{"id":2864,"date":"2023-05-16T18:12:48","date_gmt":"2023-05-16T18:12:48","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/?post_type=chapter&#038;p=2864"},"modified":"2024-10-18T20:51:38","modified_gmt":"2024-10-18T20:51:38","slug":"area-and-circumference-fresh-take","status":"web-only","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/chapter\/area-and-circumference-fresh-take\/","title":{"raw":"Area and Circumference: Fresh Take","rendered":"Area and Circumference: Fresh Take"},"content":{"raw":"<section class=\"textbox learningGoals\">\r\n<ul>\r\n\t<li>Find the area of rectangles, triangles, trapezoids, and irregular shapes<\/li>\r\n\t<li>Solve real-life area problems involving rectangles, triangles, and trapezoids<\/li>\r\n\t<li>Find the circumference and area of circular objects<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Find the Perimeter and Area of a Rectangle<\/h2>\r\n<div class=\"textbox shaded\">\r\n<p><strong>The Main Idea<\/strong><\/p>\r\n<p>The <strong>perimeter <\/strong>of a rectangle is the distance around its outer edge. To find the perimeter, you add up<br \/>\r\nthe lengths of all four sides of the rectangle. If the length of the rectangle is represented by [latex]L[\/latex] and<br \/>\r\nthe width by [latex]W[\/latex], the formula for calculating the perimeter is [latex]P = 2(L + W)[\/latex], where<br \/>\r\n[latex]P[\/latex] represents the perimeter.<\/p>\r\n<p>The <strong>area <\/strong>of a rectangle is a measurement that tells us the amount of space enclosed by the rectangle.<br \/>\r\nTo find the area, you multiply the length of the rectangle by its width. The formula for calculating the area is<br \/>\r\n[latex]A = L \u00d7 W[\/latex], where [latex]A[\/latex] represents the area.<\/p>\r\n<\/div>\r\n<section class=\"textbox watchIt\"><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=10350394&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=9HPWB8UX8GQ&amp;video_target=tpm-plugin-bqf5u7as-9HPWB8UX8GQ\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><br \/>\r\n<p>You can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Quantitative+Reasoning+-+2023+Build\/Transcriptions\/How+to+find+the+Area+and+Perimeter+of+a+Rectangle.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cHow to find the Area and Perimeter of a Rectangle\u201d here (opens in new<br \/>\r\nwindow).<\/a><\/p>\r\n<\/section>\r\n<section class=\"textbox example\">The area of a rectangular room is [latex]168[\/latex] square feet. The length is [latex]14[\/latex] feet. What is the width?<br \/>\r\n[reveal-answer q=\"247910\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"247910\"]\r\n\r\n<table id=\"eip-id1168466072798\" class=\"unnumbered unstyled\">\r\n<tbody>\r\n<tr>\r\n<td>Step 1. <strong>Read<\/strong> the problem.<\/td>\r\n<td><img class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223903\/CNX_BMath_Figure_09_04_083_img-01.png\" alt=\"A rectangle with one side labeled 14 ft and another side labeled W. The area is labeled 168 feet squared.\" width=\"337\" height=\"92\" \/><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 2. <strong>Identify<\/strong> what you are looking for.<\/td>\r\n<td>the width of a rectangular room<\/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>W<\/em> = width<\/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 and substitute in the given information.<\/p>\r\n<\/td>\r\n<td>\r\n<p>[latex]A=LW[\/latex]<\/p>\r\n<p>[latex]168=14W[\/latex]<\/p>\r\n<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 5. <strong>Solve<\/strong> the equation.<\/td>\r\n<td>\r\n<p>[latex]\\Large\\frac{168}{14}\\normalsize=<\/p>\r\n<p>\\Large\\frac{14W}{14}[\/latex]<\/p>\r\n<p>[latex]12=W[\/latex]<\/p>\r\n<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>\r\n<p>Step 6. <strong>Check:<\/strong><\/p>\r\n<p>[latex]A=LW[\/latex]<\/p>\r\n<p>[latex]168\\stackrel{?}{=}14\\cdot 12[\/latex]<\/p>\r\n<p>[latex]168=168\\checkmark[\/latex]<\/p>\r\n<\/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 width of the room is [latex]12[\/latex] feet.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">The length of a rectangle is four centimeters more than twice the width. The perimeter is [latex]32[\/latex] centimeters. Find the length and width.<br \/>\r\n[reveal-answer q=\"869311\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"869311\"]\r\n\r\n<table id=\"eip-id1168468670350\" class=\"unnumbered unstyled\">\r\n<tbody>\r\n<tr>\r\n<td>Step 1. <strong>Read<\/strong> the problem.<\/td>\r\n<td>&nbsp;<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 2. <strong>Identify<\/strong> what you are looking for.<\/td>\r\n<td>the length and width<\/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>\r\n<p>let <em>W<\/em> = width<\/p>\r\n<p>The length is four more than twice the width.<\/p>\r\n<p>[latex]2w+4[\/latex] = length<\/p>\r\n<p><img class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223858\/CNX_BMath_Figure_09_04_071_img-01.png\" alt=\"A rectangle with two sides labeled w and two sides labeled 2w + 4.\" width=\"234\" height=\"113\" \/><\/p>\r\n<\/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 and substitute in 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\/24223900\/CNX_BMath_Figure_09_04_071_img-02.png\" alt=\"The equation P = 2L + 2W. Beneath it, it is written again with 32 substituted in for P and 2w+4 substituted in for L.\" width=\"305\" height=\"53\" \/><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 5. <strong>Solve<\/strong> the equation.<\/td>\r\n<td>\r\n<p>[latex]32=4w+8+2w[\/latex]<\/p>\r\n<p>[latex]32=6w+8[\/latex]<\/p>\r\n<p>[latex]24=6w[\/latex]<\/p>\r\n<p>[latex]4=w[\/latex] (width)<\/p>\r\n<p>[latex]2w+4=[\/latex] length<\/p>\r\n<p>[latex]2(\\color{red}{4})+4[\/latex]<\/p>\r\n<p>[latex]8+4=12[\/latex]<\/p>\r\n<p>The length is [latex]12[\/latex]cm.<\/p>\r\n<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>\r\n<p>Step 6. <strong>Check:<\/strong><\/p>\r\n<p>[latex]P=2L+2W[\/latex]<\/p>\r\n<p>[latex]P\\stackrel{?}{=}2\\cdot 12+2\\cdot 4[\/latex]<\/p>\r\n<p>[latex]32=32\\quad\\checkmark[\/latex]<\/p>\r\n<\/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 length is [latex]12[\/latex] cm and the width is [latex]4[\/latex] cm.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">The perimeter of a rectangular swimming pool is [latex]150[\/latex] feet. The length is [latex]15[\/latex] feet more than the width. Find the length and width.<br \/>\r\n[reveal-answer q=\"207235\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"207235\"]\r\n\r\n<table id=\"eip-id1168469520954\" class=\"unnumbered unstyled\">\r\n<tbody>\r\n<tr>\r\n<td>Step 1. <strong>Read<\/strong> the problem. Draw the figure and label it with the given information.<\/td>\r\n<td><img class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223907\/CNX_BMath_Figure_09_04_072.img-01.png\" alt=\"A rectangular swimming pool with one side labeled W and another side labeled W + 15. Beneath it, it says P = 150 feet.\" width=\"209\" height=\"167\" \/><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 2. <strong>Identify<\/strong> what you are looking for.<\/td>\r\n<td>the length and width of the pool<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>\r\n<p>Step 3. <strong>Name.<\/strong> Choose a variable to represent it.<\/p>\r\n<p>The length is [latex]15[\/latex] feet more than the width.<\/p>\r\n<\/td>\r\n<td>\r\n<p>Let [latex]W=\\text{width}[\/latex]<\/p>\r\n<p>[latex]W+15=\\text{length}[\/latex]<\/p>\r\n<\/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 and substitute.<\/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\/24223908\/CNX_BMath_Figure_09_04_072_img-02.png\" alt=\"The equation P = 2L + 2W. Beneath it, the equation is written again with 150 substituted in for P and w + 15 substituted in for L.\" width=\"321\" height=\"68\" \/><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 5. <strong>Solve<\/strong> the equation.<\/td>\r\n<td>\r\n<p>[latex]150=2w+30+2w[\/latex]<\/p>\r\n<p>[latex]150=4w+30[\/latex]<\/p>\r\n<p>[latex]120=4w[\/latex]<\/p>\r\n<p>[latex]30=w[\/latex] the width of the pool.<\/p>\r\n<p>[latex]w+15[\/latex] the length of the pool.<\/p>\r\n<p>[latex]\\color{red}{30}+15[\/latex]<\/p>\r\n<p>[latex]45[\/latex]<\/p>\r\n<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>\r\n<p>Step 6. <strong>Check:<\/strong><\/p>\r\n<p>[latex]P=2L+2W[\/latex]<\/p>\r\n<p>[latex]150\\stackrel{?}{=}2(45)+2(30)[\/latex]<\/p>\r\n<p>[latex]150=150\\checkmark[\/latex]<\/p>\r\n<\/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 length of the pool is [latex]45[\/latex] feet and the width is [latex]30[\/latex] feet.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<p>Watch this video to see another similar example of finding area given the relationship between the length and width of a rectangle.<\/p>\r\n<section class=\"textbox watchIt\"><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/zUlU64Umnq4\" width=\"560\" height=\"315\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><br \/>\r\n<p>You can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Quantitative+Reasoning+-+2023+Build\/Transcriptions\/Ex_+Find+the+Area+of+a+Rectangle+Given+the+Perimeter.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cEx: Find the Area of a Rectangle Given the Perimeter\u201d here (opens in<br \/>\r\nnew window).<\/a><\/p>\r\n<\/section>\r\n<h2>Find the Area and Perimeter of a Triangle<\/h2>\r\n<div class=\"textbox shaded\">\r\n<p><strong>The Main Idea <\/strong><\/p>\r\n<p>The <strong>perimeter <\/strong>of a triangle is the total length of its three sides. To find the perimeter, you add up the lengths of all three sides of the triangle. If the lengths of the sides are represented by [latex]a[\/latex], [latex]b[\/latex], and [latex]c[\/latex], the formula for calculating the perimeter is [latex]P = a + b + c[\/latex], where [latex]P[\/latex] represents the perimeter.<\/p>\r\n<p>The <strong>area<\/strong> of a triangle is a measurement that tells us the amount of space enclosed by the triangle. To find the area, you multiply the base of the triangle by its corresponding height and divide the product by [latex]2[\/latex]. The formula for calculating the area is [latex]A = \\frac{1}{2} \u00d7 b \u00d7 h[\/latex], where [latex]A[\/latex] represents the area, [latex]b[\/latex] represents the base, and [latex]h[\/latex] represents the height.<\/p>\r\n<\/div>\r\n<section class=\"textbox watchIt\"><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=10350395&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=l9y6l9LxQ24&amp;video_target=tpm-plugin-w2934u9z-l9y6l9LxQ24\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><br \/>\r\n<p>You can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Quantitative+Reasoning+-+2023+Build\/Transcriptions\/How+to+Find+the+Area+and+Perimeter+of+a+Triangle.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cHow to Find the Area and Perimeter of a Triangle\u201d here (opens in new<br \/>\r\nwindow).<\/a><\/p>\r\n<\/section>\r\n<p>The following video provides another example of how to use the area formula for triangles.<\/p>\r\n<section class=\"textbox watchIt\"><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/jXbPAk2jorM\" width=\"560\" height=\"315\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><br \/>\r\n<p>You can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Quantitative+Reasoning+-+2023+Build\/Transcriptions\/Ex_+Find+the+Area+of+a+Triangle+(Whole+Number).txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cEx: Find the Area of a Triangle (Whole Number)\u201d here (opens in new<br \/>\r\nwindow).<\/a><\/p>\r\n<\/section>\r\n<p>In our next video, we show an example of how to find the height of a triangle given it\u2019s area.<\/p>\r\n<section class=\"textbox watchIt\"><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/C0vUVK_o5r0\" width=\"560\" height=\"315\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><br \/>\r\n<p>You can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Quantitative+Reasoning+-+2023+Build\/Transcriptions\/Ex_+Find+the+Height+of+a+Triangle+Given+Area+(Even+Base).txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cEx: Find the Height of a Triangle Given Area (Even Base)\u201d here (opens<br \/>\r\nin new window).<\/a><\/p>\r\n<\/section>\r\n<section class=\"textbox example\">The perimeter of an equilateral triangle is [latex]93[\/latex] inches. Find the length of each side.<br \/>\r\n[reveal-answer q=\"157458\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"157458\"]\r\n\r\n<table id=\"eip-id1168468574026\" class=\"unnumbered unstyled\" summary=\"Step 1 says, \">\r\n<tbody>\r\n<tr>\r\n<td>Step 1. <strong>Read<\/strong> the problem. Draw the figure and label it with the given information.<\/td>\r\n<td>\r\n<p><img class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223937\/CNX_BMath_Figure_09_04_076_img-01.png\" alt=\"A triangle with each side labeled s\" width=\"185\" height=\"174\" \/><\/p>\r\n<p>Perimeter = [latex]93[\/latex] in.<\/p>\r\n<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 2. <strong>Identify<\/strong> what you are looking for.<\/td>\r\n<td>length of the sides of an equilateral triangle<\/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> = length of each side<\/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.<\/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\/24223938\/CNX_BMath_Figure_09_04_076_img-02.png\" alt=\"The equation P = a + b + c. Beneath it, the equation is written again with 93 substituted in for P and s substituted in for a, b, and c.\" width=\"298\" height=\"60\" \/><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 5. <strong>Solve<\/strong> the equation.<\/td>\r\n<td>\r\n<p>[latex]93=3s[\/latex]<\/p>\r\n<p>[latex]31=s[\/latex]<\/p>\r\n<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>\r\n<p>Step 6. <strong>Check:<\/strong><\/p>\r\n<p><img class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223940\/CNX_BMath_Figure_09_04_076_img-04.png\" alt=\"A triangle with each side labeled 31\" width=\"175\" height=\"173\" \/><\/p>\r\n<p>[latex]93\\stackrel{?}{=}31+31+31[\/latex]<\/p>\r\n<p>[latex]93=93\\quad\\checkmark[\/latex]<\/p>\r\n<\/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>Each side is [latex]31[\/latex] inches.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<h2>Find the Area of a Trapezoid<\/h2>\r\n<div class=\"textbox shaded\">\r\n<p><strong>The Main Idea <\/strong><\/p>\r\n<p>A <strong>trapezoid <\/strong>is a quadrilateral with one pair of parallel sides.<\/p>\r\n<p>To find the <strong>area <\/strong>of a trapezoid, you multiply the sum of the lengths of the parallel sides (base1 ([latex]b[\/latex]) and base2 ([latex]B[\/latex])) by the height ([latex]h[\/latex]) and divide the result by [latex]2[\/latex]. The formula for calculating the area of a trapezoid is [latex]A = \\frac{(b + B) \u00d7 h}{2}[\/latex], where [latex]A[\/latex] represents the area. This can also be written as: [latex]{\\text{Area}}_{\\text{trapezoid}}=\\Large\\frac{1}{2}\\normalsize h\\left(b+B\\right)[\/latex]<\/p>\r\n<\/div>\r\n<section class=\"textbox watchIt\"><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=10350393&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=9hISqaDb6XE&amp;video_target=tpm-plugin-3w5ylzj0-9hISqaDb6XE\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><br \/>\r\n<p>You can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Quantitative+Reasoning+-+2023+Build\/Transcriptions\/Area+of+a+Trapezoid+%7C+MathHelp.com.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cArea of a Trapezoid | MathHelp.com\u201d here (opens in new window).<\/a><\/p>\r\n<\/section>\r\n<p>In the next video we show another example of how to use the formula to find the area of a trapezoid given the<br \/>\r\nlengths of it's height and bases.<\/p>\r\n<section class=\"textbox watchIt\"><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/WNo7s-XoI4w\" width=\"560\" height=\"315\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><br \/>\r\n<p>You can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Quantitative+Reasoning+-+2023+Build\/Transcriptions\/Ex_+Find+the+Area+of+a+Trapezoid.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cEx: Find the Area of a Trapezoid\u201d here (opens in new window).<\/a><\/p>\r\n<\/section>\r\n<section class=\"textbox example\">Find the area of a trapezoid whose height is [latex]5[\/latex] feet and whose bases are [latex]10.3[\/latex] and [latex]13.7[\/latex] feet.<br \/>\r\n[reveal-answer q=\"158293\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"158293\"]\r\n\r\n<table id=\"eip-id1168469577474\" class=\"unnumbered unstyled\" style=\"width: 859px;\" summary=\"Step 1 says, \">\r\n<tbody>\r\n<tr style=\"height: 178.867px;\">\r\n<td style=\"width: 417.267px; height: 178.867px;\">Step 1. <strong>Read<\/strong> the problem. Draw the figure and<br \/>\r\nlabel it with the given information.<\/td>\r\n<td style=\"width: 414.733px; height: 178.867px;\"><img class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223958\/CNX_BMath_Figure_09_04_081_img-01.png\" alt=\"A trapezoid with the smaller base labeled 10.3 feet, the larger base labeled 13.7 feet, and the height labeled 5 feet.\" width=\"421\" height=\"141\" \/><\/td>\r\n<\/tr>\r\n<tr style=\"height: 15px;\">\r\n<td style=\"width: 417.267px; height: 15px;\">Step 2. <strong>Identify<\/strong> what you are looking for.<\/td>\r\n<td style=\"width: 414.733px; height: 15px;\">the area of the trapezoid<\/td>\r\n<\/tr>\r\n<tr style=\"height: 15px;\">\r\n<td style=\"width: 417.267px; height: 15px;\">Step 3. <strong>Name.<\/strong> Choose a variable to represent it.<\/td>\r\n<td style=\"width: 414.733px; height: 15px;\">Let <em>A<\/em> = the area<\/td>\r\n<\/tr>\r\n<tr style=\"height: 102px;\">\r\n<td style=\"width: 417.267px; height: 102px;\">\r\n<p>Step 4. <strong>Translate.<\/strong><\/p>\r\n<p>Write the appropriate formula.<\/p>\r\n<p>Substitute.<\/p>\r\n<\/td>\r\n<td style=\"width: 414.733px; height: 102px;\"><img class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223959\/CNX_BMath_Figure_09_04_081_img-02.png\" alt=\"The equation A = one half times h times the quantity of little b plus big b. Beneath, the equation is rewritten with 5 substituted in for h, 10.3 substituted in for little b and 13.7 substituted in for big b.\" width=\"421\" height=\"99\" \/><\/td>\r\n<\/tr>\r\n<tr style=\"height: 90px;\">\r\n<td style=\"width: 417.267px; height: 90px;\">Step 5. <strong>Solve<\/strong> the equation.<\/td>\r\n<td style=\"width: 414.733px; height: 90px;\">\r\n<p>[latex]A={\\Large\\frac{1}{2}}\\normalsize\\cdot 5(24)[\/latex]<\/p>\r\n<p>[latex]A=12(5)[\/latex]<\/p>\r\n<p>[latex]A=60[\/latex] square feet<\/p>\r\n<\/td>\r\n<\/tr>\r\n<tr style=\"height: 305px;\">\r\n<td style=\"width: 417.267px; height: 305px;\">\r\n<p>Step 6. <strong>Check:<\/strong> Is this answer reasonable?<\/p>\r\n<p>The area of the trapezoid should be less than the area of a rectangle with base [latex]13.7[\/latex] and height<br \/>\r\n[latex]5[\/latex], but more than the area of a rectangle with base [latex]10.3[\/latex] and height [latex]5[\/latex].<\/p>\r\n<p><img class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224003\/CNX_BMath_Figure_09_04_063.png\" alt=\"An image of a trapezoid is shown with a red rectangle drawn around it. The larger base of the trapezoid is labeled 13.7 ft. and is the same as the base of the rectangle. The height of both the trapezoid and the rectangle is 5 ft. Next to this is an image of a trapezoid with a blue rectangle drawn inside it. The smaller base of the trapezoid is labeled 10.3 ft. and is the same as the base of the rectangle. Below the images is A sub red rectangle is greater than A sub trapezoid is greater than A sub blue rectangle. Below this is 68.5, 60, and 51.5.\" width=\"378\" height=\"179\" \/><\/p>\r\n<\/td>\r\n<td style=\"width: 414.733px; height: 305px;\">[latex]\\checkmark[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 15px;\">\r\n<td style=\"width: 417.267px; height: 15px;\">Step 7. <strong>Answer<\/strong> the question.<\/td>\r\n<td style=\"width: 414.733px; height: 15px;\">The area of the trapezoid is [latex]60[\/latex] square feet.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<h2>Find the Circumference and Area of Circles<\/h2>\r\n<div class=\"textbox shaded\">\r\n<p><strong>The Main Idea <\/strong><\/p>\r\n<p>The <strong>circumference <\/strong>of a circle is the distance around its outer edge or boundary. The formula for calculating the circumference of a circle is [latex]C = 2\u03c0r[\/latex] or [latex]C = \u03c0d[\/latex], where [latex]C[\/latex] represents the circumference, [latex]r[\/latex] represents the radius, and [latex]d[\/latex] represents the diameter.<\/p>\r\n<p>The <strong>area <\/strong>of a circle is a measurement that tells us the amount of space enclosed by the circle. The formula for calculating the area of a circle is [latex]A = \u03c0r^2[\/latex], where [latex]A[\/latex] represents the area, and [latex]r[\/latex] represents the radius.<\/p>\r\n<\/div>\r\n<section class=\"textbox watchIt\"><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/O-cawByg2aA\" width=\"560\" height=\"315\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><br \/>\r\n<p>You can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Quantitative+Reasoning+-+2023+Build\/Transcriptions\/Math+Antics+-+Circles%2C+Circumference+And+Area.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cMath Antics - Circles, Circumference And Area\u201d here (opens in new<br \/>\r\nwindow).<\/a><\/p>\r\n<\/section>\r\n<section class=\"textbox example\">A circle has radius [latex]10[\/latex] centimeters. Approximate its:\r\n\r\n<ol style=\"list-style-type: decimal;\">\r\n\t<li>circumference<\/li>\r\n\t<li>area<\/li>\r\n<\/ol>\r\n<p>[reveal-answer q=\"247920\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"247920\"]<\/p>\r\n<ol style=\"list-style-type: decimal;\">\r\n\t<li>\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td>Find the circumference when [latex]r=10[\/latex].<\/td>\r\n<td>&nbsp;<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Write the formula for circumference.<\/td>\r\n<td>[latex]C=2\\pi \\mathit{\\text{r}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Substitute [latex]3.14[\/latex] for [latex]\\pi [\/latex] and 10 for , [latex]r[\/latex] .<\/td>\r\n<td>[latex]C\\approx 2\\left(3.14\\right)\\left(10\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Multiply.<\/td>\r\n<td>[latex]C\\approx 62.8\\text{ centimeters}[\/latex]<\/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>Find the area when [latex]r=10[\/latex].<\/td>\r\n<td>&nbsp;<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Write the formula for area.<\/td>\r\n<td>[latex]A=\\pi {\\mathit{\\text{r}}}^{2}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Substitute [latex]3.14[\/latex] for [latex]\\pi [\/latex] and 10 for [latex]r[\/latex] .<\/td>\r\n<td>[latex]A\\approx \\left(3.14\\right){\\text{(}10\\text{)}}^{2}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Multiply.<\/td>\r\n<td>[latex]A\\approx 314\\text{ square centimeters}[\/latex]<\/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 example\">A circle has a diameter of [latex]14.6[\/latex] inches. Approximate its:\r\n\r\n<ol style=\"list-style-type: decimal;\">\r\n\t<li>circumference<\/li>\r\n\t<li>area<\/li>\r\n<\/ol>\r\n<p>[reveal-answer q=\"272573\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"272573\"]<\/p>\r\n<ol style=\"list-style-type: decimal;\">\r\n\t<li>\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td>Find the circumference when [latex]d=14.6[\/latex]<\/td>\r\n<td>&nbsp;<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Write the formula for circumference.<\/td>\r\n<td>[latex]C=\\pi \\mathit{\\text{d}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Substitute [latex]3.14[\/latex] for [latex]\\pi [\/latex] and [latex]14.6[\/latex] for [latex]d[\/latex]<\/td>\r\n<td>[latex]C\\approx \\left(3.14\\right)\\left(14.6\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Multiply.<\/td>\r\n<td>[latex]C\\approx 45.844\\text{ inches}[\/latex]<\/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 style=\"height: 15px;\">\r\n<td style=\"height: 15px;\">Find the area when [latex]d=14.6[\/latex]. We know [latex]d=2r[\/latex] so for this<br \/>\r\nproblem [latex]r=7.3[\/latex]<\/td>\r\n<td style=\"height: 15px;\">\u00a0<\/td>\r\n<\/tr>\r\n<tr style=\"height: 15px;\">\r\n<td style=\"height: 15px;\">Write the formula for area.<\/td>\r\n<td style=\"height: 15px;\">[latex]A=\\pi {\\mathit{\\text{r}}}^{2}[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 31px;\">\r\n<td style=\"height: 31px;\">Substitute [latex]3.14[\/latex] for [latex]\\pi [\/latex] and [latex]7.3[\/latex] for<br \/>\r\n[latex]r[\/latex] .<\/td>\r\n<td style=\"height: 31px;\">[latex]A\\approx \\left(3.14\\right){\\text{(}7.3\\text{)}}^{2}[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 31px;\">\r\n<td style=\"height: 31px;\">Multiply.<\/td>\r\n<td style=\"height: 31px;\">[latex]A\\approx 171.915\\text{ square inches}[\/latex]<\/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<p>Watch the following video to see another example of how to find the circumference of a circle.<\/p>\r\n<section class=\"textbox watchIt\"><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/sHtsnC2Mgnk\" width=\"560\" height=\"315\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><br \/>\r\n<p>You can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Quantitative+Reasoning+-+2023+Build\/Transcriptions\/Examples_+Determine+the+Circumference+of+a+Circle.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cExamples: Determine the Circumference of a Circle\u201d here (opens in new<br \/>\r\nwindow).<\/a><\/p>\r\n<\/section>\r\n<p>In the next video example, we find the area of a circle.<\/p>\r\n<section class=\"textbox watchIt\"><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/SIKkWLqt2mQ\" width=\"560\" height=\"315\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><br \/>\r\n<p>You can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Quantitative+Reasoning+-+2023+Build\/Transcriptions\/Example_+Determine+the+Area+of+a+Circle.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cExample: Determine the Area of a Circle\u201d here (opens in new<br \/>\r\nwindow).<\/a><\/p>\r\n<\/section>\r\n<h2>Find the Area of Irregular Figures<\/h2>\r\n<p>To find the area of an irregular figure, you break it down into simpler shapes whose areas can be calculated. For<br \/>\r\nexample, you can divide the irregular figure into rectangles, triangles, squares, circles, or other familiar shapes.<br \/>\r\nThen, calculate the area of each individual shape using the appropriate formulas. Once you have found the areas of the<br \/>\r\nsimpler shapes, add them together to find the total area of the irregular figure. Keep in mind that this method works<br \/>\r\nbest when the irregular figure can be divided into known shapes.<\/p>\r\n<p>The following video gives an example of how to find the area of an \u201cL\u201d shaped polygon using the dimensions of two<br \/>\r\nrectangles.<\/p>\r\n<section class=\"textbox watchIt\"><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/lx8vweYTPpg\" width=\"560\" height=\"315\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><br \/>\r\n<p>You can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Quantitative+Reasoning+-+2023+Build\/Transcriptions\/Ex_+Find+the+Area+of+an+L-Shaped+Polygon+Involving+Whole+Numbers.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cEx: Find the Area of an L-Shaped Polygon Involving Whole Numbers\u201d<br \/>\r\nhere (opens in new window).<\/a><\/p>\r\n<\/section>\r\n<p>The next video example find the area of an object containing one semi-circle and a rectangle.<\/p>\r\n<section class=\"textbox watchIt\"><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/085t_Fmje4o\" width=\"560\" height=\"315\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><br \/>\r\n<p>You can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Quantitative+Reasoning+-+2023+Build\/Transcriptions\/Example_+Determine+an+Area+Involving+a+Rectangle+and+Circle.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cExample: Determine an Area Involving a Rectangle and Circle\u201d here<br \/>\r\n(opens in new window).<\/a><\/p>\r\n<\/section>\r\n<p>The next video shows more example of finding the area of irregular shapes.<\/p>\r\n<section class=\"textbox watchIt\"><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=10350392&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=_e7j6rE7_Pg&amp;video_target=tpm-plugin-6t1ssnro-_e7j6rE7_Pg\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><br \/>\r\n<p>You can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Quantitative+Reasoning+-+2023+Build\/Transcriptions\/Area+and+Perimeter+of+Irregular+Shapes+-+Tons+of+Examples!.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cArea and Perimeter of Irregular Shapes - Tons of Examples!\u201d here<br \/>\r\n(opens in new window).<\/a><\/p>\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Find the area of rectangles, triangles, trapezoids, and irregular shapes<\/li>\n<li>Solve real-life area problems involving rectangles, triangles, and trapezoids<\/li>\n<li>Find the circumference and area of circular objects<\/li>\n<\/ul>\n<\/section>\n<h2>Find the Perimeter and Area of a Rectangle<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea<\/strong><\/p>\n<p>The <strong>perimeter <\/strong>of a rectangle is the distance around its outer edge. To find the perimeter, you add up<br \/>\nthe lengths of all four sides of the rectangle. If the length of the rectangle is represented by [latex]L[\/latex] and<br \/>\nthe width by [latex]W[\/latex], the formula for calculating the perimeter is [latex]P = 2(L + W)[\/latex], where<br \/>\n[latex]P[\/latex] represents the perimeter.<\/p>\n<p>The <strong>area <\/strong>of a rectangle is a measurement that tells us the amount of space enclosed by the rectangle.<br \/>\nTo find the area, you multiply the length of the rectangle by its width. The formula for calculating the area is<br \/>\n[latex]A = L \u00d7 W[\/latex], where [latex]A[\/latex] represents the area.<\/p>\n<\/div>\n<section class=\"textbox watchIt\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=10350394&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=9HPWB8UX8GQ&amp;video_target=tpm-plugin-bqf5u7as-9HPWB8UX8GQ\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/p>\n<p>You can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Quantitative+Reasoning+-+2023+Build\/Transcriptions\/How+to+find+the+Area+and+Perimeter+of+a+Rectangle.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cHow to find the Area and Perimeter of a Rectangle\u201d here (opens in new<br \/>\nwindow).<\/a><\/p>\n<\/section>\n<section class=\"textbox example\">The area of a rectangular room is [latex]168[\/latex] square feet. The length is [latex]14[\/latex] feet. What is the width?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q247910\">Show Solution<\/button><\/p>\n<div id=\"q247910\" class=\"hidden-answer\" style=\"display: none\">\n<table id=\"eip-id1168466072798\" class=\"unnumbered unstyled\">\n<tbody>\n<tr>\n<td>Step 1. <strong>Read<\/strong> the problem.<\/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\/24223903\/CNX_BMath_Figure_09_04_083_img-01.png\" alt=\"A rectangle with one side labeled 14 ft and another side labeled W. The area is labeled 168 feet squared.\" width=\"337\" height=\"92\" \/><\/td>\n<\/tr>\n<tr>\n<td>Step 2. <strong>Identify<\/strong> what you are looking for.<\/td>\n<td>the width of a rectangular room<\/td>\n<\/tr>\n<tr>\n<td>Step 3. <strong>Name.<\/strong> Choose a variable to represent it.<\/td>\n<td>Let <em>W<\/em> = width<\/td>\n<\/tr>\n<tr>\n<td>\n<p>Step 4. <strong>Translate.<\/strong><\/p>\n<p>Write the appropriate formula and substitute in the given information.<\/p>\n<\/td>\n<td>\n[latex]A=LW[\/latex]<br \/>\n[latex]168=14W[\/latex]\n<\/td>\n<\/tr>\n<tr>\n<td>Step 5. <strong>Solve<\/strong> the equation.<\/td>\n<td>\n[latex]\\Large\\frac{168}{14}\\normalsize=  <\/p>\n<p>\\Large\\frac{14W}{14}[\/latex]<br \/>\n[latex]12=W[\/latex]\n<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>Step 6. <strong>Check:<\/strong><\/p>\n<p>[latex]A=LW[\/latex]<br \/>\n[latex]168\\stackrel{?}{=}14\\cdot 12[\/latex]<br \/>\n[latex]168=168\\checkmark[\/latex]\n<\/td>\n<td>&nbsp;<\/td>\n<\/tr>\n<tr>\n<td>Step 7. <strong>Answer<\/strong> the question.<\/td>\n<td>The width of the room is [latex]12[\/latex] feet.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">The length of a rectangle is four centimeters more than twice the width. The perimeter is [latex]32[\/latex] centimeters. Find the length and width.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q869311\">Show Solution<\/button><\/p>\n<div id=\"q869311\" class=\"hidden-answer\" style=\"display: none\">\n<table id=\"eip-id1168468670350\" class=\"unnumbered unstyled\">\n<tbody>\n<tr>\n<td>Step 1. <strong>Read<\/strong> the problem.<\/td>\n<td>&nbsp;<\/td>\n<\/tr>\n<tr>\n<td>Step 2. <strong>Identify<\/strong> what you are looking for.<\/td>\n<td>the length and width<\/td>\n<\/tr>\n<tr>\n<td>Step 3. <strong>Name.<\/strong> Choose a variable to represent it.<\/td>\n<td>\n<p>let <em>W<\/em> = width<\/p>\n<p>The length is four more than twice the width.<\/p>\n<p>[latex]2w+4[\/latex] = length<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223858\/CNX_BMath_Figure_09_04_071_img-01.png\" alt=\"A rectangle with two sides labeled w and two sides labeled 2w + 4.\" width=\"234\" height=\"113\" \/><\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>Step 4. <strong>Translate.<\/strong><\/p>\n<p>Write the appropriate formula and substitute in 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\/24223900\/CNX_BMath_Figure_09_04_071_img-02.png\" alt=\"The equation P = 2L + 2W. Beneath it, it is written again with 32 substituted in for P and 2w+4 substituted in for L.\" width=\"305\" height=\"53\" \/><\/td>\n<\/tr>\n<tr>\n<td>Step 5. <strong>Solve<\/strong> the equation.<\/td>\n<td>\n[latex]32=4w+8+2w[\/latex]<br \/>\n[latex]32=6w+8[\/latex]<br \/>\n[latex]24=6w[\/latex]<\/p>\n<p>[latex]4=w[\/latex] (width)<\/p>\n<p>[latex]2w+4=[\/latex] length<\/p>\n<p>[latex]2(\\color{red}{4})+4[\/latex]<br \/>\n[latex]8+4=12[\/latex]<\/p>\n<p>The length is [latex]12[\/latex]cm.<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>Step 6. <strong>Check:<\/strong><\/p>\n<p>[latex]P=2L+2W[\/latex]<br \/>\n[latex]P\\stackrel{?}{=}2\\cdot 12+2\\cdot 4[\/latex]<br \/>\n[latex]32=32\\quad\\checkmark[\/latex]\n<\/td>\n<td>&nbsp;<\/td>\n<\/tr>\n<tr>\n<td>Step 7. <strong>Answer<\/strong> the question.<\/td>\n<td>The length is [latex]12[\/latex] cm and the width is [latex]4[\/latex] cm.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">The perimeter of a rectangular swimming pool is [latex]150[\/latex] feet. The length is [latex]15[\/latex] feet more than the width. Find the length and width.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q207235\">Show Solution<\/button><\/p>\n<div id=\"q207235\" class=\"hidden-answer\" style=\"display: none\">\n<table id=\"eip-id1168469520954\" class=\"unnumbered unstyled\">\n<tbody>\n<tr>\n<td>Step 1. <strong>Read<\/strong> the problem. Draw the figure and label it with the given information.<\/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\/24223907\/CNX_BMath_Figure_09_04_072.img-01.png\" alt=\"A rectangular swimming pool with one side labeled W and another side labeled W + 15. Beneath it, it says P = 150 feet.\" width=\"209\" height=\"167\" \/><\/td>\n<\/tr>\n<tr>\n<td>Step 2. <strong>Identify<\/strong> what you are looking for.<\/td>\n<td>the length and width of the pool<\/td>\n<\/tr>\n<tr>\n<td>\n<p>Step 3. <strong>Name.<\/strong> Choose a variable to represent it.<\/p>\n<p>The length is [latex]15[\/latex] feet more than the width.<\/p>\n<\/td>\n<td>\n<p>Let [latex]W=\\text{width}[\/latex]<\/p>\n<p>[latex]W+15=\\text{length}[\/latex]\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>Step 4. <strong>Translate.<\/strong><\/p>\n<p>Write the appropriate formula and substitute.<\/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\/24223908\/CNX_BMath_Figure_09_04_072_img-02.png\" alt=\"The equation P = 2L + 2W. Beneath it, the equation is written again with 150 substituted in for P and w + 15 substituted in for L.\" width=\"321\" height=\"68\" \/><\/td>\n<\/tr>\n<tr>\n<td>Step 5. <strong>Solve<\/strong> the equation.<\/td>\n<td>\n[latex]150=2w+30+2w[\/latex]<br \/>\n[latex]150=4w+30[\/latex]<br \/>\n[latex]120=4w[\/latex]<\/p>\n<p>[latex]30=w[\/latex] the width of the pool.<\/p>\n<p>[latex]w+15[\/latex] the length of the pool.<\/p>\n<p>[latex]\\color{red}{30}+15[\/latex]<br \/>\n[latex]45[\/latex]\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>Step 6. <strong>Check:<\/strong><\/p>\n<p>[latex]P=2L+2W[\/latex]<br \/>\n[latex]150\\stackrel{?}{=}2(45)+2(30)[\/latex]<br \/>\n[latex]150=150\\checkmark[\/latex]\n<\/td>\n<td>&nbsp;<\/td>\n<\/tr>\n<tr>\n<td>Step 7. <strong>Answer<\/strong> the question.<\/td>\n<td>The length of the pool is [latex]45[\/latex] feet and the width is [latex]30[\/latex] feet.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/section>\n<p>Watch this video to see another similar example of finding area given the relationship between the length and width of a rectangle.<\/p>\n<section class=\"textbox watchIt\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/zUlU64Umnq4\" width=\"560\" height=\"315\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>You can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Quantitative+Reasoning+-+2023+Build\/Transcriptions\/Ex_+Find+the+Area+of+a+Rectangle+Given+the+Perimeter.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cEx: Find the Area of a Rectangle Given the Perimeter\u201d here (opens in<br \/>\nnew window).<\/a><\/p>\n<\/section>\n<h2>Find the Area and Perimeter of a Triangle<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea <\/strong><\/p>\n<p>The <strong>perimeter <\/strong>of a triangle is the total length of its three sides. To find the perimeter, you add up the lengths of all three sides of the triangle. If the lengths of the sides are represented by [latex]a[\/latex], [latex]b[\/latex], and [latex]c[\/latex], the formula for calculating the perimeter is [latex]P = a + b + c[\/latex], where [latex]P[\/latex] represents the perimeter.<\/p>\n<p>The <strong>area<\/strong> of a triangle is a measurement that tells us the amount of space enclosed by the triangle. To find the area, you multiply the base of the triangle by its corresponding height and divide the product by [latex]2[\/latex]. The formula for calculating the area is [latex]A = \\frac{1}{2} \u00d7 b \u00d7 h[\/latex], where [latex]A[\/latex] represents the area, [latex]b[\/latex] represents the base, and [latex]h[\/latex] represents the height.<\/p>\n<\/div>\n<section class=\"textbox watchIt\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=10350395&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=l9y6l9LxQ24&amp;video_target=tpm-plugin-w2934u9z-l9y6l9LxQ24\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/p>\n<p>You can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Quantitative+Reasoning+-+2023+Build\/Transcriptions\/How+to+Find+the+Area+and+Perimeter+of+a+Triangle.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cHow to Find the Area and Perimeter of a Triangle\u201d here (opens in new<br \/>\nwindow).<\/a><\/p>\n<\/section>\n<p>The following video provides another example of how to use the area formula for triangles.<\/p>\n<section class=\"textbox watchIt\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/jXbPAk2jorM\" width=\"560\" height=\"315\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>You can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Quantitative+Reasoning+-+2023+Build\/Transcriptions\/Ex_+Find+the+Area+of+a+Triangle+(Whole+Number).txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cEx: Find the Area of a Triangle (Whole Number)\u201d here (opens in new<br \/>\nwindow).<\/a><\/p>\n<\/section>\n<p>In our next video, we show an example of how to find the height of a triangle given it\u2019s area.<\/p>\n<section class=\"textbox watchIt\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/C0vUVK_o5r0\" width=\"560\" height=\"315\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>You can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Quantitative+Reasoning+-+2023+Build\/Transcriptions\/Ex_+Find+the+Height+of+a+Triangle+Given+Area+(Even+Base).txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cEx: Find the Height of a Triangle Given Area (Even Base)\u201d here (opens<br \/>\nin new window).<\/a><\/p>\n<\/section>\n<section class=\"textbox example\">The perimeter of an equilateral triangle is [latex]93[\/latex] inches. Find the length of each side.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q157458\">Show Solution<\/button><\/p>\n<div id=\"q157458\" class=\"hidden-answer\" style=\"display: none\">\n<table id=\"eip-id1168468574026\" class=\"unnumbered unstyled\" summary=\"Step 1 says,\">\n<tbody>\n<tr>\n<td>Step 1. <strong>Read<\/strong> the problem. Draw the figure and label it with the given information.<\/td>\n<td>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223937\/CNX_BMath_Figure_09_04_076_img-01.png\" alt=\"A triangle with each side labeled s\" width=\"185\" height=\"174\" \/><\/p>\n<p>Perimeter = [latex]93[\/latex] in.<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>Step 2. <strong>Identify<\/strong> what you are looking for.<\/td>\n<td>length of the sides of an equilateral triangle<\/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> = length of each side<\/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.<\/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\/24223938\/CNX_BMath_Figure_09_04_076_img-02.png\" alt=\"The equation P = a + b + c. Beneath it, the equation is written again with 93 substituted in for P and s substituted in for a, b, and c.\" width=\"298\" height=\"60\" \/><\/td>\n<\/tr>\n<tr>\n<td>Step 5. <strong>Solve<\/strong> the equation.<\/td>\n<td>\n[latex]93=3s[\/latex]<br \/>\n[latex]31=s[\/latex]\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>Step 6. <strong>Check:<\/strong><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223940\/CNX_BMath_Figure_09_04_076_img-04.png\" alt=\"A triangle with each side labeled 31\" width=\"175\" height=\"173\" \/><\/p>\n<p>[latex]93\\stackrel{?}{=}31+31+31[\/latex]<br \/>\n[latex]93=93\\quad\\checkmark[\/latex]\n<\/td>\n<td>&nbsp;<\/td>\n<\/tr>\n<tr>\n<td>Step 7. <strong>Answer<\/strong> the question.<\/td>\n<td>Each side is [latex]31[\/latex] inches.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/section>\n<h2>Find the Area of a Trapezoid<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea <\/strong><\/p>\n<p>A <strong>trapezoid <\/strong>is a quadrilateral with one pair of parallel sides.<\/p>\n<p>To find the <strong>area <\/strong>of a trapezoid, you multiply the sum of the lengths of the parallel sides (base1 ([latex]b[\/latex]) and base2 ([latex]B[\/latex])) by the height ([latex]h[\/latex]) and divide the result by [latex]2[\/latex]. The formula for calculating the area of a trapezoid is [latex]A = \\frac{(b + B) \u00d7 h}{2}[\/latex], where [latex]A[\/latex] represents the area. This can also be written as: [latex]{\\text{Area}}_{\\text{trapezoid}}=\\Large\\frac{1}{2}\\normalsize h\\left(b+B\\right)[\/latex]<\/p>\n<\/div>\n<section class=\"textbox watchIt\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=10350393&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=9hISqaDb6XE&amp;video_target=tpm-plugin-3w5ylzj0-9hISqaDb6XE\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/p>\n<p>You can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Quantitative+Reasoning+-+2023+Build\/Transcriptions\/Area+of+a+Trapezoid+%7C+MathHelp.com.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cArea of a Trapezoid | MathHelp.com\u201d here (opens in new window).<\/a><\/p>\n<\/section>\n<p>In the next video we show another example of how to use the formula to find the area of a trapezoid given the<br \/>\nlengths of it&#8217;s height and bases.<\/p>\n<section class=\"textbox watchIt\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/WNo7s-XoI4w\" width=\"560\" height=\"315\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>You can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Quantitative+Reasoning+-+2023+Build\/Transcriptions\/Ex_+Find+the+Area+of+a+Trapezoid.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cEx: Find the Area of a Trapezoid\u201d here (opens in new window).<\/a><\/p>\n<\/section>\n<section class=\"textbox example\">Find the area of a trapezoid whose height is [latex]5[\/latex] feet and whose bases are [latex]10.3[\/latex] and [latex]13.7[\/latex] feet.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q158293\">Show Solution<\/button><\/p>\n<div id=\"q158293\" class=\"hidden-answer\" style=\"display: none\">\n<table id=\"eip-id1168469577474\" class=\"unnumbered unstyled\" style=\"width: 859px;\" summary=\"Step 1 says,\">\n<tbody>\n<tr style=\"height: 178.867px;\">\n<td style=\"width: 417.267px; height: 178.867px;\">Step 1. <strong>Read<\/strong> the problem. Draw the figure and<br \/>\nlabel it with the given information.<\/td>\n<td style=\"width: 414.733px; height: 178.867px;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223958\/CNX_BMath_Figure_09_04_081_img-01.png\" alt=\"A trapezoid with the smaller base labeled 10.3 feet, the larger base labeled 13.7 feet, and the height labeled 5 feet.\" width=\"421\" height=\"141\" \/><\/td>\n<\/tr>\n<tr style=\"height: 15px;\">\n<td style=\"width: 417.267px; height: 15px;\">Step 2. <strong>Identify<\/strong> what you are looking for.<\/td>\n<td style=\"width: 414.733px; height: 15px;\">the area of the trapezoid<\/td>\n<\/tr>\n<tr style=\"height: 15px;\">\n<td style=\"width: 417.267px; height: 15px;\">Step 3. <strong>Name.<\/strong> Choose a variable to represent it.<\/td>\n<td style=\"width: 414.733px; height: 15px;\">Let <em>A<\/em> = the area<\/td>\n<\/tr>\n<tr style=\"height: 102px;\">\n<td style=\"width: 417.267px; height: 102px;\">\n<p>Step 4. <strong>Translate.<\/strong><\/p>\n<p>Write the appropriate formula.<\/p>\n<p>Substitute.<\/p>\n<\/td>\n<td style=\"width: 414.733px; height: 102px;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223959\/CNX_BMath_Figure_09_04_081_img-02.png\" alt=\"The equation A = one half times h times the quantity of little b plus big b. Beneath, the equation is rewritten with 5 substituted in for h, 10.3 substituted in for little b and 13.7 substituted in for big b.\" width=\"421\" height=\"99\" \/><\/td>\n<\/tr>\n<tr style=\"height: 90px;\">\n<td style=\"width: 417.267px; height: 90px;\">Step 5. <strong>Solve<\/strong> the equation.<\/td>\n<td style=\"width: 414.733px; height: 90px;\">\n[latex]A={\\Large\\frac{1}{2}}\\normalsize\\cdot 5(24)[\/latex]<br \/>\n[latex]A=12(5)[\/latex]<\/p>\n<p>[latex]A=60[\/latex] square feet<\/p>\n<\/td>\n<\/tr>\n<tr style=\"height: 305px;\">\n<td style=\"width: 417.267px; height: 305px;\">\n<p>Step 6. <strong>Check:<\/strong> Is this answer reasonable?<\/p>\n<p>The area of the trapezoid should be less than the area of a rectangle with base [latex]13.7[\/latex] and height<br \/>\n[latex]5[\/latex], but more than the area of a rectangle with base [latex]10.3[\/latex] and height [latex]5[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224003\/CNX_BMath_Figure_09_04_063.png\" alt=\"An image of a trapezoid is shown with a red rectangle drawn around it. The larger base of the trapezoid is labeled 13.7 ft. and is the same as the base of the rectangle. The height of both the trapezoid and the rectangle is 5 ft. Next to this is an image of a trapezoid with a blue rectangle drawn inside it. The smaller base of the trapezoid is labeled 10.3 ft. and is the same as the base of the rectangle. Below the images is A sub red rectangle is greater than A sub trapezoid is greater than A sub blue rectangle. Below this is 68.5, 60, and 51.5.\" width=\"378\" height=\"179\" \/><\/p>\n<\/td>\n<td style=\"width: 414.733px; height: 305px;\">[latex]\\checkmark[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 15px;\">\n<td style=\"width: 417.267px; height: 15px;\">Step 7. <strong>Answer<\/strong> the question.<\/td>\n<td style=\"width: 414.733px; height: 15px;\">The area of the trapezoid is [latex]60[\/latex] square feet.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/section>\n<h2>Find the Circumference and Area of Circles<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea <\/strong><\/p>\n<p>The <strong>circumference <\/strong>of a circle is the distance around its outer edge or boundary. The formula for calculating the circumference of a circle is [latex]C = 2\u03c0r[\/latex] or [latex]C = \u03c0d[\/latex], where [latex]C[\/latex] represents the circumference, [latex]r[\/latex] represents the radius, and [latex]d[\/latex] represents the diameter.<\/p>\n<p>The <strong>area <\/strong>of a circle is a measurement that tells us the amount of space enclosed by the circle. The formula for calculating the area of a circle is [latex]A = \u03c0r^2[\/latex], where [latex]A[\/latex] represents the area, and [latex]r[\/latex] represents the radius.<\/p>\n<\/div>\n<section class=\"textbox watchIt\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/O-cawByg2aA\" width=\"560\" height=\"315\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>You can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Quantitative+Reasoning+-+2023+Build\/Transcriptions\/Math+Antics+-+Circles%2C+Circumference+And+Area.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cMath Antics &#8211; Circles, Circumference And Area\u201d here (opens in new<br \/>\nwindow).<\/a><\/p>\n<\/section>\n<section class=\"textbox example\">A circle has radius [latex]10[\/latex] centimeters. Approximate its:<\/p>\n<ol style=\"list-style-type: decimal;\">\n<li>circumference<\/li>\n<li>area<\/li>\n<\/ol>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q247920\">Show Solution<\/button><\/p>\n<div id=\"q247920\" class=\"hidden-answer\" style=\"display: none\">\n<ol style=\"list-style-type: decimal;\">\n<li>\n<table>\n<tbody>\n<tr>\n<td>Find the circumference when [latex]r=10[\/latex].<\/td>\n<td>&nbsp;<\/td>\n<\/tr>\n<tr>\n<td>Write the formula for circumference.<\/td>\n<td>[latex]C=2\\pi \\mathit{\\text{r}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Substitute [latex]3.14[\/latex] for [latex]\\pi[\/latex] and 10 for , [latex]r[\/latex] .<\/td>\n<td>[latex]C\\approx 2\\left(3.14\\right)\\left(10\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Multiply.<\/td>\n<td>[latex]C\\approx 62.8\\text{ centimeters}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/li>\n<li>\n<table>\n<tbody>\n<tr>\n<td>Find the area when [latex]r=10[\/latex].<\/td>\n<td>&nbsp;<\/td>\n<\/tr>\n<tr>\n<td>Write the formula for area.<\/td>\n<td>[latex]A=\\pi {\\mathit{\\text{r}}}^{2}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Substitute [latex]3.14[\/latex] for [latex]\\pi[\/latex] and 10 for [latex]r[\/latex] .<\/td>\n<td>[latex]A\\approx \\left(3.14\\right){\\text{(}10\\text{)}}^{2}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Multiply.<\/td>\n<td>[latex]A\\approx 314\\text{ square centimeters}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">A circle has a diameter of [latex]14.6[\/latex] inches. Approximate its:<\/p>\n<ol style=\"list-style-type: decimal;\">\n<li>circumference<\/li>\n<li>area<\/li>\n<\/ol>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q272573\">Show Solution<\/button><\/p>\n<div id=\"q272573\" class=\"hidden-answer\" style=\"display: none\">\n<ol style=\"list-style-type: decimal;\">\n<li>\n<table>\n<tbody>\n<tr>\n<td>Find the circumference when [latex]d=14.6[\/latex]<\/td>\n<td>&nbsp;<\/td>\n<\/tr>\n<tr>\n<td>Write the formula for circumference.<\/td>\n<td>[latex]C=\\pi \\mathit{\\text{d}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Substitute [latex]3.14[\/latex] for [latex]\\pi[\/latex] and [latex]14.6[\/latex] for [latex]d[\/latex]<\/td>\n<td>[latex]C\\approx \\left(3.14\\right)\\left(14.6\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Multiply.<\/td>\n<td>[latex]C\\approx 45.844\\text{ inches}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/li>\n<li>\n<table>\n<tbody>\n<tr style=\"height: 15px;\">\n<td style=\"height: 15px;\">Find the area when [latex]d=14.6[\/latex]. We know [latex]d=2r[\/latex] so for this<br \/>\nproblem [latex]r=7.3[\/latex]<\/td>\n<td style=\"height: 15px;\">\u00a0<\/td>\n<\/tr>\n<tr style=\"height: 15px;\">\n<td style=\"height: 15px;\">Write the formula for area.<\/td>\n<td style=\"height: 15px;\">[latex]A=\\pi {\\mathit{\\text{r}}}^{2}[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 31px;\">\n<td style=\"height: 31px;\">Substitute [latex]3.14[\/latex] for [latex]\\pi[\/latex] and [latex]7.3[\/latex] for<br \/>\n[latex]r[\/latex] .<\/td>\n<td style=\"height: 31px;\">[latex]A\\approx \\left(3.14\\right){\\text{(}7.3\\text{)}}^{2}[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 31px;\">\n<td style=\"height: 31px;\">Multiply.<\/td>\n<td style=\"height: 31px;\">[latex]A\\approx 171.915\\text{ square inches}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n<p>Watch the following video to see another example of how to find the circumference of a circle.<\/p>\n<section class=\"textbox watchIt\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/sHtsnC2Mgnk\" width=\"560\" height=\"315\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>You can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Quantitative+Reasoning+-+2023+Build\/Transcriptions\/Examples_+Determine+the+Circumference+of+a+Circle.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cExamples: Determine the Circumference of a Circle\u201d here (opens in new<br \/>\nwindow).<\/a><\/p>\n<\/section>\n<p>In the next video example, we find the area of a circle.<\/p>\n<section class=\"textbox watchIt\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/SIKkWLqt2mQ\" width=\"560\" height=\"315\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>You can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Quantitative+Reasoning+-+2023+Build\/Transcriptions\/Example_+Determine+the+Area+of+a+Circle.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cExample: Determine the Area of a Circle\u201d here (opens in new<br \/>\nwindow).<\/a><\/p>\n<\/section>\n<h2>Find the Area of Irregular Figures<\/h2>\n<p>To find the area of an irregular figure, you break it down into simpler shapes whose areas can be calculated. For<br \/>\nexample, you can divide the irregular figure into rectangles, triangles, squares, circles, or other familiar shapes.<br \/>\nThen, calculate the area of each individual shape using the appropriate formulas. Once you have found the areas of the<br \/>\nsimpler shapes, add them together to find the total area of the irregular figure. Keep in mind that this method works<br \/>\nbest when the irregular figure can be divided into known shapes.<\/p>\n<p>The following video gives an example of how to find the area of an \u201cL\u201d shaped polygon using the dimensions of two<br \/>\nrectangles.<\/p>\n<section class=\"textbox watchIt\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/lx8vweYTPpg\" width=\"560\" height=\"315\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>You can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Quantitative+Reasoning+-+2023+Build\/Transcriptions\/Ex_+Find+the+Area+of+an+L-Shaped+Polygon+Involving+Whole+Numbers.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cEx: Find the Area of an L-Shaped Polygon Involving Whole Numbers\u201d<br \/>\nhere (opens in new window).<\/a><\/p>\n<\/section>\n<p>The next video example find the area of an object containing one semi-circle and a rectangle.<\/p>\n<section class=\"textbox watchIt\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/085t_Fmje4o\" width=\"560\" height=\"315\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>You can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Quantitative+Reasoning+-+2023+Build\/Transcriptions\/Example_+Determine+an+Area+Involving+a+Rectangle+and+Circle.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cExample: Determine an Area Involving a Rectangle and Circle\u201d here<br \/>\n(opens in new window).<\/a><\/p>\n<\/section>\n<p>The next video shows more example of finding the area of irregular shapes.<\/p>\n<section class=\"textbox watchIt\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=10350392&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=_e7j6rE7_Pg&amp;video_target=tpm-plugin-6t1ssnro-_e7j6rE7_Pg\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/p>\n<p>You can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Quantitative+Reasoning+-+2023+Build\/Transcriptions\/Area+and+Perimeter+of+Irregular+Shapes+-+Tons+of+Examples!.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cArea and Perimeter of Irregular Shapes &#8211; Tons of Examples!\u201d here<br \/>\n(opens in new window).<\/a><\/p>\n<\/section>\n","protected":false},"author":15,"menu_order":25,"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":"fresh_take","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\/2864"}],"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":44,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters\/2864\/revisions"}],"predecessor-version":[{"id":15333,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters\/2864\/revisions\/15333"}],"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\/2864\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/media?parent=2864"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapter-type?post=2864"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/contributor?post=2864"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/license?post=2864"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}