{"id":1500,"date":"2017-02-10T19:08:48","date_gmt":"2017-02-10T19:08:48","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/waymakermath4libarts\/?post_type=chapter&#038;p=1500"},"modified":"2019-05-30T16:27:56","modified_gmt":"2019-05-30T16:27:56","slug":"geometry","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/waymakermath4libarts\/chapter\/geometry\/","title":{"raw":"A Bit of Geometry","rendered":"A Bit of Geometry"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Given the part and the whole, write a percent<\/li>\r\n \t<li>Calculate both relative\u00a0and absolute change of a quantity<\/li>\r\n \t<li>Calculate tax on a purchase<\/li>\r\n<\/ul>\r\n<\/div>\r\nGeometric shapes, as well as area and volumes, can often be important in problem solving.\r\n\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/11\/14214054\/10131639494_43480a0a1f_z.jpg\"><img class=\"aligncenter size-full wp-image-501\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/11\/14214054\/10131639494_43480a0a1f_z.jpg\" alt=\"terraced hillside of rice paddies\" width=\"640\" height=\"427\" \/><\/a>\r\n\r\nLet's start things off with an example, rather than trying to explain geometric concepts to you.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\n<div>\r\n\r\nYou are curious how tall a tree is, but don\u2019t have any way to climb it. Describe a method for determining the height.\r\n[reveal-answer q=\"924148\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"924148\"]\r\n\r\nThere are several approaches we could take. We\u2019ll use one based on triangles, which requires that it\u2019s a sunny day. Suppose the tree is casting a shadow, say 15 ft long. I can then have a friend help me measure my own shadow. Suppose I am 6 ft tall, and cast a 1.5 ft shadow. Since the triangle formed by the tree and its shadow has the same angles as the triangle formed by me and my shadow, these triangles are called <strong>similar triangles<\/strong> and their sides will scale proportionally. In other words, the ratio of height to width will be the same in both triangles. Using this, we can find the height of the tree, which we\u2019ll denote by <em>h<\/em>:\r\n\r\n[latex]\\frac{6\\text{ ft. tall}}{1.5\\text{ ft. shadow}}=\\frac{h\\text{ ft. tall}}{15\\text{ ft. shadow}}[\/latex]\r\n<div>\r\n\r\nMultiplying both sides by 15, we get <em>h<\/em> = 60. The tree is about 60 ft tall.\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<h2>Similar Triangles<\/h2>\r\nWe introduced the idea of similar triangles in the previous example. One property of geometric shapes that we have learned is a helpful problem-solving tool is that of similarity. \u00a0If two triangles are the same, meaning the angles between the sides are all the same, we can find an unknown length or height as in the last example. This idea of similarity holds for other geometric shapes as well.\r\n<div class=\"textbox exercises\">\r\n<h3>Guided Example<\/h3>\r\nMary was out in the yard one day and had her two daughters with her. She\u00a0was doing some renovations and wanted to know how tall the house was.\u00a0She noticed a shadow 3 feet long when her daughter was standing 12 feet\u00a0from the house and used it to set up figure 1.\r\n\r\n[caption id=\"attachment_447\" align=\"alignnone\" width=\"358\"]<img class=\"wp-image-447 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/282\/2016\/01\/20155003\/Fig2_3_1.png\" alt=\"Fig2_3_1\" width=\"358\" height=\"339\" \/> Figure 1.[\/caption]\r\n\r\nWe can take that drawing and separate the two triangles as follows allowing us to\u00a0focus on the numbers and the shapes.\r\n\r\nThese triangles are what are called <strong>similar triangles<\/strong>. They have the\u00a0same angles and sides in <em>proportion<\/em>\u00a0to each other. We can use that\u00a0information to determine the height\u00a0of the house as seen in figure 2.\r\n\r\n[caption id=\"attachment_448\" align=\"alignnone\" width=\"419\"]<img class=\"size-full wp-image-448\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/282\/2016\/01\/20155004\/Fig2_3_2.png\" alt=\"Figure 2.\" width=\"419\" height=\"268\" \/> Figure 2.[\/caption]\r\n\r\nTo determine the height of the house, we set up the following proportion:\r\n\r\n[latex]\\displaystyle\\frac{x}{15}=\\frac{5}{3}\\\\[\/latex]\r\n\r\nThen, we solve for the unknown <em>x<\/em> by using cross products as we have done before:\r\n\r\n[latex]\\displaystyle{x}=\\frac{5\\times{15}}{3}=\\frac{75}{3}=25\\\\[\/latex]\r\n\r\nTherefore, we can conclude that the house is 25 feet high.\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<iframe id=\"mom5\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=42493&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"460\"><\/iframe>\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\nIt may be helpful to recall some formulas for areas and volumes of a few basic shapes:\r\n<div class=\"textbox\">\r\n<h3>Areas<\/h3>\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><strong>Rectangle<\/strong><\/td>\r\n<td><strong>Circle<\/strong>, radius <em>r<\/em><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Area: [latex]L\\times{W}[\/latex]<\/td>\r\n<td>Area: [latex]\\pi{r^2}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Perimeter: [latex]2l+2W[\/latex]<\/td>\r\n<td>\u00a0Circumference[latex]2\\pi{r}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2017\/02\/10191723\/Screen-Shot-2017-02-10-at-11.17.08-AM.png\">\r\n<img class=\" wp-image-1502 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2017\/02\/10191723\/Screen-Shot-2017-02-10-at-11.17.08-AM-300x82.png\" alt=\"\" width=\"531\" height=\"146\" \/><\/a>\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox\">\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><strong>Rectangular Box<\/strong><\/td>\r\n<td>Cylinder<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Volume: [latex]L\\times{W}\\times{H}[\/latex]<em>\u00a0 \u00a0<\/em><\/td>\r\n<td>\u00a0<em>\u00a0<\/em>Volume: [latex]\\pi{r^2}h[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<em>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2017\/02\/10192318\/Screen-Shot-2017-02-10-at-11.23.02-AM.png\"><img class=\" wp-image-1503 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2017\/02\/10192318\/Screen-Shot-2017-02-10-at-11.23.02-AM-300x82.png\" alt=\"\" width=\"512\" height=\"140\" \/><\/a><\/em>\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\nIn our next two examples, we will combine the ideas we have explored about ratios with the geometry of some basic shapes to answer questions. \u00a0In the first example, we will predict how much dough will be needed for a pizza that is 16 inches in diameter given that we know how much dough it takes for a pizza with a diameter of 12 inches. The second example uses the volume of a cylinder to determine the number of calories in a marshmallow.\r\n<div class=\"textbox exercises\">\r\n<h3>Examples<\/h3>\r\nIf a 12 inch diameter pizza requires 10 ounces of dough, how much dough is needed for a 16 inch pizza?\r\n[reveal-answer q=\"967082\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"967082\"]\r\n\r\nTo answer this question, we need to consider how the weight of the dough will scale. The weight will be based on the volume of the dough. However, since both pizzas will be about the same thickness, the weight will scale with the area of the top of the pizza. We can find the area of each pizza using the formula for area of a circle, [latex]A=\\pi{r}^2[\/latex]:\r\n\r\nA 12\" pizza has radius 6 inches, so the area will be [latex]\\pi6^2[\/latex]\u00a0= about 113 square inches.\r\n\r\nA 16\" pizza has radius 8 inches, so the area will be [latex]\\pi8^2[\/latex]\u00a0= about 201 square inches.\r\n\r\nNotice that if both pizzas were 1 inch thick, the volumes would be 113 in<sup>3<\/sup> and 201 in<sup>3<\/sup> respectively, which are at the same ratio as the areas. As mentioned earlier, since the thickness is the same for both pizzas, we can safely ignore it.\r\n\r\nWe can now set up a proportion to find the weight of the dough for a 16\" pizza:\r\n\r\n[latex]\\displaystyle\\frac{10\\text{ ounces}}{113\\text{in}^2}=\\frac{x\\text{ ounces}}{201\\text{in}^2}[\/latex]\r\n\r\nMultiply both sides by 201\r\n\r\n[latex]\\displaystyle{x}=201\\cdot\\frac{10}{113}[\/latex] = about 17.8 ounces of dough for a 16\"\u00a0pizza.\r\n\r\nIt is interesting to note that while the diameter is [latex]\\displaystyle\\frac{16}{12}[\/latex] = 1.33 times larger, the dough required, which scales with area, is 1.33<sup>2<\/sup> = 1.78 times larger.\r\n\r\n[\/hidden-answer]\r\n\r\nThe following video illustrates how to solve this problem.\r\n\r\nhttps:\/\/youtu.be\/e75bk1qCsUE\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nA company makes regular and jumbo marshmallows. The regular marshmallow has 25 calories. How many calories will the jumbo marshmallow have?\r\n[reveal-answer q=\"353804\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"353804\"]\r\n\r\nWe would expect the calories to scale with volume. Since the marshmallows have cylindrical shapes, we can use that formula to find the volume. From the grid in the image, we can estimate the radius and height of each marshmallow.\r\n\r\nThe regular marshmallow appears to have a diameter of about 3.5 units, giving a radius of 1.75 units, and a height of about 3.5 units. The volume is about \u03c0(1.75)<sup>2<\/sup>(3.5) = 33.7 units<sup>3<\/sup>.\r\n\r\nThe jumbo marshmallow appears to have a diameter of about 5.5 units, giving a radius of 2.75 units, and a height of about 5 units. The volume is about \u03c0(2.75)<sup>2<\/sup>(5) = 118.8\u00a0units<sup>3<\/sup>.\r\n\r\nWe could now set up a proportion, or use rates. The regular marshmallow has 25 calories for 33.7 cubic units of volume. The jumbo marshmallow will have:\r\n\r\n[latex]\\displaystyle{118.8}\\text{ units}^3\\cdot\\frac{25\\text{ calories}}{33.7\\text{ units}^3}=88.1\\text{ calories}[\/latex]\r\n\r\nIt is interesting to note that while the diameter and height are about 1.5 times larger for the jumbo marshmallow, the volume and calories are about 1.5<sup>3<\/sup> = 3.375 times larger.\r\n\r\n[\/hidden-answer]\r\n\r\nFor more about the marshmallow example, watch this video.\r\n\r\nhttps:\/\/youtu.be\/QJgGpxzRt6Y\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nA website says that you\u2019ll need 48 fifty-pound bags of sand to fill a sandbox that measure 8ft by 8ft by 1ft. How many bags would you need for a sandbox 6ft by 4ft by 1ft?\r\n[reveal-answer q=\"953701\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"953701\"]\r\n<div>\r\n\r\nThe original sandbox has volume [latex]64\\text{ft}^3[\/latex]. The smaller sandbox has volume [latex]24\\text{ft}^3[\/latex].\r\n\r\n[latex]\\displaystyle\\frac{48\\text{bags}}{64\\text{ft}^2}=\\frac{x\\text{ bags}}{24\\text{in}^3}[\/latex] results in <em>x<\/em> = 18 bags.\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\nMary (from the application that started this topic), decides to use what she knows about the height of the roof to measure the height of her second daughter. If her second daughter casts a shadow that is 1.5 feet long when she is 13.5 feet from the house, what is the height of the second daughter? Draw an accurate diagram and use similar triangles to solve.\r\n[reveal-answer q=\"434062\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"434062\"]2.5 ft[\/hidden-answer]\r\n<iframe id=\"mom2\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=17456&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"250\"><\/iframe>\r\n\r\n<\/div>\r\nIn the next section, we will explore the process of combining different types of information to answer questions.","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Given the part and the whole, write a percent<\/li>\n<li>Calculate both relative\u00a0and absolute change of a quantity<\/li>\n<li>Calculate tax on a purchase<\/li>\n<\/ul>\n<\/div>\n<p>Geometric shapes, as well as area and volumes, can often be important in problem solving.<\/p>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/11\/14214054\/10131639494_43480a0a1f_z.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-501\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/11\/14214054\/10131639494_43480a0a1f_z.jpg\" alt=\"terraced hillside of rice paddies\" width=\"640\" height=\"427\" \/><\/a><\/p>\n<p>Let&#8217;s start things off with an example, rather than trying to explain geometric concepts to you.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<div>\n<p>You are curious how tall a tree is, but don\u2019t have any way to climb it. Describe a method for determining the height.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q924148\">Show Solution<\/span><\/p>\n<div id=\"q924148\" class=\"hidden-answer\" style=\"display: none\">\n<p>There are several approaches we could take. We\u2019ll use one based on triangles, which requires that it\u2019s a sunny day. Suppose the tree is casting a shadow, say 15 ft long. I can then have a friend help me measure my own shadow. Suppose I am 6 ft tall, and cast a 1.5 ft shadow. Since the triangle formed by the tree and its shadow has the same angles as the triangle formed by me and my shadow, these triangles are called <strong>similar triangles<\/strong> and their sides will scale proportionally. In other words, the ratio of height to width will be the same in both triangles. Using this, we can find the height of the tree, which we\u2019ll denote by <em>h<\/em>:<\/p>\n<p>[latex]\\frac{6\\text{ ft. tall}}{1.5\\text{ ft. shadow}}=\\frac{h\\text{ ft. tall}}{15\\text{ ft. shadow}}[\/latex]<\/p>\n<div>\n<p>Multiplying both sides by 15, we get <em>h<\/em> = 60. The tree is about 60 ft tall.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<h2>Similar Triangles<\/h2>\n<p>We introduced the idea of similar triangles in the previous example. One property of geometric shapes that we have learned is a helpful problem-solving tool is that of similarity. \u00a0If two triangles are the same, meaning the angles between the sides are all the same, we can find an unknown length or height as in the last example. This idea of similarity holds for other geometric shapes as well.<\/p>\n<div class=\"textbox exercises\">\n<h3>Guided Example<\/h3>\n<p>Mary was out in the yard one day and had her two daughters with her. She\u00a0was doing some renovations and wanted to know how tall the house was.\u00a0She noticed a shadow 3 feet long when her daughter was standing 12 feet\u00a0from the house and used it to set up figure 1.<\/p>\n<div id=\"attachment_447\" style=\"width: 368px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-447\" class=\"wp-image-447 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/282\/2016\/01\/20155003\/Fig2_3_1.png\" alt=\"Fig2_3_1\" width=\"358\" height=\"339\" \/><\/p>\n<p id=\"caption-attachment-447\" class=\"wp-caption-text\">Figure 1.<\/p>\n<\/div>\n<p>We can take that drawing and separate the two triangles as follows allowing us to\u00a0focus on the numbers and the shapes.<\/p>\n<p>These triangles are what are called <strong>similar triangles<\/strong>. They have the\u00a0same angles and sides in <em>proportion<\/em>\u00a0to each other. We can use that\u00a0information to determine the height\u00a0of the house as seen in figure 2.<\/p>\n<div id=\"attachment_448\" style=\"width: 429px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-448\" class=\"size-full wp-image-448\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/282\/2016\/01\/20155004\/Fig2_3_2.png\" alt=\"Figure 2.\" width=\"419\" height=\"268\" \/><\/p>\n<p id=\"caption-attachment-448\" class=\"wp-caption-text\">Figure 2.<\/p>\n<\/div>\n<p>To determine the height of the house, we set up the following proportion:<\/p>\n<p>[latex]\\displaystyle\\frac{x}{15}=\\frac{5}{3}\\\\[\/latex]<\/p>\n<p>Then, we solve for the unknown <em>x<\/em> by using cross products as we have done before:<\/p>\n<p>[latex]\\displaystyle{x}=\\frac{5\\times{15}}{3}=\\frac{75}{3}=25\\\\[\/latex]<\/p>\n<p>Therefore, we can conclude that the house is 25 feet high.<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom5\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=42493&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"460\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p>It may be helpful to recall some formulas for areas and volumes of a few basic shapes:<\/p>\n<div class=\"textbox\">\n<h3>Areas<\/h3>\n<table>\n<tbody>\n<tr>\n<td><strong>Rectangle<\/strong><\/td>\n<td><strong>Circle<\/strong>, radius <em>r<\/em><\/td>\n<\/tr>\n<tr>\n<td>Area: [latex]L\\times{W}[\/latex]<\/td>\n<td>Area: [latex]\\pi{r^2}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Perimeter: [latex]2l+2W[\/latex]<\/td>\n<td>\u00a0Circumference[latex]2\\pi{r}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2017\/02\/10191723\/Screen-Shot-2017-02-10-at-11.17.08-AM.png\"><br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1502 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2017\/02\/10191723\/Screen-Shot-2017-02-10-at-11.17.08-AM-300x82.png\" alt=\"\" width=\"531\" height=\"146\" \/><\/a><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox\">\n<table>\n<tbody>\n<tr>\n<td><strong>Rectangular Box<\/strong><\/td>\n<td>Cylinder<\/td>\n<\/tr>\n<tr>\n<td>Volume: [latex]L\\times{W}\\times{H}[\/latex]<em>\u00a0 \u00a0<\/em><\/td>\n<td>\u00a0<em>\u00a0<\/em>Volume: [latex]\\pi{r^2}h[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><em>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2017\/02\/10192318\/Screen-Shot-2017-02-10-at-11.23.02-AM.png\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1503 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2017\/02\/10192318\/Screen-Shot-2017-02-10-at-11.23.02-AM-300x82.png\" alt=\"\" width=\"512\" height=\"140\" \/><\/a><\/em><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p>In our next two examples, we will combine the ideas we have explored about ratios with the geometry of some basic shapes to answer questions. \u00a0In the first example, we will predict how much dough will be needed for a pizza that is 16 inches in diameter given that we know how much dough it takes for a pizza with a diameter of 12 inches. The second example uses the volume of a cylinder to determine the number of calories in a marshmallow.<\/p>\n<div class=\"textbox exercises\">\n<h3>Examples<\/h3>\n<p>If a 12 inch diameter pizza requires 10 ounces of dough, how much dough is needed for a 16 inch pizza?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q967082\">Show Solution<\/span><\/p>\n<div id=\"q967082\" class=\"hidden-answer\" style=\"display: none\">\n<p>To answer this question, we need to consider how the weight of the dough will scale. The weight will be based on the volume of the dough. However, since both pizzas will be about the same thickness, the weight will scale with the area of the top of the pizza. We can find the area of each pizza using the formula for area of a circle, [latex]A=\\pi{r}^2[\/latex]:<\/p>\n<p>A 12&#8243; pizza has radius 6 inches, so the area will be [latex]\\pi6^2[\/latex]\u00a0= about 113 square inches.<\/p>\n<p>A 16&#8243; pizza has radius 8 inches, so the area will be [latex]\\pi8^2[\/latex]\u00a0= about 201 square inches.<\/p>\n<p>Notice that if both pizzas were 1 inch thick, the volumes would be 113 in<sup>3<\/sup> and 201 in<sup>3<\/sup> respectively, which are at the same ratio as the areas. As mentioned earlier, since the thickness is the same for both pizzas, we can safely ignore it.<\/p>\n<p>We can now set up a proportion to find the weight of the dough for a 16&#8243; pizza:<\/p>\n<p>[latex]\\displaystyle\\frac{10\\text{ ounces}}{113\\text{in}^2}=\\frac{x\\text{ ounces}}{201\\text{in}^2}[\/latex]<\/p>\n<p>Multiply both sides by 201<\/p>\n<p>[latex]\\displaystyle{x}=201\\cdot\\frac{10}{113}[\/latex] = about 17.8 ounces of dough for a 16&#8243;\u00a0pizza.<\/p>\n<p>It is interesting to note that while the diameter is [latex]\\displaystyle\\frac{16}{12}[\/latex] = 1.33 times larger, the dough required, which scales with area, is 1.33<sup>2<\/sup> = 1.78 times larger.<\/p>\n<\/div>\n<\/div>\n<p>The following video illustrates how to solve this problem.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Scaling with area\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/e75bk1qCsUE?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>A company makes regular and jumbo marshmallows. The regular marshmallow has 25 calories. How many calories will the jumbo marshmallow have?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q353804\">Show Solution<\/span><\/p>\n<div id=\"q353804\" class=\"hidden-answer\" style=\"display: none\">\n<p>We would expect the calories to scale with volume. Since the marshmallows have cylindrical shapes, we can use that formula to find the volume. From the grid in the image, we can estimate the radius and height of each marshmallow.<\/p>\n<p>The regular marshmallow appears to have a diameter of about 3.5 units, giving a radius of 1.75 units, and a height of about 3.5 units. The volume is about \u03c0(1.75)<sup>2<\/sup>(3.5) = 33.7 units<sup>3<\/sup>.<\/p>\n<p>The jumbo marshmallow appears to have a diameter of about 5.5 units, giving a radius of 2.75 units, and a height of about 5 units. The volume is about \u03c0(2.75)<sup>2<\/sup>(5) = 118.8\u00a0units<sup>3<\/sup>.<\/p>\n<p>We could now set up a proportion, or use rates. The regular marshmallow has 25 calories for 33.7 cubic units of volume. The jumbo marshmallow will have:<\/p>\n<p>[latex]\\displaystyle{118.8}\\text{ units}^3\\cdot\\frac{25\\text{ calories}}{33.7\\text{ units}^3}=88.1\\text{ calories}[\/latex]<\/p>\n<p>It is interesting to note that while the diameter and height are about 1.5 times larger for the jumbo marshmallow, the volume and calories are about 1.5<sup>3<\/sup> = 3.375 times larger.<\/p>\n<\/div>\n<\/div>\n<p>For more about the marshmallow example, watch this video.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Scaling with volume\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/QJgGpxzRt6Y?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>A website says that you\u2019ll need 48 fifty-pound bags of sand to fill a sandbox that measure 8ft by 8ft by 1ft. How many bags would you need for a sandbox 6ft by 4ft by 1ft?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q953701\">Show Solution<\/span><\/p>\n<div id=\"q953701\" class=\"hidden-answer\" style=\"display: none\">\n<div>\n<p>The original sandbox has volume [latex]64\\text{ft}^3[\/latex]. The smaller sandbox has volume [latex]24\\text{ft}^3[\/latex].<\/p>\n<p>[latex]\\displaystyle\\frac{48\\text{bags}}{64\\text{ft}^2}=\\frac{x\\text{ bags}}{24\\text{in}^3}[\/latex] results in <em>x<\/em> = 18 bags.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Mary (from the application that started this topic), decides to use what she knows about the height of the roof to measure the height of her second daughter. If her second daughter casts a shadow that is 1.5 feet long when she is 13.5 feet from the house, what is the height of the second daughter? Draw an accurate diagram and use similar triangles to solve.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q434062\">Show Solution<\/span><\/p>\n<div id=\"q434062\" class=\"hidden-answer\" style=\"display: none\">2.5 ft<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom2\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=17456&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\n<p>In the next section, we will explore the process of combining different types of information to answer questions.<\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1500\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Math in Society. <strong>Authored by<\/strong>: Lippman, David. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/www.opentextbookstore.com\/mathinsociety\/\">http:\/\/www.opentextbookstore.com\/mathinsociety\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-sa\/4.0\/\">CC BY-SA: Attribution-ShareAlike<\/a><\/em><\/li><li>Rice Terraces. <strong>Authored by<\/strong>: Hoang Giang Hai . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/www.stockpholio.net\/view\/image\/id\/10131639494#!Rice+Terraces\">http:\/\/www.stockpholio.net\/view\/image\/id\/10131639494#!Rice+Terraces<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":21,"menu_order":5,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Math in Society\",\"author\":\"Lippman, David\",\"organization\":\"\",\"url\":\"http:\/\/www.opentextbookstore.com\/mathinsociety\/\",\"project\":\"\",\"license\":\"cc-by-sa\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Rice Terraces\",\"author\":\"Hoang Giang Hai 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