible error to each measurement before calculating.<\/p>\n\n\n\n

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

\n
\n

The rectangular prism below is labelled with its measured dimensions. Taking measurement error into account, what are the minimum and maximum possible volumes?<\/strong><\/p>\n\n\n

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

Minimum possible volume = ________ cm\u00b3<\/p>\n\n\n\n

Maximum possible volume = _______ cm\u00b3<\/p>\n<\/div><\/div>\n\n\n\n

You want to find the minimum and maximum volume, taking measurement error into account.<\/p>\n\n\n\n

First find the greatest possible error. Each measurement was made to the nearest whole centimetre, so the greatest possible error is half of 1 centimetre, which is 0.5 centimetres.<\/p>\n\n\n\n

To find the maximum possible volume, add the greatest possible error to each measurement, then multiply.<\/p>\n\n\n\n

Vmax<\/sub><\/strong> = Lmax<\/sub>Wmax<\/sub>Hmax<\/sub><\/strong><\/p>\n\n\n\n

= (11 + 0.5)(20 + 0.5)(8 + 0.5) Add the greatest possible error, 0.5<\/p>\n\n\n\n

= (11.5)(20.5)(8.5)<\/p>\n\n\n\n

= 2,003.875<\/p>\n\n\n\n

To find the minimum possible volume, subtract the greatest possible error from each measurement, then multiply.<\/p>\n\n\n\n

Vmin<\/span><\/span><\/div>
\u00a0=\u00a0<\/div>
Lmin<\/span><\/span>Wmin<\/span><\/span>Hmin<\/span><\/span><\/div><\/p>\n\n\n\n

= (11 \u2212 0.5)(20 \u2212 0.5)(8 \u2212 0.5) Subtract the greatest possible error, 0.5<\/p>\n\n\n\n

= (10.5)(19.5)(7.5)<\/p>\n\n\n\n

= 1,535.625<\/p>\n\n\n\n

The minimum possible volume is 1,535.625 cubic centimetres, and the maximum possible volume is 2,003.875 cubic centimetres.<\/p>\n<\/div><\/div>\n\n\n\n

\n
\n

The rectangular prism below is labelled with its measured dimensions. Taking measurement error into account, what are the minimum and maximum possible volumes?<\/strong><\/p>\n\n\n

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

Minimum possible volume = ________ m\u00b3<\/p>\n\n\n\n

Maximum possible volume = ________ m\u00b3<\/p>\n<\/div><\/div>\n\n\n\n

You want to find the minimum and maximum volume, taking measurement error into account.<\/p>\n\n\n\n

First find the greatest possible error. Each measurement was made to the nearest whole metre, so the greatest possible error is half of 1 metre, which is 0.5 metres.<\/p>\n\n\n\n

To find the maximum possible volume, add the greatest possible error to each measurement, then multiply.<\/p>\n\n\n\n

Vmax<\/sub><\/strong> = Lmax<\/sub>Wmax<\/sub>Hmax<\/sub><\/strong><\/p>\n\n\n\n

= (7 + 0.5)(12 + 0.5)(18 + 0.5) Add the greatest possible error, 0.5<\/p>\n\n\n\n

= (7.5)(12.5)(18.5)<\/p>\n\n\n\n

= 1,734.375<\/p>\n\n\n\n

To find the minimum possible volume, subtract the greatest possible error from each measurement, then multiply.<\/p>\n\n\n\n

Vmin<\/sub><\/strong> = Lmin<\/sub>Wmin<\/sub>Hmin<\/sub><\/strong><\/p>\n\n\n\n

= (7 \u2212 0.5)(12 \u2212 0.5)(18 \u2212 0.5) Subtract the greatest possible error, 0.5<\/p>\n\n\n\n

= (6.5)(11.5)(17.5)<\/p>\n\n\n\n

= 1,308.125<\/p>\n\n\n\n

The minimum possible volume is 1,308.125 cubic metres, and the maximum possible volume is 1,734.375 cubic metres.<\/p>\n<\/div><\/div>\n\n\n\n

\n
\n

The rectangle below is labelled with its measured dimensions. Taking measurement error into account, what are the minimum and maximum possible areas?<\/strong><\/p>\n\n\n

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

Minimum possible area = ________ km\u00b2<\/p>\n\n\n\n

Maximum possible area = ________ km\u00b2<\/p>\n<\/div><\/div>\n\n\n\n

You want to find the minimum and maximum area, taking measurement error into account.<\/p>\n\n\n\n

First find the greatest possible error. Each measurement was made to the nearest whole kilometre, so the greatest possible error is half of 1 kilometre, which is 0.5 kilometres.<\/p>\n\n\n\n

To find the maximum possible area, add the greatest possible error to each measurement, then multiply.<\/p>\n\n\n\n

Amax<\/sub><\/strong> = Lmax<\/sub>Wmax<\/sub><\/strong><\/p>\n\n\n\n

= (2 + 0.5)(9 + 0.5) Add the greatest possible error, 0.5<\/p>\n\n\n\n

= (2.5)(9.5)<\/p>\n\n\n\n

= 23.75<\/p>\n\n\n\n

To find the minimum possible area, subtract the greatest possible error from each measurement, then multiply.<\/p>\n\n\n\n

Amin<\/sub> = Lmin<\/sub>Wmin<\/sub><\/strong><\/p>\n\n\n\n

= (2 \u2212 0.5)(9 \u2212 0.5) Subtract the greatest possible error, 0 <\/p>\n\n\n\n

= (1.5)(8.5)<\/p>\n\n\n\n

= 12.75<\/p>\n\n\n\n

The minimum possible area is 12.75 square kilometres, and the maximum possible area is 23.75 square kilometres.<\/p>\n<\/div><\/div>\n\n\n\n

Let’s practice!\ud83d\udd8a\ufe0f<\/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":"

Minimum and maximum area and volume key notes : \ud83c\udfaf To find the minimum possible volume, subtract the greatest possible error from each measurement before calculating. \ud83c\udfaf To find the maximum possible volume, add the greatest possible error to each measurement before calculating. Learn with an example The rectangular prism below is labelled with itsContinue reading “U.4 Minimum and maximum area and volume”<\/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-403","page","type-page","status-publish","hentry","entry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages\/403","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=403"}],"version-history":[{"count":13,"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages\/403\/revisions"}],"predecessor-version":[{"id":17509,"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages\/403\/revisions\/17509"}],"wp:attachment":[{"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/media?parent=403"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}