{"id":2862,"date":"2023-05-16T18:12:19","date_gmt":"2023-05-16T18:12:19","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/?post_type=chapter&#038;p=2862"},"modified":"2025-08-26T03:33:42","modified_gmt":"2025-08-26T03:33:42","slug":"area-and-circumference-learn-it-1","status":"web-only","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/chapter\/area-and-circumference-learn-it-1\/","title":{"raw":"Area and Circumference: Learn It 1","rendered":"Area and Circumference: Learn It 1"},"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<p>A rectangle has four sides and four right angles. The opposite sides of a rectangle are the same length. We refer to one side of the rectangle as the length, [latex]L[\/latex], and the adjacent side as the width, [latex]W[\/latex].<\/p>\r\n<center>\r\n[caption id=\"\" align=\"aligncenter\" width=\"189\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223837\/CNX_BMath_Figure_09_04_012.png\" alt=\"A rectangle is shown. Each angle is marked with a square. The top and bottom are labeled L, the sides are labeled W.\" width=\"189\" height=\"123\" \/> Figure 1. The right angles, lengths, and widths of the rectangle are labeled[\/caption]\r\n<\/center>\r\n<p>&nbsp;<\/p>\r\n<p>The <strong>perimeter<\/strong>, [latex]P[\/latex], of the rectangle is the distance around the rectangle. If you started at one corner and walked around the rectangle, you would walk [latex]L+W+L+W[\/latex] units, or two lengths and two widths. The perimeter then is:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}P=L+W+L+W\\hfill \\\\ \\hfill \\text {or} \\hfill \\\\ P=2L+2W\\hfill \\end{array}[\/latex]<\/p>\r\n<p>What about the <strong>area <\/strong>of a rectangle? Below is a rectangular rug. It is [latex]2[\/latex] feet long by [latex]3[\/latex] feet wide, and its area is [latex]6[\/latex] square feet. Since [latex]A=2\\cdot 3[\/latex], we see that the area, [latex]A[\/latex], is the length, [latex]L[\/latex], times the width, [latex]W[\/latex], so the area of a rectangle is [latex]A=L\\cdot W[\/latex].<\/p>\r\n<center>\r\n[caption id=\"\" align=\"aligncenter\" width=\"241\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223838\/CNX_BMath_Figure_09_04_013.png\" alt=\"A rectangle made up of 6 squares. The bottom is 2 squares across and marked as 2, the side is 3 squares long and marked as 3.\" width=\"241\" height=\"178\" \/> Figure 2. The area is calculated 2 x 3 = 6[\/caption]\r\n<\/center>\r\n<p>&nbsp;<\/p>\r\n<section class=\"textbox keyTakeaway\">\r\n<div>\r\n<h3>properties of rectangles<\/h3>\r\n<ul>\r\n\t<li>Rectangles have four sides and four right [latex]\\left(\\text{90}^ \\circ\\right)[\/latex] angles.<\/li>\r\n\t<li>The lengths of opposite sides are equal.<\/li>\r\n\t<li>The <strong>perimeter<\/strong>, [latex]P[\/latex], of a rectangle is the sum of twice the length and twice the width. See the first image.<\/li>\r\n<\/ul>\r\n<center>[latex]P=2L+2W \\text{ or } P = 2(L+W)[\/latex]<\/center>\r\n<p>&nbsp;<\/p>\r\n<ul>\r\n\t<li>The <strong>area<\/strong>, [latex]A[\/latex], of a rectangle is the length times the width. The area will be expressed in square units.<\/li>\r\n<\/ul>\r\n<center>[latex]A=L\\cdot W[\/latex]<\/center><\/div>\r\n<\/section>\r\n<section class=\"textbox proTip\">Remember to use the Problem-Solving Strategy for Geometry Applications when working on problems in this section.\r\n\r\n<ol id=\"eip-id1170325029073\" class=\"stepwise\">\r\n\t<li><strong>Read<\/strong> the problem and make sure you understand all the words and ideas. Draw the figure and label it with the given information.<\/li>\r\n\t<li><strong>Identify<\/strong> what you are looking for.<\/li>\r\n\t<li><strong>Name<\/strong> what you are looking for. Choose a variable to represent that quantity.<\/li>\r\n\t<li><strong>Translate<\/strong> into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.<\/li>\r\n\t<li><strong>Solve<\/strong> the equation using good algebra techniques.<\/li>\r\n\t<li><strong>Check<\/strong> the answer in the problem and make sure it makes sense.<\/li>\r\n\t<li><strong>Answer<\/strong> the question with a complete sentence.<\/li>\r\n<\/ol>\r\n<\/section>\r\n<section class=\"textbox example\">The length of a rectangle is [latex]32[\/latex] meters and the width is [latex]20[\/latex] meters. Find the:\r\n\r\n<ol style=\"list-style-type: decimal;\">\r\n\t<li>Perimeter<\/li>\r\n\t<li>Area<\/li>\r\n<\/ol>\r\n\r\n[reveal-answer q=\"172561\"]Show Solution[\/reveal-answer] [hidden-answer a=\"172561\"]\r\n\r\n<ol>\r\n\t<li>\r\n<table>\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\/24223840\/CNX_BMath_Figure_09_04_067_img_MW-01.png\" alt=\"A rectangle with the top and bottom labeled 32 m and the sides labeled 20 m\" width=\"303\" height=\"174\" \/><\/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 perimeter of a rectangle<\/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 [latex]P[\/latex] = the perimeter<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 4. <strong>Translate.<\/strong> Write the appropriate formula. Substitute.<\/td>\r\n<td><img class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223841\/CNX_BMath_Figure_09_04_067_img_MW-02.png\" alt=\"The formula P = 2L + 2W. The formula is then written again with 32 substituted in for L and 20 substituted in for W\" width=\"524\" height=\"100\" \/><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 5. <strong>Solve<\/strong> the equation.<\/td>\r\n<td>[latex]P=64+40[\/latex] [latex]P=104[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 6. <strong>Check.<\/strong><\/td>\r\n<td>\r\n<p>[latex]p\\stackrel{?}{=}104[\/latex]<\/p>\r\n<p>[latex]20+32+20+32\\stackrel{?}{=}104[\/latex]<\/p>\r\n<p>[latex]104=104\\checkmark[\/latex]<\/p>\r\n<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 7. <strong>Answer<\/strong> the question.<\/td>\r\n<td>The perimeter of the rectangle is [latex]104[\/latex] meters.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/li>\r\n\t<li>\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td>Step 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\/24223845\/CNX_BMath_Figure_09_04_068_img_MW-01.png\" alt=\"A rectangle with the top and bottom labeled 32 m and the sides labeled 20 m\" width=\"310\" height=\"176\" \/><\/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 area of a rectangle<\/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>A<\/em> = the area<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 4. <strong>Translate.<\/strong> Write the appropriate formula. Substitute.<\/td>\r\n<td><img class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223846\/CNX_BMath_Figure_09_04_068_img_MW-02.png\" alt=\"The formula A = L times W. The formula is then written again with 32 substituted in for L and 20 substituted in for W\" width=\"310\" height=\"64\" \/><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 5. <strong>Solve<\/strong> the equation.<\/td>\r\n<td>[latex]A=640[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 6. <strong>Check.<\/strong><\/td>\r\n<td>\r\n<p>[latex]A\\stackrel{?}{=}640[\/latex]<\/p>\r\n<p>[latex]32\\cdot 20\\stackrel{?}{=}640[\/latex]<\/p>\r\n<p>[latex]640=640\\checkmark[\/latex]<\/p>\r\n<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 7. <strong>Answer<\/strong> the question.<\/td>\r\n<td>The area of the rectangle is [latex]640[\/latex] square meters.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/li>\r\n<\/ol>\r\n\r\n[\/hidden-answer]<\/section>\r\n<section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]6988[\/ohm2_question]<\/section>\r\n<section class=\"textbox example\">Find the length of a rectangle with perimeter [latex]50[\/latex] inches and width [latex]10[\/latex] inches. [reveal-answer q=\"172562\"]Show Solution[\/reveal-answer] [hidden-answer a=\"172562\"]\r\n\r\n<table id=\"eip-id1168469470101\" 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\/24223850\/CNX_BMath_Figure_09_04_069_img-01.png\" alt=\"A rectangle with the top and bottom labeled L and the sides labeled 10 in\" width=\"296\" height=\"177\" \/><\/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 of the rectangle<\/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\u00a0[latex]L[\/latex] = the length<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 4. <strong>Translate.<\/strong> Write the appropriate formula. Substitute.<\/td>\r\n<td><img class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223852\/CNX_BMath_Figure_09_04_069_img-02.png\" alt=\"The formula P = 2L + 2W. The formula is then written again with 50 substituted in for P and 10 substituted in for W.\" width=\"272\" height=\"55\" \/><\/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]50\\color{red}{- 20}\\color{black}=2L+20\\color{red}{- 20}[\/latex]<\/p>\r\n<p>[latex]30=2L[\/latex]<\/p>\r\n<p>[latex]{\\Large\\frac{30}{\\color{red}{2}}}\\color{black}={\\Large\\frac{2L}{\\color{red}{2}}}[\/latex]<\/p>\r\n<p>[latex]15=L[\/latex]<\/p>\r\n<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 6. <strong>Check.<\/strong><\/td>\r\n<td>\r\n<p>[latex]p=50[\/latex]<\/p>\r\n<p>[latex]15+10+15+10\\stackrel{?}{=}50[\/latex]<\/p>\r\n<p>[latex]50=50\\checkmark[\/latex]<\/p>\r\n<\/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]15[\/latex] inches.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n\r\n[\/hidden-answer]<\/section>\r\n<section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]6989[\/ohm2_question]<\/section>\r\n<p>In the next example, the width is defined in terms of the length. We\u2019ll wait to draw the figure until we write an expression for the width so that we can label one side with that expression.<\/p>\r\n<section class=\"textbox example\">The width of a rectangle is two inches less than the length. The perimeter is [latex]52[\/latex] inches. Find the length and width. [reveal-answer q=\"886120\"]Show Solution[\/reveal-answer] [hidden-answer a=\"886120\"]Solution\r\n\r\n<table id=\"eip-id1168466427903\" 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 of the rectangle<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 3. <strong>Name.<\/strong> Choose a variable to represent it. Now we can draw a figure using these expressions for the length and width.<\/td>\r\n<td>\r\n<p>Since the width is defined in terms of the length, we let <em>L<\/em> = length. The width is two feet less that the length, so we let <em>L<\/em> \u2212 2 = width.<\/p>\r\n<p><img class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223855\/CNX_BMath_Figure_09_04_070_img-01.png\" alt=\"A rectangle with top and bottom labeled L and the sides labeled L - 2.\" width=\"299\" height=\"212\" \/><\/p>\r\n<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 4.<strong>Translate.<\/strong> Write the appropriate formula. The formula for the perimeter of a rectangle relates all the information. Substitute in 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\/24223856\/CNX_BMath_Figure_09_04_070_img-02.png\" alt=\"The formula P = 2L + 2W. The formula is then written again with 52 substituted in for P and L - 2 substituted in for W.\" width=\"299\" height=\"55\" \/><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 5. <strong>Solve<\/strong> the equation.<\/td>\r\n<td>[latex]52=2L+2L - 4[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Combine like terms.<\/td>\r\n<td>[latex]52=4L - 4[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Add [latex]4[\/latex] to each side.<\/td>\r\n<td>[latex]56=4L[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Divide by [latex]4[\/latex].<\/td>\r\n<td>[latex]{\\Large\\frac{56}{4}}={\\Large\\frac{4L}{4}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>&nbsp;<\/td>\r\n<td>[latex]14=L[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>&nbsp;<\/td>\r\n<td>The length is [latex]14[\/latex] inches.<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Now we need to find the width.<\/td>\r\n<td>\r\n<p>[latex]L-2[\/latex]<\/p>\r\n<p>[latex]\\color{red}{14}\\color{black}-2[\/latex]<\/p>\r\n<p>[latex]12[\/latex]<\/p>\r\n<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>The width is [latex]L\u22122[\/latex].<\/td>\r\n<td>\u00a0 The width is [latex]12[\/latex] inches.<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 6. <strong>Check.<\/strong><\/td>\r\n<td>Since [latex]14+12+14+12=52[\/latex] , this works!<\/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]14[\/latex] feet and the width is [latex]12[\/latex] feet.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n\r\n[\/hidden-answer]<\/section>\r\n<section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]6991[\/ohm2_question]<\/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<p>A rectangle has four sides and four right angles. The opposite sides of a rectangle are the same length. We refer to one side of the rectangle as the length, [latex]L[\/latex], and the adjacent side as the width, [latex]W[\/latex].<\/p>\n<div style=\"text-align: center;\">\n<figure style=\"width: 189px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223837\/CNX_BMath_Figure_09_04_012.png\" alt=\"A rectangle is shown. Each angle is marked with a square. The top and bottom are labeled L, the sides are labeled W.\" width=\"189\" height=\"123\" \/><figcaption class=\"wp-caption-text\">Figure 1. The right angles, lengths, and widths of the rectangle are labeled<\/figcaption><\/figure>\n<\/div>\n<p>&nbsp;<\/p>\n<p>The <strong>perimeter<\/strong>, [latex]P[\/latex], of the rectangle is the distance around the rectangle. If you started at one corner and walked around the rectangle, you would walk [latex]L+W+L+W[\/latex] units, or two lengths and two widths. The perimeter then is:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}P=L+W+L+W\\hfill \\\\ \\hfill \\text {or} \\hfill \\\\ P=2L+2W\\hfill \\end{array}[\/latex]<\/p>\n<p>What about the <strong>area <\/strong>of a rectangle? Below is a rectangular rug. It is [latex]2[\/latex] feet long by [latex]3[\/latex] feet wide, and its area is [latex]6[\/latex] square feet. Since [latex]A=2\\cdot 3[\/latex], we see that the area, [latex]A[\/latex], is the length, [latex]L[\/latex], times the width, [latex]W[\/latex], so the area of a rectangle is [latex]A=L\\cdot W[\/latex].<\/p>\n<div style=\"text-align: center;\">\n<figure style=\"width: 241px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223838\/CNX_BMath_Figure_09_04_013.png\" alt=\"A rectangle made up of 6 squares. The bottom is 2 squares across and marked as 2, the side is 3 squares long and marked as 3.\" width=\"241\" height=\"178\" \/><figcaption class=\"wp-caption-text\">Figure 2. The area is calculated 2 x 3 = 6<\/figcaption><\/figure>\n<\/div>\n<p>&nbsp;<\/p>\n<section class=\"textbox keyTakeaway\">\n<div>\n<h3>properties of rectangles<\/h3>\n<ul>\n<li>Rectangles have four sides and four right [latex]\\left(\\text{90}^ \\circ\\right)[\/latex] angles.<\/li>\n<li>The lengths of opposite sides are equal.<\/li>\n<li>The <strong>perimeter<\/strong>, [latex]P[\/latex], of a rectangle is the sum of twice the length and twice the width. See the first image.<\/li>\n<\/ul>\n<div style=\"text-align: center;\">[latex]P=2L+2W \\text{ or } P = 2(L+W)[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<ul>\n<li>The <strong>area<\/strong>, [latex]A[\/latex], of a rectangle is the length times the width. The area will be expressed in square units.<\/li>\n<\/ul>\n<div style=\"text-align: center;\">[latex]A=L\\cdot W[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox proTip\">Remember to use the Problem-Solving Strategy for Geometry Applications when working on problems in this section.<\/p>\n<ol id=\"eip-id1170325029073\" class=\"stepwise\">\n<li><strong>Read<\/strong> the problem and make sure you understand all the words and ideas. Draw the figure and label it with the given information.<\/li>\n<li><strong>Identify<\/strong> what you are looking for.<\/li>\n<li><strong>Name<\/strong> what you are looking for. Choose a variable to represent that quantity.<\/li>\n<li><strong>Translate<\/strong> into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.<\/li>\n<li><strong>Solve<\/strong> the equation using good algebra techniques.<\/li>\n<li><strong>Check<\/strong> the answer in the problem and make sure it makes sense.<\/li>\n<li><strong>Answer<\/strong> the question with a complete sentence.<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\">The length of a rectangle is [latex]32[\/latex] meters and the width is [latex]20[\/latex] meters. Find the:<\/p>\n<ol style=\"list-style-type: decimal;\">\n<li>Perimeter<\/li>\n<li>Area<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q172561\">Show Solution<\/button> <\/p>\n<div id=\"q172561\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>\n<table>\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\/24223840\/CNX_BMath_Figure_09_04_067_img_MW-01.png\" alt=\"A rectangle with the top and bottom labeled 32 m and the sides labeled 20 m\" width=\"303\" height=\"174\" \/><\/td>\n<\/tr>\n<tr>\n<td>Step 2. <strong>Identify<\/strong> what you are looking for.<\/td>\n<td>the perimeter of a rectangle<\/td>\n<\/tr>\n<tr>\n<td>Step 3. <strong>Name.<\/strong> Choose a variable to represent it.<\/td>\n<td>Let [latex]P[\/latex] = the perimeter<\/td>\n<\/tr>\n<tr>\n<td>Step 4. <strong>Translate.<\/strong> Write the appropriate formula. Substitute.<\/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\/24223841\/CNX_BMath_Figure_09_04_067_img_MW-02.png\" alt=\"The formula P = 2L + 2W. The formula is then written again with 32 substituted in for L and 20 substituted in for W\" width=\"524\" height=\"100\" \/><\/td>\n<\/tr>\n<tr>\n<td>Step 5. <strong>Solve<\/strong> the equation.<\/td>\n<td>[latex]P=64+40[\/latex] [latex]P=104[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Step 6. <strong>Check.<\/strong><\/td>\n<td>\n[latex]p\\stackrel{?}{=}104[\/latex]<br \/>\n[latex]20+32+20+32\\stackrel{?}{=}104[\/latex]<br \/>\n[latex]104=104\\checkmark[\/latex]\n<\/td>\n<\/tr>\n<tr>\n<td>Step 7. <strong>Answer<\/strong> the question.<\/td>\n<td>The perimeter of the rectangle is [latex]104[\/latex] meters.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/li>\n<li>\n<table>\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\/24223845\/CNX_BMath_Figure_09_04_068_img_MW-01.png\" alt=\"A rectangle with the top and bottom labeled 32 m and the sides labeled 20 m\" width=\"310\" height=\"176\" \/><\/td>\n<\/tr>\n<tr>\n<td>Step 2. <strong>Identify<\/strong> what you are looking for.<\/td>\n<td>the area of a rectangle<\/td>\n<\/tr>\n<tr>\n<td>Step 3. <strong>Name.<\/strong> Choose a variable to represent it.<\/td>\n<td>Let <em>A<\/em> = the area<\/td>\n<\/tr>\n<tr>\n<td>Step 4. <strong>Translate.<\/strong> Write the appropriate formula. Substitute.<\/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\/24223846\/CNX_BMath_Figure_09_04_068_img_MW-02.png\" alt=\"The formula A = L times W. The formula is then written again with 32 substituted in for L and 20 substituted in for W\" width=\"310\" height=\"64\" \/><\/td>\n<\/tr>\n<tr>\n<td>Step 5. <strong>Solve<\/strong> the equation.<\/td>\n<td>[latex]A=640[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Step 6. <strong>Check.<\/strong><\/td>\n<td>\n[latex]A\\stackrel{?}{=}640[\/latex]<br \/>\n[latex]32\\cdot 20\\stackrel{?}{=}640[\/latex]<br \/>\n[latex]640=640\\checkmark[\/latex]\n<\/td>\n<\/tr>\n<tr>\n<td>Step 7. <strong>Answer<\/strong> the question.<\/td>\n<td>The area of the rectangle is [latex]640[\/latex] square meters.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm6988\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=6988&theme=lumen&iframe_resize_id=ohm6988&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox example\">Find the length of a rectangle with perimeter [latex]50[\/latex] inches and width [latex]10[\/latex] inches. <\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q172562\">Show Solution<\/button> <\/p>\n<div id=\"q172562\" class=\"hidden-answer\" style=\"display: none\">\n<table id=\"eip-id1168469470101\" 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\/24223850\/CNX_BMath_Figure_09_04_069_img-01.png\" alt=\"A rectangle with the top and bottom labeled L and the sides labeled 10 in\" width=\"296\" height=\"177\" \/><\/td>\n<\/tr>\n<tr>\n<td>Step 2. <strong>Identify<\/strong> what you are looking for.<\/td>\n<td>the length of the rectangle<\/td>\n<\/tr>\n<tr>\n<td>Step 3. <strong>Name.<\/strong> Choose a variable to represent it.<\/td>\n<td>Let\u00a0[latex]L[\/latex] = the length<\/td>\n<\/tr>\n<tr>\n<td>Step 4. <strong>Translate.<\/strong> Write the appropriate formula. Substitute.<\/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\/24223852\/CNX_BMath_Figure_09_04_069_img-02.png\" alt=\"The formula P = 2L + 2W. The formula is then written again with 50 substituted in for P and 10 substituted in for W.\" width=\"272\" height=\"55\" \/><\/td>\n<\/tr>\n<tr>\n<td>Step 5. <strong>Solve<\/strong> the equation.<\/td>\n<td>\n[latex]50\\color{red}{- 20}\\color{black}=2L+20\\color{red}{- 20}[\/latex]<br \/>\n[latex]30=2L[\/latex]<br \/>\n[latex]{\\Large\\frac{30}{\\color{red}{2}}}\\color{black}={\\Large\\frac{2L}{\\color{red}{2}}}[\/latex]<br \/>\n[latex]15=L[\/latex]\n<\/td>\n<\/tr>\n<tr>\n<td>Step 6. <strong>Check.<\/strong><\/td>\n<td>\n[latex]p=50[\/latex]<br \/>\n[latex]15+10+15+10\\stackrel{?}{=}50[\/latex]<br \/>\n[latex]50=50\\checkmark[\/latex]\n<\/td>\n<\/tr>\n<tr>\n<td>Step 7. <strong>Answer<\/strong> the question.<\/td>\n<td>The length is [latex]15[\/latex] inches.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm6989\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=6989&theme=lumen&iframe_resize_id=ohm6989&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<p>In the next example, the width is defined in terms of the length. We\u2019ll wait to draw the figure until we write an expression for the width so that we can label one side with that expression.<\/p>\n<section class=\"textbox example\">The width of a rectangle is two inches less than the length. The perimeter is [latex]52[\/latex] inches. Find the length and width. <\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q886120\">Show Solution<\/button> <\/p>\n<div id=\"q886120\" class=\"hidden-answer\" style=\"display: none\">Solution<\/p>\n<table id=\"eip-id1168466427903\" 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 of the rectangle<\/td>\n<\/tr>\n<tr>\n<td>Step 3. <strong>Name.<\/strong> Choose a variable to represent it. Now we can draw a figure using these expressions for the length and width.<\/td>\n<td>\n<p>Since the width is defined in terms of the length, we let <em>L<\/em> = length. The width is two feet less that the length, so we let <em>L<\/em> \u2212 2 = width.<\/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\/24223855\/CNX_BMath_Figure_09_04_070_img-01.png\" alt=\"A rectangle with top and bottom labeled L and the sides labeled L - 2.\" width=\"299\" height=\"212\" \/><\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>Step 4.<strong>Translate.<\/strong> Write the appropriate formula. The formula for the perimeter of a rectangle relates all the information. Substitute in 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\/24223856\/CNX_BMath_Figure_09_04_070_img-02.png\" alt=\"The formula P = 2L + 2W. The formula is then written again with 52 substituted in for P and L - 2 substituted in for W.\" width=\"299\" height=\"55\" \/><\/td>\n<\/tr>\n<tr>\n<td>Step 5. <strong>Solve<\/strong> the equation.<\/td>\n<td>[latex]52=2L+2L - 4[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Combine like terms.<\/td>\n<td>[latex]52=4L - 4[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Add [latex]4[\/latex] to each side.<\/td>\n<td>[latex]56=4L[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Divide by [latex]4[\/latex].<\/td>\n<td>[latex]{\\Large\\frac{56}{4}}={\\Large\\frac{4L}{4}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>&nbsp;<\/td>\n<td>[latex]14=L[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>&nbsp;<\/td>\n<td>The length is [latex]14[\/latex] inches.<\/td>\n<\/tr>\n<tr>\n<td>Now we need to find the width.<\/td>\n<td>\n[latex]L-2[\/latex]<br \/>\n[latex]\\color{red}{14}\\color{black}-2[\/latex]<br \/>\n[latex]12[\/latex]\n<\/td>\n<\/tr>\n<tr>\n<td>The width is [latex]L\u22122[\/latex].<\/td>\n<td>\u00a0 The width is [latex]12[\/latex] inches.<\/td>\n<\/tr>\n<tr>\n<td>Step 6. <strong>Check.<\/strong><\/td>\n<td>Since [latex]14+12+14+12=52[\/latex] , this works!<\/td>\n<\/tr>\n<tr>\n<td>Step 7. <strong>Answer<\/strong> the question.<\/td>\n<td>The length is [latex]14[\/latex] feet and the width is [latex]12[\/latex] feet.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm6991\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=6991&theme=lumen&iframe_resize_id=ohm6991&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":15,"menu_order":18,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":71,"module-header":"learn_it","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters\/2862"}],"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":56,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters\/2862\/revisions"}],"predecessor-version":[{"id":15648,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters\/2862\/revisions\/15648"}],"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\/2862\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/media?parent=2862"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapter-type?post=2862"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/contributor?post=2862"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/license?post=2862"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}