{"id":1710,"date":"2024-04-24T17:05:51","date_gmt":"2024-04-24T17:05:51","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=1710"},"modified":"2025-08-17T23:53:47","modified_gmt":"2025-08-17T23:53:47","slug":"contextual-applications-of-derivatives-background-youll-need-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/contextual-applications-of-derivatives-background-youll-need-3\/","title":{"raw":"Contextual Applications of Derivatives: Background You'll Need 3","rendered":"Contextual Applications of Derivatives: Background You&#8217;ll Need 3"},"content":{"raw":"<section class=\"textbox learningGoals\">\r\n<ul>\r\n\t<li>Write formulas to calculate the area, perimeter, and volume of different shapes<\/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\" \/> Rectangle with all sides 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\" \/> Rug with a length of 2 and a width of 3[\/caption]\r\n<\/center>\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 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>\r\n[caption id=\"\" align=\"alignnone\" width=\"303\"]<img 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\" \/> Rectangle with all sides labeled[\/caption]\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>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>\r\n[caption id=\"\" align=\"alignnone\" width=\"524\"]<img 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\" \/> Formula for perimeter[\/caption]\r\n<\/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>\r\n[caption id=\"\" align=\"alignnone\" width=\"310\"]<img 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\" \/> Rectangle with all sides labeled[\/caption]\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>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 [latex]A[\/latex] = 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>\r\n[caption id=\"\" align=\"alignnone\" width=\"310\"]<img 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\" \/> Formula for Area[\/caption]\r\n<\/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\">[ohm_question hide_question_numbers=1]288400[\/ohm_question]<\/section>\r\n<section>\r\n<h2>Find the Area and Perimeter of a Triangle<\/h2>\r\n<p>We now know how to find the area of a rectangle. We can use this fact to help us visualize the formula for the area of a triangle. In the rectangle below, we\u2019ve labeled the length [latex]b[\/latex] and the width [latex]h[\/latex], so its area is [latex]bh[\/latex].<\/p>\r\n<center>\r\n[caption id=\"\" align=\"aligncenter\" width=\"151\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223912\/CNX_BMath_Figure_09_04_035.png\" alt=\"A rectangle with the side labeled h and the bottom labeled b. The center says A equals bh.\" width=\"151\" height=\"89\" \/> Rectangle with with height, base, and area labeled[\/caption]\r\n<\/center>\r\n<p>&nbsp;<\/p>\r\n<p><br \/>\r\nWe can divide this rectangle into two congruent triangles (see the image below). Triangles that are congruent have identical side lengths and angles, and so their areas are equal. The area of each triangle is one-half the area of the rectangle, or [latex]\\Large\\frac{1}{2}\\normalsize bh[\/latex]. This example helps us see why the formula for the area of a triangle is [latex]A=\\Large\\frac{1}{2}\\normalsize bh[\/latex].<\/p>\r\n<center>\r\n[caption id=\"\" align=\"aligncenter\" width=\"323\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223913\/CNX_BMath_Figure_09_04_036.png\" alt=\"A rectangle with a diagonal line drawn from the upper left corner to the bottom right corner. The side of the rectangle is labeled h and the bottom is labeled b. Each triangle says one-half bh. To the right of the rectangle, it says &quot;Area of each triangle A = one-half bh&quot;. \" width=\"323\" height=\"107\" \/> Rectangle divided into two triangles with height, base, and area labeled[\/caption]\r\n<\/center>\r\n<p>&nbsp;<\/p>\r\n<p>To find the area of the triangle, you need to know its base and height. The base is the length of one side of the triangle, usually the side at the bottom. The height is the length of the line that connects the base to the opposite vertex, and makes a [latex]\\text{90}^ \\circ[\/latex] angle with the base. The image below\u00a0shows three triangles with the base and height of each marked.<\/p>\r\n<center>\r\n[caption id=\"\" align=\"aligncenter\" width=\"563\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223914\/CNX_BMath_Figure_09_04_037.png\" alt=\"Three triangles. The triangle on the left is a right triangle. The bottom is labeled b and the side is labeled h. The middle triangle is an acute triangle. The bottom is labeled b. There is a dotted line from the top vertex to the base of the triangle, forming a right angle with the base. That line is labeled h. The triangle on the right is an obtuse triangle. The bottom of the triangle is labeled b. The base has a dotted line extended out and forms a right angle with a dotted line to the top of the triangle. The vertical line is labeled h.\" width=\"563\" height=\"107\" \/> Examples of how the height of a triangle can be represented relative to its base[\/caption]\r\n<\/center>\r\n<section class=\"textbox keyTakeaway\">\r\n<div>\r\n<h3>triangle properties<\/h3>\r\n<p>For any triangle [latex]\\Delta ABC[\/latex], the sum of the measures of the angles is [latex]\\text{180}^ \\circ[\/latex].<\/p>\r\n<p>&nbsp;<\/p>\r\n<p style=\"text-align: center;\">[latex]m\\angle{A}+m\\angle{B}+m\\angle{C}=180^\\circ [\/latex]<\/p>\r\n<p>&nbsp;<\/p>\r\n<p>The <strong>perimeter<\/strong> of a triangle is the sum of the lengths of the sides.<\/p>\r\n<p>&nbsp;<\/p>\r\n<p style=\"text-align: center;\">[latex]P=a+b+c[\/latex]<\/p>\r\n<p>&nbsp;<\/p>\r\n<p>The <strong>area<\/strong> of a triangle is one-half the base, [latex]b[\/latex], times the height, [latex]h[\/latex].<\/p>\r\n<p>&nbsp;<\/p>\r\n<p style=\"text-align: center;\">[latex]A={\\Large\\frac{1}{2}}bh[\/latex]<\/p>\r\n<p>&nbsp;<\/p>\r\n<br \/>\r\n<center>\r\n[caption id=\"\" align=\"aligncenter\" width=\"190\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223917\/CNX_BMath_Figure_09_04_038_img.png\" alt=\"A triangle, with vertices labeled A, B, and C. The sides are labeled a, b, and c. There is a vertical dotted line from vertex B at the top of the triangle to the base of the triangle, meeting the base at a right angle. The dotted line is labeled h.\" width=\"190\" height=\"160\" \/> Triangle with key features labeled[\/caption]\r\n<\/center>\r\n<p>&nbsp;<\/p>\r\n<\/div>\r\n<\/section>\r\n<section class=\"textbox example\">Find the area of a triangle whose base is [latex]11[\/latex] inches and whose height is [latex]8[\/latex] inches.<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-id1168468457178\" class=\"unnumbered unstyled\" style=\"width: 42.0764%; height: 557px;\" summary=\"Step 1 says, \">\r\n<tbody>\r\n<tr style=\"height: 207px;\">\r\n<td style=\"height: 207px;\">Step 1. <strong>Read<\/strong> the problem. Draw the figure and label it with the given information.<\/td>\r\n<td style=\"height: 207px;\">\r\n[caption id=\"\" align=\"alignnone\" width=\"318\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223918\/CNX_BMath_Figure_09_04_073_img-01.png\" alt=\"A triangle with the base labeled 11 in and a dotted vertical line from the top vertex to the base to form a right angle. This dotted line is labeled 8 in.\" width=\"318\" height=\"202\" \/> Triangle with base and height labeled[\/caption]\r\n<\/td>\r\n<\/tr>\r\n<tr style=\"height: 22px;\">\r\n<td style=\"height: 22px;\">Step 2. <strong>Identify<\/strong> what you are looking for.<\/td>\r\n<td style=\"height: 22px;\">the area of the triangle<\/td>\r\n<\/tr>\r\n<tr style=\"height: 22px;\">\r\n<td style=\"height: 22px;\">Step 3. <strong>Name.<\/strong> Choose a variable to represent it.<\/td>\r\n<td style=\"height: 22px;\">Let [latex]A[\/latex] = area of the triangle<\/td>\r\n<\/tr>\r\n<tr style=\"height: 131px;\">\r\n<td style=\"height: 131px;\">\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=\"height: 131px;\">\r\n[caption id=\"\" align=\"alignnone\" width=\"318\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223920\/CNX_BMath_Figure_09_04_073_img-02.png\" alt=\"The equation A = one half times b times h. The equation is written again with 11 substituted for b and 8 substituted for h.\" width=\"318\" height=\"110\" \/> Formula for Area[\/caption]\r\n<\/td>\r\n<\/tr>\r\n<tr style=\"height: 22px;\">\r\n<td style=\"height: 22px;\">Step 5. <strong>Solve<\/strong> the equation.<\/td>\r\n<td style=\"height: 22px;\">[latex]A=44[\/latex] square inches.<\/td>\r\n<\/tr>\r\n<tr style=\"height: 131px;\">\r\n<td style=\"height: 131px;\">\r\n<p>Step 6. <strong>Check.<\/strong><\/p>\r\n<\/td>\r\n<td style=\"height: 131px;\">\r\n<p>[latex]A=\\frac{1}{2}bh[\/latex]<\/p>\r\n<p>[latex]44\\stackrel{?}{=}\\frac{1}{2}(11)8[\/latex]<\/p>\r\n<p>[latex]44=44\\quad\\checkmark[\/latex]<\/p>\r\n<\/td>\r\n<\/tr>\r\n<tr style=\"height: 22px;\">\r\n<td style=\"height: 22px;\">Step 7. <strong>Answer<\/strong> the question.<\/td>\r\n<td style=\"height: 22px;\">The area is [latex]44[\/latex] square inches.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox tryIt\">\r\n<p>[ohm_question hide_question_numbers=1]288402[\/ohm_question]<\/p>\r\n<\/section>\r\n<h2>Find the Circumference and Area of Circles<\/h2>\r\n<p>The properties of circles have been studied for over [latex]2,000[\/latex] years. All circles have exactly the same shape, but their sizes are affected by the length of the <strong>radius<\/strong>, a line segment from the center to any point on the circle. A line segment that passes through a circle\u2019s center connecting two points on the circle is called a <strong>diameter<\/strong>. The diameter is twice as long as the radius. The distance around a circle is called its <strong>circumference<\/strong>.<\/p>\r\n<center>\r\n[caption id=\"\" align=\"aligncenter\" width=\"212\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221636\/CNX_BMath_Figure_05_03_008.png\" alt=\"A circle is shown. A dotted line running through the widest portion of the circle is labeled as a diameter. A dotted line from the center of the circle to a point on the circle is labeled as a radius. Along the edge of the circle is the circumference.\" width=\"212\" height=\"212\" \/> Diagram of a circle with radius, diameter, and circumference labeled[\/caption]\r\n<\/center>\r\n<p>&nbsp;<\/p>\r\n<p>Archimedes discovered that for circles of all different sizes, dividing the circumference by the diameter always gives the same number. The value of this number is pi, symbolized by Greek letter [latex]\\pi [\/latex] (pronounced \"pie\"). We approximate [latex]\\pi [\/latex] with [latex]3.14[\/latex] or [latex]\\Large\\frac{22}{7}[\/latex] depending on whether the radius of the circle is given as a decimal or a fraction.<\/p>\r\n<section class=\"textbox proTip\">If you use the [latex]\\pi [\/latex] key on your calculator to do the calculations in this section, your answers will be slightly different from the answers shown. That is because the [latex]\\pi [\/latex] key uses more than two decimal places.<\/section>\r\n<section class=\"textbox keyTakeaway\">\r\n<div>\r\n<h3>properties of circles<\/h3>\r\n<center>\r\n[caption id=\"\" align=\"aligncenter\" width=\"156\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224027\/CNX_BMath_Figure_09_05_001.png\" alt=\"An image of a circle is shown. There is a line drawn through the widest part at the center of the circle with a red dot indicating the center of the circle. The line is labeled d. The two segments from the center of the circle to the outside of the circle are each labeled r.\" width=\"156\" height=\"156\" \/> Diagram of a circle[\/caption]\r\n<\/center>\r\n<p>&nbsp;<\/p>\r\n<br \/>\r\n<ul id=\"fs-id1489165\">\r\n\t<li>[latex]r[\/latex] is the length of the <strong>radius<\/strong><\/li>\r\n\t<li>[latex]d[\/latex] is the length of the <strong>diameter<\/strong><\/li>\r\n\t<li>[latex]d=2r[\/latex]<\/li>\r\n\t<li><strong>Circumference <\/strong>is the perimeter of a circle. The formula for circumference is [latex]C=2\\pi r[\/latex]<\/li>\r\n\t<li>The formula for <strong>area <\/strong>of a circle is [latex]A=\\pi {r}^{2}[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p>A circular sandbox has a radius of [latex]2.5[\/latex] feet. Find the<\/p>\r\n<ol style=\"list-style-type: decimal;\">\r\n\t<li>Circumference of the sandbox<\/li>\r\n\t<li>Area of the sandbox<\/li>\r\n<\/ol>\r\n<p>[reveal-answer q=\"247919\"]Show Solution[\/reveal-answer] [hidden-answer a=\"247919\"]<\/p>\r\n<table style=\"width: 100%;\">\r\n<tbody>\r\n<tr>\r\n<th style=\"width: 29.1063%;\">1. Circumference of the sandbox<\/th>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 29.1063%;\">Step 1. <strong>Read<\/strong> the problem. Draw the figure and label it with the given information.<\/td>\r\n<td style=\"width: 69.9275%;\">\r\n[caption id=\"\" align=\"alignnone\" width=\"159\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224028\/CNX_BMath_Figure_09_05_029_img-01.png\" alt=\"A circle with radius labeled as 2.5 feet\" width=\"159\" height=\"159\" \/> Circle with radius of 2.5 ft[\/caption]\r\n<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 29.1063%;\">Step 2. <strong>Identify<\/strong> what you are looking for.<\/td>\r\n<td style=\"width: 69.9275%;\">the circumference of the circle<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 29.1063%;\">Step 3. <strong>Name.<\/strong> Choose a variable to represent it.<\/td>\r\n<td style=\"width: 69.9275%;\">Let [latex]C [\/latex] = circumference of the circle<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 29.1063%;\">Step 4. <strong>Translate.<\/strong> Write the appropriate formula Substitute<\/td>\r\n<td style=\"width: 69.9275%;\">[latex]C=2\\pi r[\/latex] [latex]C=2\\pi \\left(2.5\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 29.1063%;\">Step 5. <strong>Solve<\/strong> the equation.<\/td>\r\n<td style=\"width: 69.9275%;\">[latex]C\\approx 2\\left(3.14\\right)\\left(2.5\\right)[\/latex] [latex]C\\approx 15\\text{ft}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 29.1063%;\">Step 6. <strong>Check.<\/strong> Does this answer make sense?<\/td>\r\n<td style=\"width: 69.9275%;\">Yes. If we draw a square around the circle, its sides would be [latex]5[\/latex] ft (twice the radius), so its perimeter would be [latex]20[\/latex] ft. This is slightly more than the circle's circumference, [latex]15.7[\/latex] ft.\r\n[caption id=\"\" align=\"alignnone\" width=\"206\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224029\/CNX_BMath_Figure_09_05_029_img-02.png\" alt=\"A circle in a red square. The circle's radius is shown as 2.5 feet and the sides of the square are each labeled as 5 feet.\" width=\"206\" height=\"188\" \/> Square of side 5 ft containing a circle of radius 2.5 ft[\/caption]\r\n<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 29.1063%;\">Step 7. <strong>Answer<\/strong> the question.<\/td>\r\n<td style=\"width: 69.9275%;\">The circumference of the sandbox is [latex]15.7[\/latex] feet.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<table style=\"width: 100%; height: 340px;\">\r\n<tbody>\r\n<tr style=\"height: 22px;\">\r\n<th style=\"height: 22px;\">2. Area of the sandbox<\/th>\r\n<\/tr>\r\n<tr style=\"height: 164px;\">\r\n<td style=\"height: 164px;\">Step 1. <strong>Read<\/strong> the problem. Draw the figure and label it with the given information.<\/td>\r\n<td style=\"height: 164px;\">\r\n[caption id=\"\" align=\"alignnone\" width=\"159\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224028\/CNX_BMath_Figure_09_05_029_img-01.png\" alt=\"A circle with radius labeled as 2.5 feet\" width=\"159\" height=\"159\" \/> Circle with radius of 2.5 ft[\/caption]\r\n<\/td>\r\n<\/tr>\r\n<tr style=\"height: 22px;\">\r\n<td style=\"height: 22px;\">Step 2. <strong>Identify<\/strong> what you are looking for.<\/td>\r\n<td style=\"height: 22px;\">the area of the circle<\/td>\r\n<\/tr>\r\n<tr style=\"height: 22px;\">\r\n<td style=\"height: 22px;\">Step 3. <strong>Name.<\/strong> Choose a variable to represent it.<\/td>\r\n<td style=\"height: 22px;\">Let [latex]A[\/latex] = the area of the circle<\/td>\r\n<\/tr>\r\n<tr style=\"height: 22px;\">\r\n<td style=\"height: 22px;\">Step 4. <strong>Translate.<\/strong> Write the appropriate formula Substitute<\/td>\r\n<td style=\"height: 22px;\">[latex]A=\\pi {r}^{2}[\/latex] [latex]A=\\pi{\\left(2.5\\right)}^{2}[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 22px;\">\r\n<td style=\"height: 22px;\">Step 5. <strong>Solve<\/strong> the equation.<\/td>\r\n<td style=\"height: 22px;\">[latex]A\\approx \\left(3.14\\right){\\left(2.5\\right)}^{2}[\/latex] [latex]A\\approx 19.625\\text{ sq. ft}[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 44px;\">\r\n<td style=\"height: 44px;\">Step 6. <strong>Check.<\/strong> Does this answer make sense?<\/td>\r\n<td style=\"height: 44px;\">Yes. If we draw a square around the circle, its sides would be [latex]5[\/latex] ft, as shown in part 1. So the area of the square would be [latex]25[\/latex] sq. ft. This is slightly more than the circle's area, [latex]19.625[\/latex] sq. ft.<\/td>\r\n<\/tr>\r\n<tr style=\"height: 22px;\">\r\n<td style=\"height: 22px;\">Step 7. <strong>Answer<\/strong> the question.<\/td>\r\n<td style=\"height: 22px;\">The area of the circle is [latex]19.625[\/latex] square 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 tryIt\">\r\n<p>[ohm_question hide_question_numbers=1]146563[\/ohm_question]<\/p>\r\n<\/section>\r\n<h2>Find the Volume and Surface Area of Rectangular Solids<\/h2>\r\n<p>Volume measures the space a shape occupies, while surface area describes the total area of all the surfaces of a three-dimensional object. For rectangular solids, which include cubes and rectangular prisms, these measurements are based on the object's length, width, and height.<\/p>\r\n<section class=\"textbox keyTakeaway\">\r\n<div>\r\n<h3>volume and surface area of a rectangular solid<\/h3>\r\n<p>For a rectangular solid with length [latex]L[\/latex], width [latex]W[\/latex], and height [latex]H[\/latex]:<\/p>\r\n<p>&nbsp;<\/p>\r\n<center>\r\n[caption id=\"\" align=\"aligncenter\" width=\"439\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224139\/CNX_BMath_Figure_09_06_006_img.png\" alt=\"A rectangular solid, with sides labeled L, W, and H. Beside it is Volume: V equals LWH equals BH. Below that is Surface Area: S equals 2LH plus 2LW plus 2WH.\" width=\"439\" height=\"174\" \/> Rectangular solid with formulas for volume and surface area[\/caption]\r\n<\/center><\/div>\r\n<\/section>\r\n<section class=\"textbox example\">For a rectangular solid with length [latex]14[\/latex] cm, height [latex]17[\/latex] cm, and width [latex]9[\/latex] cm. Find the\r\n\r\n<ol style=\"list-style-type: decimal;\">\r\n\t<li>Volume<\/li>\r\n\t<li>Surface area<\/li>\r\n<\/ol>\r\n<p>[reveal-answer q=\"4331\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"4331\"]<br \/>\r\nStep 1 is the same for both 1. and 2., so we will show it just once.<\/p>\r\n<table id=\"eip-id1168468779989\" class=\"unnumbered unstyled\" summary=\"The text reads, \">\r\n<tbody>\r\n<tr>\r\n<td>\r\n<p>Step 1. <strong>Read<\/strong> the problem. Draw the figure and<\/p>\r\n<p>label it with the given information.<\/p>\r\n<\/td>\r\n<td>\r\n[caption id=\"\" align=\"alignnone\" width=\"170\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224140\/CNX_BMath_Figure_09_06_038_img-01.png\" alt=\"A rectangular prism with one side labeled 14, one labeled 9, and another labeled 17\" width=\"170\" height=\"117\" \/> Rectangular prism with sides labeled[\/caption]\r\n<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<ol style=\"list-style-type: decimal;\">\r\n\t<li>\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td>Step 2. <strong>Identify<\/strong> what you are looking for.<\/td>\r\n<td>the volume of the rectangular solid<\/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]V[\/latex] = volume<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>\r\n<p>Step 4. <strong>Translate.<\/strong><\/p>\r\n<p>Write the appropriate formula.<\/p>\r\n<p>Substitute.<\/p>\r\n<\/td>\r\n<td>\r\n<p>[latex]V=LWH[\/latex]<\/p>\r\n<p>[latex]V=\\mathrm{14}\\cdot 9\\cdot 17[\/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>[latex]V=2,142[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>\r\n<p>Step 6. <strong>Check<\/strong><\/p>\r\n<p>We leave it to you to check your calculations.<\/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 volume is [latex]2,142[\/latex] cubic centimeters.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/li>\r\n\t<li>\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td>Step 2. <strong>Identify<\/strong> what you are looking for.<\/td>\r\n<td>the surface area of the solid<\/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]S[\/latex] = surface area<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>\r\n<p>Step 4. <strong>Translate.<\/strong><\/p>\r\n<p>Write the appropriate formula.<\/p>\r\n<p>Substitute.<\/p>\r\n<\/td>\r\n<td>\r\n<p>[latex]S=2LH+2LW+2WH[\/latex]<\/p>\r\n<p>[latex]S=2\\left(14\\cdot 17\\right)+2\\left(14\\cdot 9\\right)+2\\left(9\\cdot 17\\right)[\/latex]<\/p>\r\n<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 5. <strong>Solve the equation.<\/strong><\/td>\r\n<td>[latex]S=1,034[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 6. <strong>Check:<\/strong> Double-check with a calculator.<\/td>\r\n<td>&nbsp;<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 7. <strong>Answer<\/strong> the question.<\/td>\r\n<td>The surface area is [latex]1,034[\/latex] square centimeters.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/li>\r\n<\/ol>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox tryIt\">\r\n<p>[ohm_question hide_question_numbers=1]146789[\/ohm_question]<\/p>\r\n<\/section>\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Write formulas to calculate the area, perimeter, and volume of different shapes<\/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\">Rectangle with all sides 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\">Rug with a length of 2 and a width of 3<\/figcaption><\/figure>\n<\/div>\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 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>\n<figure style=\"width: 303px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" 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\" \/><figcaption class=\"wp-caption-text\">Rectangle with all sides labeled<\/figcaption><\/figure>\n<\/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>\n<figure style=\"width: 524px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" 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\" \/><figcaption class=\"wp-caption-text\">Formula for perimeter<\/figcaption><\/figure>\n<\/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>\n<figure style=\"width: 310px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" 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\" \/><figcaption class=\"wp-caption-text\">Rectangle with all sides labeled<\/figcaption><\/figure>\n<\/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 [latex]A[\/latex] = the area<\/td>\n<\/tr>\n<tr>\n<td>Step 4. <strong>Translate.<\/strong> Write the appropriate formula. Substitute.<\/td>\n<td>\n<figure style=\"width: 310px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" 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\" \/><figcaption class=\"wp-caption-text\">Formula for Area<\/figcaption><\/figure>\n<\/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=\"ohm288400\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=288400&theme=lumen&iframe_resize_id=ohm288400&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section>\n<h2>Find the Area and Perimeter of a Triangle<\/h2>\n<p>We now know how to find the area of a rectangle. We can use this fact to help us visualize the formula for the area of a triangle. In the rectangle below, we\u2019ve labeled the length [latex]b[\/latex] and the width [latex]h[\/latex], so its area is [latex]bh[\/latex].<\/p>\n<div style=\"text-align: center;\">\n<figure style=\"width: 151px\" 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\/24223912\/CNX_BMath_Figure_09_04_035.png\" alt=\"A rectangle with the side labeled h and the bottom labeled b. The center says A equals bh.\" width=\"151\" height=\"89\" \/><figcaption class=\"wp-caption-text\">Rectangle with with height, base, and area labeled<\/figcaption><\/figure>\n<\/div>\n<p>&nbsp;<\/p>\n<p>\nWe can divide this rectangle into two congruent triangles (see the image below). Triangles that are congruent have identical side lengths and angles, and so their areas are equal. The area of each triangle is one-half the area of the rectangle, or [latex]\\Large\\frac{1}{2}\\normalsize bh[\/latex]. This example helps us see why the formula for the area of a triangle is [latex]A=\\Large\\frac{1}{2}\\normalsize bh[\/latex].<\/p>\n<div style=\"text-align: center;\">\n<figure style=\"width: 323px\" 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\/24223913\/CNX_BMath_Figure_09_04_036.png\" alt=\"A rectangle with a diagonal line drawn from the upper left corner to the bottom right corner. The side of the rectangle is labeled h and the bottom is labeled b. Each triangle says one-half bh. To the right of the rectangle, it says &quot;Area of each triangle A = one-half bh&quot;.\" width=\"323\" height=\"107\" \/><figcaption class=\"wp-caption-text\">Rectangle divided into two triangles with height, base, and area labeled<\/figcaption><\/figure>\n<\/div>\n<p>&nbsp;<\/p>\n<p>To find the area of the triangle, you need to know its base and height. The base is the length of one side of the triangle, usually the side at the bottom. The height is the length of the line that connects the base to the opposite vertex, and makes a [latex]\\text{90}^ \\circ[\/latex] angle with the base. The image below\u00a0shows three triangles with the base and height of each marked.<\/p>\n<div style=\"text-align: center;\">\n<figure style=\"width: 563px\" 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\/24223914\/CNX_BMath_Figure_09_04_037.png\" alt=\"Three triangles. The triangle on the left is a right triangle. The bottom is labeled b and the side is labeled h. The middle triangle is an acute triangle. The bottom is labeled b. There is a dotted line from the top vertex to the base of the triangle, forming a right angle with the base. That line is labeled h. The triangle on the right is an obtuse triangle. The bottom of the triangle is labeled b. The base has a dotted line extended out and forms a right angle with a dotted line to the top of the triangle. The vertical line is labeled h.\" width=\"563\" height=\"107\" \/><figcaption class=\"wp-caption-text\">Examples of how the height of a triangle can be represented relative to its base<\/figcaption><\/figure>\n<\/div>\n<section class=\"textbox keyTakeaway\">\n<div>\n<h3>triangle properties<\/h3>\n<p>For any triangle [latex]\\Delta ABC[\/latex], the sum of the measures of the angles is [latex]\\text{180}^ \\circ[\/latex].<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: center;\">[latex]m\\angle{A}+m\\angle{B}+m\\angle{C}=180^\\circ[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>The <strong>perimeter<\/strong> of a triangle is the sum of the lengths of the sides.<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: center;\">[latex]P=a+b+c[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>The <strong>area<\/strong> of a triangle is one-half the base, [latex]b[\/latex], times the height, [latex]h[\/latex].<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: center;\">[latex]A={\\Large\\frac{1}{2}}bh[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<div class=\"wp-nocaption \"><\/div>\n<div style=\"text-align: center;\">\n<figure style=\"width: 190px\" 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\/24223917\/CNX_BMath_Figure_09_04_038_img.png\" alt=\"A triangle, with vertices labeled A, B, and C. The sides are labeled a, b, and c. There is a vertical dotted line from vertex B at the top of the triangle to the base of the triangle, meeting the base at a right angle. The dotted line is labeled h.\" width=\"190\" height=\"160\" \/><figcaption class=\"wp-caption-text\">Triangle with key features labeled<\/figcaption><\/figure>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/section>\n<section class=\"textbox example\">Find the area of a triangle whose base is [latex]11[\/latex] inches and whose height is [latex]8[\/latex] inches.<\/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-id1168468457178\" class=\"unnumbered unstyled\" style=\"width: 42.0764%; height: 557px;\" summary=\"Step 1 says,\">\n<tbody>\n<tr style=\"height: 207px;\">\n<td style=\"height: 207px;\">Step 1. <strong>Read<\/strong> the problem. Draw the figure and label it with the given information.<\/td>\n<td style=\"height: 207px;\">\n<figure style=\"width: 318px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223918\/CNX_BMath_Figure_09_04_073_img-01.png\" alt=\"A triangle with the base labeled 11 in and a dotted vertical line from the top vertex to the base to form a right angle. This dotted line is labeled 8 in.\" width=\"318\" height=\"202\" \/><figcaption class=\"wp-caption-text\">Triangle with base and height labeled<\/figcaption><\/figure>\n<\/td>\n<\/tr>\n<tr style=\"height: 22px;\">\n<td style=\"height: 22px;\">Step 2. <strong>Identify<\/strong> what you are looking for.<\/td>\n<td style=\"height: 22px;\">the area of the triangle<\/td>\n<\/tr>\n<tr style=\"height: 22px;\">\n<td style=\"height: 22px;\">Step 3. <strong>Name.<\/strong> Choose a variable to represent it.<\/td>\n<td style=\"height: 22px;\">Let [latex]A[\/latex] = area of the triangle<\/td>\n<\/tr>\n<tr style=\"height: 131px;\">\n<td style=\"height: 131px;\">\n<p>Step 4.<strong>Translate.<\/strong><\/p>\n<p>Write the appropriate formula.<\/p>\n<p>Substitute.<\/p>\n<\/td>\n<td style=\"height: 131px;\">\n<figure style=\"width: 318px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223920\/CNX_BMath_Figure_09_04_073_img-02.png\" alt=\"The equation A = one half times b times h. The equation is written again with 11 substituted for b and 8 substituted for h.\" width=\"318\" height=\"110\" \/><figcaption class=\"wp-caption-text\">Formula for Area<\/figcaption><\/figure>\n<\/td>\n<\/tr>\n<tr style=\"height: 22px;\">\n<td style=\"height: 22px;\">Step 5. <strong>Solve<\/strong> the equation.<\/td>\n<td style=\"height: 22px;\">[latex]A=44[\/latex] square inches.<\/td>\n<\/tr>\n<tr style=\"height: 131px;\">\n<td style=\"height: 131px;\">\n<p>Step 6. <strong>Check.<\/strong><\/p>\n<\/td>\n<td style=\"height: 131px;\">\n[latex]A=\\frac{1}{2}bh[\/latex]<br \/>\n[latex]44\\stackrel{?}{=}\\frac{1}{2}(11)8[\/latex]<br \/>\n[latex]44=44\\quad\\checkmark[\/latex]\n<\/td>\n<\/tr>\n<tr style=\"height: 22px;\">\n<td style=\"height: 22px;\">Step 7. <strong>Answer<\/strong> the question.<\/td>\n<td style=\"height: 22px;\">The area is [latex]44[\/latex] square inches.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\">\n<iframe loading=\"lazy\" id=\"ohm288402\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=288402&theme=lumen&iframe_resize_id=ohm288402&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><br \/>\n<\/section>\n<h2>Find the Circumference and Area of Circles<\/h2>\n<p>The properties of circles have been studied for over [latex]2,000[\/latex] years. All circles have exactly the same shape, but their sizes are affected by the length of the <strong>radius<\/strong>, a line segment from the center to any point on the circle. A line segment that passes through a circle\u2019s center connecting two points on the circle is called a <strong>diameter<\/strong>. The diameter is twice as long as the radius. The distance around a circle is called its <strong>circumference<\/strong>.<\/p>\n<div style=\"text-align: center;\">\n<figure style=\"width: 212px\" 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\/24221636\/CNX_BMath_Figure_05_03_008.png\" alt=\"A circle is shown. A dotted line running through the widest portion of the circle is labeled as a diameter. A dotted line from the center of the circle to a point on the circle is labeled as a radius. Along the edge of the circle is the circumference.\" width=\"212\" height=\"212\" \/><figcaption class=\"wp-caption-text\">Diagram of a circle with radius, diameter, and circumference labeled<\/figcaption><\/figure>\n<\/div>\n<p>&nbsp;<\/p>\n<p>Archimedes discovered that for circles of all different sizes, dividing the circumference by the diameter always gives the same number. The value of this number is pi, symbolized by Greek letter [latex]\\pi[\/latex] (pronounced &#8220;pie&#8221;). We approximate [latex]\\pi[\/latex] with [latex]3.14[\/latex] or [latex]\\Large\\frac{22}{7}[\/latex] depending on whether the radius of the circle is given as a decimal or a fraction.<\/p>\n<section class=\"textbox proTip\">If you use the [latex]\\pi[\/latex] key on your calculator to do the calculations in this section, your answers will be slightly different from the answers shown. That is because the [latex]\\pi[\/latex] key uses more than two decimal places.<\/section>\n<section class=\"textbox keyTakeaway\">\n<div>\n<h3>properties of circles<\/h3>\n<div style=\"text-align: center;\">\n<figure style=\"width: 156px\" 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\/24224027\/CNX_BMath_Figure_09_05_001.png\" alt=\"An image of a circle is shown. There is a line drawn through the widest part at the center of the circle with a red dot indicating the center of the circle. The line is labeled d. The two segments from the center of the circle to the outside of the circle are each labeled r.\" width=\"156\" height=\"156\" \/><figcaption class=\"wp-caption-text\">Diagram of a circle<\/figcaption><\/figure>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"wp-nocaption \"><\/div>\n<ul id=\"fs-id1489165\">\n<li>[latex]r[\/latex] is the length of the <strong>radius<\/strong><\/li>\n<li>[latex]d[\/latex] is the length of the <strong>diameter<\/strong><\/li>\n<li>[latex]d=2r[\/latex]<\/li>\n<li><strong>Circumference <\/strong>is the perimeter of a circle. The formula for circumference is [latex]C=2\\pi r[\/latex]<\/li>\n<li>The formula for <strong>area <\/strong>of a circle is [latex]A=\\pi {r}^{2}[\/latex]<\/li>\n<\/ul>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p>A circular sandbox has a radius of [latex]2.5[\/latex] feet. Find the<\/p>\n<ol style=\"list-style-type: decimal;\">\n<li>Circumference of the sandbox<\/li>\n<li>Area of the sandbox<\/li>\n<\/ol>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q247919\">Show Solution<\/button> <\/p>\n<div id=\"q247919\" class=\"hidden-answer\" style=\"display: none\">\n<table style=\"width: 100%;\">\n<tbody>\n<tr>\n<th style=\"width: 29.1063%;\">1. Circumference of the sandbox<\/th>\n<\/tr>\n<tr>\n<td style=\"width: 29.1063%;\">Step 1. <strong>Read<\/strong> the problem. Draw the figure and label it with the given information.<\/td>\n<td style=\"width: 69.9275%;\">\n<figure style=\"width: 159px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224028\/CNX_BMath_Figure_09_05_029_img-01.png\" alt=\"A circle with radius labeled as 2.5 feet\" width=\"159\" height=\"159\" \/><figcaption class=\"wp-caption-text\">Circle with radius of 2.5 ft<\/figcaption><\/figure>\n<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 29.1063%;\">Step 2. <strong>Identify<\/strong> what you are looking for.<\/td>\n<td style=\"width: 69.9275%;\">the circumference of the circle<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 29.1063%;\">Step 3. <strong>Name.<\/strong> Choose a variable to represent it.<\/td>\n<td style=\"width: 69.9275%;\">Let [latex]C[\/latex] = circumference of the circle<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 29.1063%;\">Step 4. <strong>Translate.<\/strong> Write the appropriate formula Substitute<\/td>\n<td style=\"width: 69.9275%;\">[latex]C=2\\pi r[\/latex] [latex]C=2\\pi \\left(2.5\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 29.1063%;\">Step 5. <strong>Solve<\/strong> the equation.<\/td>\n<td style=\"width: 69.9275%;\">[latex]C\\approx 2\\left(3.14\\right)\\left(2.5\\right)[\/latex] [latex]C\\approx 15\\text{ft}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 29.1063%;\">Step 6. <strong>Check.<\/strong> Does this answer make sense?<\/td>\n<td style=\"width: 69.9275%;\">Yes. If we draw a square around the circle, its sides would be [latex]5[\/latex] ft (twice the radius), so its perimeter would be [latex]20[\/latex] ft. This is slightly more than the circle&#8217;s circumference, [latex]15.7[\/latex] ft.<\/p>\n<figure style=\"width: 206px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224029\/CNX_BMath_Figure_09_05_029_img-02.png\" alt=\"A circle in a red square. The circle's radius is shown as 2.5 feet and the sides of the square are each labeled as 5 feet.\" width=\"206\" height=\"188\" \/><figcaption class=\"wp-caption-text\">Square of side 5 ft containing a circle of radius 2.5 ft<\/figcaption><\/figure>\n<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 29.1063%;\">Step 7. <strong>Answer<\/strong> the question.<\/td>\n<td style=\"width: 69.9275%;\">The circumference of the sandbox is [latex]15.7[\/latex] feet.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<table style=\"width: 100%; height: 340px;\">\n<tbody>\n<tr style=\"height: 22px;\">\n<th style=\"height: 22px;\">2. Area of the sandbox<\/th>\n<\/tr>\n<tr style=\"height: 164px;\">\n<td style=\"height: 164px;\">Step 1. <strong>Read<\/strong> the problem. Draw the figure and label it with the given information.<\/td>\n<td style=\"height: 164px;\">\n<figure style=\"width: 159px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224028\/CNX_BMath_Figure_09_05_029_img-01.png\" alt=\"A circle with radius labeled as 2.5 feet\" width=\"159\" height=\"159\" \/><figcaption class=\"wp-caption-text\">Circle with radius of 2.5 ft<\/figcaption><\/figure>\n<\/td>\n<\/tr>\n<tr style=\"height: 22px;\">\n<td style=\"height: 22px;\">Step 2. <strong>Identify<\/strong> what you are looking for.<\/td>\n<td style=\"height: 22px;\">the area of the circle<\/td>\n<\/tr>\n<tr style=\"height: 22px;\">\n<td style=\"height: 22px;\">Step 3. <strong>Name.<\/strong> Choose a variable to represent it.<\/td>\n<td style=\"height: 22px;\">Let [latex]A[\/latex] = the area of the circle<\/td>\n<\/tr>\n<tr style=\"height: 22px;\">\n<td style=\"height: 22px;\">Step 4. <strong>Translate.<\/strong> Write the appropriate formula Substitute<\/td>\n<td style=\"height: 22px;\">[latex]A=\\pi {r}^{2}[\/latex] [latex]A=\\pi{\\left(2.5\\right)}^{2}[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 22px;\">\n<td style=\"height: 22px;\">Step 5. <strong>Solve<\/strong> the equation.<\/td>\n<td style=\"height: 22px;\">[latex]A\\approx \\left(3.14\\right){\\left(2.5\\right)}^{2}[\/latex] [latex]A\\approx 19.625\\text{ sq. ft}[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 44px;\">\n<td style=\"height: 44px;\">Step 6. <strong>Check.<\/strong> Does this answer make sense?<\/td>\n<td style=\"height: 44px;\">Yes. If we draw a square around the circle, its sides would be [latex]5[\/latex] ft, as shown in part 1. So the area of the square would be [latex]25[\/latex] sq. ft. This is slightly more than the circle&#8217;s area, [latex]19.625[\/latex] sq. ft.<\/td>\n<\/tr>\n<tr style=\"height: 22px;\">\n<td style=\"height: 22px;\">Step 7. <strong>Answer<\/strong> the question.<\/td>\n<td style=\"height: 22px;\">The area of the circle is [latex]19.625[\/latex] square feet.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\">\n<iframe loading=\"lazy\" id=\"ohm146563\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=146563&theme=lumen&iframe_resize_id=ohm146563&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><br \/>\n<\/section>\n<h2>Find the Volume and Surface Area of Rectangular Solids<\/h2>\n<p>Volume measures the space a shape occupies, while surface area describes the total area of all the surfaces of a three-dimensional object. For rectangular solids, which include cubes and rectangular prisms, these measurements are based on the object&#8217;s length, width, and height.<\/p>\n<section class=\"textbox keyTakeaway\">\n<div>\n<h3>volume and surface area of a rectangular solid<\/h3>\n<p>For a rectangular solid with length [latex]L[\/latex], width [latex]W[\/latex], and height [latex]H[\/latex]:<\/p>\n<p>&nbsp;<\/p>\n<div style=\"text-align: center;\">\n<figure style=\"width: 439px\" 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\/24224139\/CNX_BMath_Figure_09_06_006_img.png\" alt=\"A rectangular solid, with sides labeled L, W, and H. Beside it is Volume: V equals LWH equals BH. Below that is Surface Area: S equals 2LH plus 2LW plus 2WH.\" width=\"439\" height=\"174\" \/><figcaption class=\"wp-caption-text\">Rectangular solid with formulas for volume and surface area<\/figcaption><\/figure>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">For a rectangular solid with length [latex]14[\/latex] cm, height [latex]17[\/latex] cm, and width [latex]9[\/latex] cm. Find the<\/p>\n<ol style=\"list-style-type: decimal;\">\n<li>Volume<\/li>\n<li>Surface area<\/li>\n<\/ol>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q4331\">Show Solution<\/button><\/p>\n<div id=\"q4331\" class=\"hidden-answer\" style=\"display: none\">\nStep 1 is the same for both 1. and 2., so we will show it just once.<\/p>\n<table id=\"eip-id1168468779989\" class=\"unnumbered unstyled\" summary=\"The text reads,\">\n<tbody>\n<tr>\n<td>\n<p>Step 1. <strong>Read<\/strong> the problem. Draw the figure and<\/p>\n<p>label it with the given information.<\/p>\n<\/td>\n<td>\n<figure style=\"width: 170px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224140\/CNX_BMath_Figure_09_06_038_img-01.png\" alt=\"A rectangular prism with one side labeled 14, one labeled 9, and another labeled 17\" width=\"170\" height=\"117\" \/><figcaption class=\"wp-caption-text\">Rectangular prism with sides labeled<\/figcaption><\/figure>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<ol style=\"list-style-type: decimal;\">\n<li>\n<table>\n<tbody>\n<tr>\n<td>Step 2. <strong>Identify<\/strong> what you are looking for.<\/td>\n<td>the volume of the rectangular solid<\/td>\n<\/tr>\n<tr>\n<td>Step 3. <strong>Name.<\/strong> Choose a variable to represent it.<\/td>\n<td>Let [latex]V[\/latex] = volume<\/td>\n<\/tr>\n<tr>\n<td>\n<p>Step 4. <strong>Translate.<\/strong><\/p>\n<p>Write the appropriate formula.<\/p>\n<p>Substitute.<\/p>\n<\/td>\n<td>\n[latex]V=LWH[\/latex]<br \/>\n[latex]V=\\mathrm{14}\\cdot 9\\cdot 17[\/latex]\n<\/td>\n<\/tr>\n<tr>\n<td>Step 5. <strong>Solve<\/strong> the equation.<\/td>\n<td>[latex]V=2,142[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>\n<p>Step 6. <strong>Check<\/strong><\/p>\n<p>We leave it to you to check your calculations.<\/p>\n<\/td>\n<td>&nbsp;<\/td>\n<\/tr>\n<tr>\n<td>Step 7. <strong>Answer<\/strong> the question.<\/td>\n<td>The volume is [latex]2,142[\/latex] cubic centimeters.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/li>\n<li>\n<table>\n<tbody>\n<tr>\n<td>Step 2. <strong>Identify<\/strong> what you are looking for.<\/td>\n<td>the surface area of the solid<\/td>\n<\/tr>\n<tr>\n<td>Step 3. <strong>Name.<\/strong> Choose a variable to represent it.<\/td>\n<td>Let [latex]S[\/latex] = surface area<\/td>\n<\/tr>\n<tr>\n<td>\n<p>Step 4. <strong>Translate.<\/strong><\/p>\n<p>Write the appropriate formula.<\/p>\n<p>Substitute.<\/p>\n<\/td>\n<td>\n[latex]S=2LH+2LW+2WH[\/latex]<br \/>\n[latex]S=2\\left(14\\cdot 17\\right)+2\\left(14\\cdot 9\\right)+2\\left(9\\cdot 17\\right)[\/latex]\n<\/td>\n<\/tr>\n<tr>\n<td>Step 5. <strong>Solve the equation.<\/strong><\/td>\n<td>[latex]S=1,034[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Step 6. <strong>Check:<\/strong> Double-check with a calculator.<\/td>\n<td>&nbsp;<\/td>\n<\/tr>\n<tr>\n<td>Step 7. <strong>Answer<\/strong> the question.<\/td>\n<td>The surface area is [latex]1,034[\/latex] square centimeters.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\">\n<iframe loading=\"lazy\" id=\"ohm146789\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=146789&theme=lumen&iframe_resize_id=ohm146789&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><br \/>\n<\/section>\n<\/section>\n","protected":false},"author":15,"menu_order":4,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":292,"module-header":"background_you_need","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\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/1710"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":15,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/1710\/revisions"}],"predecessor-version":[{"id":4781,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/1710\/revisions\/4781"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/292"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/1710\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=1710"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=1710"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=1710"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=1710"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}