{"id":2812,"date":"2023-05-15T18:41:05","date_gmt":"2023-05-15T18:41:05","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/?post_type=chapter&#038;p=2812"},"modified":"2025-08-26T03:56:20","modified_gmt":"2025-08-26T03:56:20","slug":"volume-and-surface-area-learn-it-3","status":"web-only","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/chapter\/volume-and-surface-area-learn-it-3\/","title":{"raw":"Volume and Surface Area: Learn It 3","rendered":"Volume and Surface Area: Learn It 3"},"content":{"raw":"<h2>Finding the Volume and Surface Area of a Cylinder<\/h2>\r\n<p>If you have ever seen a can of vegetables, you know what a cylinder looks like. A <strong>cylinder <\/strong>is a solid figure with two parallel circles of the same size at the top and bottom. The top and bottom of a cylinder are called the <strong>bases<\/strong>. The <strong>height <\/strong>[latex]h[\/latex] of a cylinder is the distance between the two bases. For all the cylinders we will work with here, the sides and the height, [latex]h[\/latex] , will be perpendicular to the bases.<\/p>\r\n<center>\r\n[caption id=\"\" align=\"aligncenter\" width=\"170\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224148\/CNX_BMath_Figure_09_06_020_img.png\" alt=\"A cylinder with an arrow pointing to the radius of the top labeling it r, radius. There is an arrow pointing to the height of the cylinder labeling it h, height.\" width=\"170\" height=\"156\" \/> Figure 1. The cylinder's radius, r, and height, h, are labeled[\/caption]\r\n<\/center>\r\n<p>&nbsp;<\/p>\r\n<section class=\"textbox keyTakeaway\">\r\n<div>\r\n<h3>volume and surface area of a cylinder<\/h3>\r\n<p>For a cylinder with radius [latex]r[\/latex] and height [latex]h[\/latex]:<\/p>\r\n<p>&nbsp;<\/p>\r\n<center><img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224155\/CNX_BMath_Figure_09_06_024.png\" alt=\"A cylinder, with the height labeled h and the radius of the top labeled r. Beside it is Volume: V equals pi times r squared times h or V equals capital B times h. Below this is Surface Area: S equals 2 times pi times r squared plus 2 times pi times r times h.\" width=\"329\" height=\"136\" \/><\/center>\r\n<p>&nbsp;<\/p>\r\n<p>For a cylinder, the area of the base, [latex]B[\/latex], is the area of its circular base, [latex]\\pi {r}^{2}[\/latex]. <em>This is different from the area of the base for rectangular solids.<\/em><\/p>\r\n<\/div>\r\n<\/section>\r\n<p>How did we come up with the two equations above?<\/p>\r\n<p>Seeing how a cylinder is similar to a rectangular solid may make it easier to understand the formula for the volume of a cylinder.<\/p>\r\n<p>For the rectangular solid, the area of the base, [latex]B[\/latex] , is the area of the rectangular base, length \u00d7 width. For a cylinder, the area of the base, [latex]B[\/latex], is the area of its circular base, [latex]\\pi {r}^{2}[\/latex]. The image below compares how the formula [latex]V=Bh[\/latex] is used for rectangular solids and cylinders.<\/p>\r\n<center>\r\n[caption id=\"\" align=\"aligncenter\" width=\"246\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224149\/CNX_BMath_Figure_09_06_021.png\" alt=\"In (a), a rectangular solid is shown. The sides are labeled L, W, and H. Below this is V equals capital Bh, then V equals Base times h, then V equals parentheses lw times h, then V equals lwh. In (b), a cylinder is shown. The radius of the top is labeled r, the height is labeled h. Below this is V equals capital Bh, then V equals Base times h, then V equals parentheses pi r squared times h, then V equals pi times r squared times h.\" width=\"246\" height=\"286\" \/> Figure 2. Similar to the rectangular solid, the cylinder's volume can be calculated using the base and height[\/caption]\r\n<\/center>\r\n<p>&nbsp;<\/p>\r\n<p>To understand the formula for the surface area of a cylinder, think of a can of vegetables. It has three surfaces: the top, the bottom, and the piece that forms the sides of the can. If you carefully cut the label off the side of the can and unroll it, you will see that it is a rectangle. (See the image below.) The length of the rectangle is the circumference of the cylinder\u2019s base, and the width is the height of the cylinder.<\/p>\r\n<center>\r\n[caption id=\"\" align=\"aligncenter\" width=\"431\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224151\/CNX_BMath_Figure_09_06_022.png\" alt=\"A cylindrical can of green beans. The height is labeled h. Beside this are pictures of circles for the top and bottom of the can and a rectangle for the other portion of the can. Above the circles is C equals 2 times pi times r. The top of the rectangle says l equals 2 times pi times r. The left side of the rectangle is labeled h, the right side is labeled w.\" width=\"431\" height=\"153\" \/> Figure 3. To understand the cylinder's surface area, imagine a rectangle[\/caption]\r\n<\/center>\r\n<p>&nbsp;<\/p>\r\n<p>The distance around the edge of the can is the circumference of the cylinder\u2019s base it is also the length [latex]L[\/latex] of the rectangular label. The height of the cylinder is the width [latex]W[\/latex] of the rectangular label. So the area of the label can be represented as:<\/p>\r\n<center>\r\n[caption id=\"\" align=\"aligncenter\" width=\"79\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224152\/CNX_BMath_Figure_09_06_023_img.png\" alt=\"The top line says A equals l times red w. Below the l is 2 times pi times r. Below the w is a red h.\" width=\"79\" height=\"74\" \/> Figure 3. The cylinder's surface area formula[\/caption]\r\n<\/center>\r\n<p>&nbsp;<\/p>\r\n<p>To find the total surface area of the cylinder, we add the areas of the two circles to the area of the rectangle.<\/p>\r\n<center>\r\n[caption id=\"\" align=\"aligncenter\" width=\"326\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224153\/CNX_BMath_Figure_09_06_044_img.png\" alt=\"A rectangle with circles coming off the top and bottom. One side of the rectangle is labeled h. The circles are labeled 2 pi r. Beneath these, it reads S = A top circle plus A bottom circle plus A rectangle. Beneath this, it says S = pi r squared plus pi r squared plus 2 pi r times h. Beneath this, it says S = 2 times pi r squared plus 2 pi r h. Beneath this, it says S = 2 pi r squared plus 2 pi r h.\" width=\"326\" height=\"236\" \/> Figure 4. The areas of the two circles and rectangle are added to determine the surface area of the cylinder[\/caption]\r\n<\/center>\r\n<p>&nbsp;<\/p>\r\n<p>The surface area of a cylinder with radius [latex]r[\/latex] and height [latex]h[\/latex], is:<\/p>\r\n<center>[latex]S=2\\pi {r}^{2}+2\\pi rh[\/latex]<\/center>\r\n<section class=\"textbox example\">A cylinder has height [latex]5[\/latex] centimeters and radius [latex]3[\/latex] centimeters. Find its:\r\n\r\n<ol style=\"list-style-type: decimal;\">\r\n\t<li>volume<\/li>\r\n\t<li>surface area<\/li>\r\n<\/ol>\r\n<p>[reveal-answer q=\"57883\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"57883\"]<br \/>\r\nStep 1 is the same for both 1. and 2., so we will show it just once.<\/p>\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td>\r\n<p>Step 1. <strong>Read<\/strong> the problem. Draw the figure and label <br \/>\r\nit with the given information.<\/p>\r\n<\/td>\r\n<td><img class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224156\/CNX_BMath_Figure_09_06_046_img-01.png\" alt=\"A cylinder with height 5 and radius 3.\" width=\"155\" height=\"191\" \/><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<ol style=\"list-style-type: decimal;\">\r\n\t<li>\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td>Step 2. <strong>Identify<\/strong> what you are looking for.<\/td>\r\n<td>the volume of the cylinder<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 3. <strong>Name.<\/strong> Choose a variable to represent it.<\/td>\r\n<td>let <em>V<\/em> = volume<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>\r\n<p>Step 4. <strong>Translate.<\/strong><\/p>\r\n<p>Write the appropriate formula.<\/p>\r\n<p>Substitute. (Use [latex]3.14[\/latex] for [latex]\\pi [\/latex] )<\/p>\r\n<\/td>\r\n<td>\r\n<p>[latex]V=\\pi {r}^{2}h[\/latex]<\/p>\r\n<p>[latex]V\\approx \\left(3.14\\right){3}^{2}\\cdot 5[\/latex]<\/p>\r\n<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 5. <strong>Solve.<\/strong><\/td>\r\n<td>[latex]V\\approx 141.3[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 6. <strong>Check:<\/strong> We leave it to you to check your calculations.<\/td>\r\n<td>&nbsp;<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 7. <strong>Answer<\/strong> the question.<\/td>\r\n<td>The volume is approximately [latex]141.3[\/latex] cubic inches.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/li>\r\n\t<li>\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td>Step 2. <strong>Identify<\/strong> what you are looking for.<\/td>\r\n<td>the surface area of the cylinder<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 3. <strong>Name.<\/strong> Choose a variable to represent it.<\/td>\r\n<td>let <em>S<\/em> = surface area<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>\r\n<p>Step 4. <strong>Translate.<\/strong><\/p>\r\n<p>Write the appropriate formula.<\/p>\r\n<p>Substitute. (Use [latex]3.14[\/latex] for [latex]\\pi [\/latex] )<\/p>\r\n<\/td>\r\n<td>\r\n<p>[latex]S=2\\pi {r}^{2}+2\\pi rh[\/latex]<\/p>\r\n<p>[latex]S\\approx 2\\left(3.14\\right){3}^{2}+2\\left(3.14\\right)\\left(3\\right)5[\/latex]<\/p>\r\n<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 5. <strong>Solve.<\/strong><\/td>\r\n<td>[latex]S\\approx 150.72[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 6. <strong>Check:<\/strong> We leave it to you to check your calculations.<\/td>\r\n<td>&nbsp;<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 7. <strong>Answer<\/strong> the question.<\/td>\r\n<td>The surface area is approximately [latex]150.72[\/latex] square inches.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/li>\r\n<\/ol>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]6985[\/ohm2_question]<\/section>","rendered":"<h2>Finding the Volume and Surface Area of a Cylinder<\/h2>\n<p>If you have ever seen a can of vegetables, you know what a cylinder looks like. A <strong>cylinder <\/strong>is a solid figure with two parallel circles of the same size at the top and bottom. The top and bottom of a cylinder are called the <strong>bases<\/strong>. The <strong>height <\/strong>[latex]h[\/latex] of a cylinder is the distance between the two bases. For all the cylinders we will work with here, the sides and the height, [latex]h[\/latex] , will be perpendicular to the bases.<\/p>\n<div style=\"text-align: center;\">\n<figure style=\"width: 170px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224148\/CNX_BMath_Figure_09_06_020_img.png\" alt=\"A cylinder with an arrow pointing to the radius of the top labeling it r, radius. There is an arrow pointing to the height of the cylinder labeling it h, height.\" width=\"170\" height=\"156\" \/><figcaption class=\"wp-caption-text\">Figure 1. The cylinder&#8217;s radius, r, and height, h, are labeled<\/figcaption><\/figure>\n<\/div>\n<p>&nbsp;<\/p>\n<section class=\"textbox keyTakeaway\">\n<div>\n<h3>volume and surface area of a cylinder<\/h3>\n<p>For a cylinder with radius [latex]r[\/latex] and height [latex]h[\/latex]:<\/p>\n<p>&nbsp;<\/p>\n<div style=\"text-align: center;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224155\/CNX_BMath_Figure_09_06_024.png\" alt=\"A cylinder, with the height labeled h and the radius of the top labeled r. Beside it is Volume: V equals pi times r squared times h or V equals capital B times h. Below this is Surface Area: S equals 2 times pi times r squared plus 2 times pi times r times h.\" width=\"329\" height=\"136\" \/><\/div>\n<p>&nbsp;<\/p>\n<p>For a cylinder, the area of the base, [latex]B[\/latex], is the area of its circular base, [latex]\\pi {r}^{2}[\/latex]. <em>This is different from the area of the base for rectangular solids.<\/em><\/p>\n<\/div>\n<\/section>\n<p>How did we come up with the two equations above?<\/p>\n<p>Seeing how a cylinder is similar to a rectangular solid may make it easier to understand the formula for the volume of a cylinder.<\/p>\n<p>For the rectangular solid, the area of the base, [latex]B[\/latex] , is the area of the rectangular base, length \u00d7 width. For a cylinder, the area of the base, [latex]B[\/latex], is the area of its circular base, [latex]\\pi {r}^{2}[\/latex]. The image below compares how the formula [latex]V=Bh[\/latex] is used for rectangular solids and cylinders.<\/p>\n<div style=\"text-align: center;\">\n<figure style=\"width: 246px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224149\/CNX_BMath_Figure_09_06_021.png\" alt=\"In (a), a rectangular solid is shown. The sides are labeled L, W, and H. Below this is V equals capital Bh, then V equals Base times h, then V equals parentheses lw times h, then V equals lwh. In (b), a cylinder is shown. The radius of the top is labeled r, the height is labeled h. Below this is V equals capital Bh, then V equals Base times h, then V equals parentheses pi r squared times h, then V equals pi times r squared times h.\" width=\"246\" height=\"286\" \/><figcaption class=\"wp-caption-text\">Figure 2. Similar to the rectangular solid, the cylinder&#8217;s volume can be calculated using the base and height<\/figcaption><\/figure>\n<\/div>\n<p>&nbsp;<\/p>\n<p>To understand the formula for the surface area of a cylinder, think of a can of vegetables. It has three surfaces: the top, the bottom, and the piece that forms the sides of the can. If you carefully cut the label off the side of the can and unroll it, you will see that it is a rectangle. (See the image below.) The length of the rectangle is the circumference of the cylinder\u2019s base, and the width is the height of the cylinder.<\/p>\n<div style=\"text-align: center;\">\n<figure style=\"width: 431px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224151\/CNX_BMath_Figure_09_06_022.png\" alt=\"A cylindrical can of green beans. The height is labeled h. Beside this are pictures of circles for the top and bottom of the can and a rectangle for the other portion of the can. Above the circles is C equals 2 times pi times r. The top of the rectangle says l equals 2 times pi times r. The left side of the rectangle is labeled h, the right side is labeled w.\" width=\"431\" height=\"153\" \/><figcaption class=\"wp-caption-text\">Figure 3. To understand the cylinder&#8217;s surface area, imagine a rectangle<\/figcaption><\/figure>\n<\/div>\n<p>&nbsp;<\/p>\n<p>The distance around the edge of the can is the circumference of the cylinder\u2019s base it is also the length [latex]L[\/latex] of the rectangular label. The height of the cylinder is the width [latex]W[\/latex] of the rectangular label. So the area of the label can be represented as:<\/p>\n<div style=\"text-align: center;\">\n<figure style=\"width: 79px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224152\/CNX_BMath_Figure_09_06_023_img.png\" alt=\"The top line says A equals l times red w. Below the l is 2 times pi times r. Below the w is a red h.\" width=\"79\" height=\"74\" \/><figcaption class=\"wp-caption-text\">Figure 3. The cylinder&#8217;s surface area formula<\/figcaption><\/figure>\n<\/div>\n<p>&nbsp;<\/p>\n<p>To find the total surface area of the cylinder, we add the areas of the two circles to the area of the rectangle.<\/p>\n<div style=\"text-align: center;\">\n<figure style=\"width: 326px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224153\/CNX_BMath_Figure_09_06_044_img.png\" alt=\"A rectangle with circles coming off the top and bottom. One side of the rectangle is labeled h. The circles are labeled 2 pi r. Beneath these, it reads S = A top circle plus A bottom circle plus A rectangle. Beneath this, it says S = pi r squared plus pi r squared plus 2 pi r times h. Beneath this, it says S = 2 times pi r squared plus 2 pi r h. Beneath this, it says S = 2 pi r squared plus 2 pi r h.\" width=\"326\" height=\"236\" \/><figcaption class=\"wp-caption-text\">Figure 4. The areas of the two circles and rectangle are added to determine the surface area of the cylinder<\/figcaption><\/figure>\n<\/div>\n<p>&nbsp;<\/p>\n<p>The surface area of a cylinder with radius [latex]r[\/latex] and height [latex]h[\/latex], is:<\/p>\n<div style=\"text-align: center;\">[latex]S=2\\pi {r}^{2}+2\\pi rh[\/latex]<\/div>\n<section class=\"textbox example\">A cylinder has height [latex]5[\/latex] centimeters and radius [latex]3[\/latex] centimeters. Find its:<\/p>\n<ol style=\"list-style-type: decimal;\">\n<li>volume<\/li>\n<li>surface area<\/li>\n<\/ol>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q57883\">Show Solution<\/button><\/p>\n<div id=\"q57883\" class=\"hidden-answer\" style=\"display: none\">\nStep 1 is the same for both 1. and 2., so we will show it just once.<\/p>\n<table>\n<tbody>\n<tr>\n<td>\n<p>Step 1. <strong>Read<\/strong> the problem. Draw the figure and label <br \/>\nit with the given information.<\/p>\n<\/td>\n<td><img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224156\/CNX_BMath_Figure_09_06_046_img-01.png\" alt=\"A cylinder with height 5 and radius 3.\" width=\"155\" height=\"191\" \/><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<ol style=\"list-style-type: decimal;\">\n<li>\n<table>\n<tbody>\n<tr>\n<td>Step 2. <strong>Identify<\/strong> what you are looking for.<\/td>\n<td>the volume of the cylinder<\/td>\n<\/tr>\n<tr>\n<td>Step 3. <strong>Name.<\/strong> Choose a variable to represent it.<\/td>\n<td>let <em>V<\/em> = volume<\/td>\n<\/tr>\n<tr>\n<td>\n<p>Step 4. <strong>Translate.<\/strong><\/p>\n<p>Write the appropriate formula.<\/p>\n<p>Substitute. (Use [latex]3.14[\/latex] for [latex]\\pi[\/latex] )<\/p>\n<\/td>\n<td>\n[latex]V=\\pi {r}^{2}h[\/latex]<br \/>\n[latex]V\\approx \\left(3.14\\right){3}^{2}\\cdot 5[\/latex]\n<\/td>\n<\/tr>\n<tr>\n<td>Step 5. <strong>Solve.<\/strong><\/td>\n<td>[latex]V\\approx 141.3[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Step 6. <strong>Check:<\/strong> We leave it to you to check your calculations.<\/td>\n<td>&nbsp;<\/td>\n<\/tr>\n<tr>\n<td>Step 7. <strong>Answer<\/strong> the question.<\/td>\n<td>The volume is approximately [latex]141.3[\/latex] cubic inches.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/li>\n<li>\n<table>\n<tbody>\n<tr>\n<td>Step 2. <strong>Identify<\/strong> what you are looking for.<\/td>\n<td>the surface area of the cylinder<\/td>\n<\/tr>\n<tr>\n<td>Step 3. <strong>Name.<\/strong> Choose a variable to represent it.<\/td>\n<td>let <em>S<\/em> = surface area<\/td>\n<\/tr>\n<tr>\n<td>\n<p>Step 4. <strong>Translate.<\/strong><\/p>\n<p>Write the appropriate formula.<\/p>\n<p>Substitute. (Use [latex]3.14[\/latex] for [latex]\\pi[\/latex] )<\/p>\n<\/td>\n<td>\n[latex]S=2\\pi {r}^{2}+2\\pi rh[\/latex]<br \/>\n[latex]S\\approx 2\\left(3.14\\right){3}^{2}+2\\left(3.14\\right)\\left(3\\right)5[\/latex]\n<\/td>\n<\/tr>\n<tr>\n<td>Step 5. <strong>Solve.<\/strong><\/td>\n<td>[latex]S\\approx 150.72[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Step 6. <strong>Check:<\/strong> We leave it to you to check your calculations.<\/td>\n<td>&nbsp;<\/td>\n<\/tr>\n<tr>\n<td>Step 7. <strong>Answer<\/strong> the question.<\/td>\n<td>The surface area is approximately [latex]150.72[\/latex] square inches.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm6985\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=6985&theme=lumen&iframe_resize_id=ohm6985&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":15,"menu_order":28,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":71,"module-header":"learn_it","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters\/2812"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":31,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters\/2812\/revisions"}],"predecessor-version":[{"id":15660,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters\/2812\/revisions\/15660"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/parts\/71"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters\/2812\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/media?parent=2812"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapter-type?post=2812"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/contributor?post=2812"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/license?post=2812"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}