{"id":257,"date":"2022-04-13T10:29:37","date_gmt":"2022-04-13T10:29:37","guid":{"rendered":"http:\/\/10thclass.deltapublications.in\/?page_id=257"},"modified":"2025-02-12T10:57:02","modified_gmt":"2025-02-12T10:57:02","slug":"m-8-area-between-two-shapes","status":"publish","type":"page","link":"https:\/\/10thclass.deltapublications.in\/index.php\/m-8-area-between-two-shapes\/","title":{"rendered":"M.8 Area between two shapes"},"content":{"rendered":"\n

Area between two shapes<\/strong><\/h2>\n\n\n\n

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

To find the area of a figure with a hole in it, subtract the area of the inner shape from the area of the outer shape.<\/p>\n\n\n\n

Area of a triangle:<\/p>\n\n\n\n

area = 1\/2 . base . height<\/p>\n\n\n\n

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

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What is the area of the shaded region?<\/strong><\/p>\n\n\n

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

Write your answer as a whole number or a decimal rounded to the nearest hundredth.  <\/p>\n\n\n\n

_______ square meters<\/p>\n<\/div><\/div>\n\n\n\n

To find the area of the shaded region, subtract the area of the inner shape from the area of the outer shape. Start by finding the area of the inner shape.<\/p>\n\n\n\n

Find the base and height of the inner triangle.<\/p>\n\n\n\n

base: 8 m<\/p>\n\n\n\n

height: 12 m<\/p>\n\n\n\n

Use these numbers in the formula.<\/p>\n\n\n\n

inner area = 1\/2 . base . height<\/p>\n\n\n\n

= 1\/2 . 8 . 12<\/p>\n\n\n\n

= 48<\/p>\n\n\n\n

Now find the units. The lengths are measured in meters, so the area is measured in square meters.<\/p>\n\n\n\n

The area of the inner triangle is 48 square meters.<\/p>\n\n\n\n

Next, find the area of the outer shape.<\/p>\n\n\n\n

Find the base and height of the outer triangle.<\/p>\n\n\n\n

base: 13 m<\/p>\n\n\n\n

height: 17 m<\/p>\n\n\n\n

Use these numbers in the formula.<\/p>\n\n\n\n

outer area = 1\/2 base . height<\/p>\n\n\n\n

= 1\/2 . 13 . 17<\/p>\n\n\n\n

= 110.5<\/p>\n\n\n\n

As with the inner triangle, the lengths of the outer triangle are measured in meters. So, the area is measured in square meters.<\/p>\n\n\n\n

The area of the outer triangle is 110.5 square meters.<\/p>\n\n\n\n

Finally, subtract the inner area from the outer area to find the area of the shaded region.<\/p>\n\n\n\n

shaded area = 110.5 \u2013 48<\/p>\n\n\n\n

= 62.5<\/p>\n\n\n\n

The area of the shaded region is 62.5 square meters.<\/p>\n<\/div><\/div>\n\n\n\n

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\n

What is the area of the shaded region?<\/strong><\/p>\n\n\n

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

_____  square centimeters<\/p>\n<\/div><\/div>\n\n\n\n

To find the area of the shaded region, subtract the area of the inner shape from the area of the outer shape. Start by finding the area of the inner shape.<\/p>\n\n\n\n

Find the side length of the inner square.<\/p>\n\n\n\n

side: 10 cm<\/p>\n\n\n\n

Use this number in the formula.<\/p>\n\n\n\n

inner area=side<\/p>\n\n\n\n

side = 10 . 10<\/p>\n\n\n\n

=100<\/p>\n\n\n\n

Now find the units. The lengths are measured in centimeters, so the area is measured in square centimeters.<\/p>\n\n\n\n

The area of the inner square is 100 square centimeters.<\/p>\n\n\n\n

Next, find the area of the outer shape.<\/p>\n\n\n\n

Find the side length of the outer square.<\/p>\n\n\n\n

side: 19 cm<\/p>\n\n\n\n

Use this number in the formula.<\/p>\n\n\n\n

outer area=side . side<\/p>\n\n\n\n

=19 . 19<\/p>\n\n\n\n

=361<\/p>\n\n\n\n

As with the inner square, the lengths of the outer square are measured in centimeters. So, the area is measured in square centimeters.<\/p>\n\n\n\n

The area of the outer square is 361 square centimeters.<\/p>\n\n\n\n

Finally, subtract the inner area from the outer area to find the area of the shaded region.<\/p>\n\n\n\n

shaded area = 361 \u2013 100<\/p>\n\n\n\n

= 261<\/p>\n\n\n\n

The area of the shaded region is 261 square centimeters.<\/p>\n<\/div><\/div>\n\n\n\n

\n
\n

What is the area of the shaded region?<\/strong><\/p>\n\n\n

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

______  square meters<\/p>\n<\/div><\/div>\n\n\n\n

To find the area of the shaded region, subtract the area of the inner shape from the area of the outer shape. Start by finding the area of the inner shape.<\/p>\n\n\n\n

Find the side length of the inner square.<\/p>\n\n\n\n

side: 6 m<\/p>\n\n\n\n

Use this number in the formula.<\/p>\n\n\n\n

inner area = side . side<\/p>\n\n\n\n

= 6 . 6<\/p>\n\n\n\n

= 36<\/p>\n\n\n\n

Now find the units. The lengths are measured in meters, so the area is measured in square meters.<\/p>\n\n\n\n

The area of the inner square is 36 square meters.<\/p>\n\n\n\n

Next, find the area of the outer shape.<\/p>\n\n\n\n

Find the side length of the outer square.<\/p>\n\n\n\n

side: 17 m<\/p>\n\n\n\n

Use this number in the formula.<\/p>\n\n\n\n

outer area=side . side<\/p>\n\n\n\n

= 17 . 17<\/p>\n\n\n\n

= 289<\/p>\n\n\n\n

As with the inner square, the lengths of the outer square are measured in meters. So, the area is measured in square meters.<\/p>\n\n\n\n

The area of the outer square is 289 square meters.<\/p>\n\n\n\n

Finally, subtract the inner area from the outer area to find the area of the shaded region.<\/p>\n\n\n\n

shaded area = 289 \u2013 36<\/p>\n\n\n\n

= 253<\/p>\n\n\n\n

The area of the shaded region is 253 square meters.<\/p>\n<\/div><\/div>\n\n\n\n

let’s practice!\ud83d\udd8a\ufe0f<\/p>\n<\/div>\n<\/div>\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 between two shapes Key Notes : To find the area of a figure with a hole in it, subtract the area of the inner shape from the area of the outer shape. Area of a triangle: area = 1\/2 . base . height Learn with an example What is the area of the shaded region? Write your answer as a wholeContinue reading “M.8 Area between two shapes”<\/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-257","page","type-page","status-publish","hentry","entry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages\/257","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=257"}],"version-history":[{"count":10,"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages\/257\/revisions"}],"predecessor-version":[{"id":17522,"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages\/257\/revisions\/17522"}],"wp:attachment":[{"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/media?parent=257"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}