{"id":1795,"date":"2025-07-28T19:39:51","date_gmt":"2025-07-28T19:39:51","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&#038;p=1795"},"modified":"2025-08-13T03:03:15","modified_gmt":"2025-08-13T03:03:15","slug":"arcs-and-sectors-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/arcs-and-sectors-learn-it-2\/","title":{"raw":"Arcs and Sectors: Learn It 2","rendered":"Arcs and Sectors: Learn It 2"},"content":{"raw":"<h2>Finding the Area of a Sector of a Circle<\/h2>\r\n<section><section>In addition to arc length, we can also use angles to find the area of a <strong>sector of a circle<\/strong>. A sector is a region of a circle bounded by two radii and the intercepted arc, like a slice of pizza or pie.<\/section><section><section class=\"textbox recall\" aria-label=\"Recall\">The area of a circle with radius [latex]r[\/latex] can be found using the formula [latex]A=\\pi {r}^{2}[\/latex].<\/section><span style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">If the two radii form an angle of [latex]\\theta [\/latex], measured in radians, then [latex]\\frac{\\theta }{2\\pi }[\/latex] is the ratio of the angle measure to the measure of a full rotation and is also, therefore, the ratio of the area of the sector to the area of the circle. Thus, the <\/span><strong style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">area of a sector<\/strong><span style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\"> is the fraction [latex]\\frac{\\theta }{2\\pi }[\/latex]\u00a0multiplied by the entire area. (Always remember that this formula only applies if [latex]\\theta [\/latex] is in radians.)<\/span>\r\n\r\n<\/section><section>\r\n<div style=\"text-align: center;\">[latex]\\begin{align}\\text{Area of sector}&amp;=\\left(\\frac{\\theta }{2\\pi }\\right)\\pi {r}^{2} \\\\ &amp;=\\frac{\\theta \\pi {r}^{2}}{2\\pi } \\\\ &amp;=\\frac{1}{2}\\theta {r}^{2} \\end{align}[\/latex]<\/div>\r\n<\/section>\r\n<div><section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>area of a sector<\/h3>\r\nThe <strong>area of a sector<\/strong> of a circle with radius [latex]r[\/latex] subtended by an angle [latex]\\theta [\/latex], measured in radians, is\r\n<div>\r\n<p style=\"text-align: center;\">[latex]A=\\frac{1}{2}\\theta {r}^{2}[\/latex]<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"442\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003452\/CNX_Precalc_Figure_05_01_026F2.jpg\" alt=\"Graph showing a circle with angle theta and radius r, and the area of the slice of circle created by the initial side and terminal side of the angle.\" width=\"442\" height=\"394\" \/> The area of the sector equals half the square of the radius times the central angle measured in radians.[\/caption]\r\n\r\n<\/div>\r\n<\/section><section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given a circle of radius [latex]r[\/latex], find the area of a sector defined by a given angle [latex]\\theta [\/latex].<\/strong>\r\n<ol>\r\n \t<li>If necessary, convert [latex]\\theta [\/latex] to radians.<\/li>\r\n \t<li>Multiply half the radian measure of [latex]\\theta [\/latex] by the square of the radius [latex]r:\\text{ } A=\\frac{1}{2}\\theta {r}^{2}[\/latex].<\/li>\r\n<\/ol>\r\n<\/section><\/div>\r\n<section><section class=\"textbox example\" aria-label=\"Example\">An automatic lawn sprinkler sprays a distance of 20 feet while rotating 30 degrees. What is the area of the sector of grass the sprinkler waters?\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003454\/CNX_Precalc_Figure_05_01_0272.jpg\" alt=\"Illustration of a 30 degree ange with a terminal and initial side with length of 20 feet.\" width=\"487\" height=\"169\" \/> The sprinkler sprays 20 ft within an arc of 30\u00b0.[\/caption]\r\n\r\n[reveal-answer q=\"785904\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"785904\"]\r\n\r\nFirst, we need to convert the angle measure into radians. Because 30 degrees is one of our special angles, we already know the equivalent radian measure, but we can also convert:\r\n<p style=\"text-align: center;\">[latex]\\begin{align}30\\text{ degrees}&amp;=30\\cdot \\frac{\\pi }{180} \\\\ &amp;=\\frac{\\pi }{6}\\text{ radians} \\end{align}[\/latex]<\/p>\r\nThe area of the sector is then\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\text{Area} &amp;= \\frac{1}{2}\\left(\\frac{\\pi }{6}\\right){\\left(20\\right)}^{2} \\\\ &amp;\\approx 104.72 \\end{align}[\/latex]<\/p>\r\nSo the area is about [latex]104.72{\\text{ ft}}^{2}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">In central pivot irrigation, a large irrigation pipe on wheels rotates around a center point. A farmer has a central pivot system with a radius of 400 meters. If water restrictions only allow her to water 150 thousand square meters a day, what angle should she set the system to cover? Write the answer in radian measure to two decimal places.[reveal-answer q=\"495928\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"495928\"]1.88[\/hidden-answer]<\/section><\/section><\/section>","rendered":"<h2>Finding the Area of a Sector of a Circle<\/h2>\n<section>\n<section>In addition to arc length, we can also use angles to find the area of a <strong>sector of a circle<\/strong>. A sector is a region of a circle bounded by two radii and the intercepted arc, like a slice of pizza or pie.<\/section>\n<section>\n<section class=\"textbox recall\" aria-label=\"Recall\">The area of a circle with radius [latex]r[\/latex] can be found using the formula [latex]A=\\pi {r}^{2}[\/latex].<\/section>\n<p><span style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">If the two radii form an angle of [latex]\\theta[\/latex], measured in radians, then [latex]\\frac{\\theta }{2\\pi }[\/latex] is the ratio of the angle measure to the measure of a full rotation and is also, therefore, the ratio of the area of the sector to the area of the circle. Thus, the <\/span><strong style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">area of a sector<\/strong><span style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\"> is the fraction [latex]\\frac{\\theta }{2\\pi }[\/latex]\u00a0multiplied by the entire area. (Always remember that this formula only applies if [latex]\\theta[\/latex] is in radians.)<\/span><\/p>\n<\/section>\n<section>\n<div style=\"text-align: center;\">[latex]\\begin{align}\\text{Area of sector}&=\\left(\\frac{\\theta }{2\\pi }\\right)\\pi {r}^{2} \\\\ &=\\frac{\\theta \\pi {r}^{2}}{2\\pi } \\\\ &=\\frac{1}{2}\\theta {r}^{2} \\end{align}[\/latex]<\/div>\n<\/section>\n<div>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>area of a sector<\/h3>\n<p>The <strong>area of a sector<\/strong> of a circle with radius [latex]r[\/latex] subtended by an angle [latex]\\theta[\/latex], measured in radians, is<\/p>\n<div>\n<p style=\"text-align: center;\">[latex]A=\\frac{1}{2}\\theta {r}^{2}[\/latex]<\/p>\n<figure style=\"width: 442px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003452\/CNX_Precalc_Figure_05_01_026F2.jpg\" alt=\"Graph showing a circle with angle theta and radius r, and the area of the slice of circle created by the initial side and terminal side of the angle.\" width=\"442\" height=\"394\" \/><figcaption class=\"wp-caption-text\">The area of the sector equals half the square of the radius times the central angle measured in radians.<\/figcaption><\/figure>\n<\/div>\n<\/section>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given a circle of radius [latex]r[\/latex], find the area of a sector defined by a given angle [latex]\\theta[\/latex].<\/strong><\/p>\n<ol>\n<li>If necessary, convert [latex]\\theta[\/latex] to radians.<\/li>\n<li>Multiply half the radian measure of [latex]\\theta[\/latex] by the square of the radius [latex]r:\\text{ } A=\\frac{1}{2}\\theta {r}^{2}[\/latex].<\/li>\n<\/ol>\n<\/section>\n<\/div>\n<section>\n<section class=\"textbox example\" aria-label=\"Example\">An automatic lawn sprinkler sprays a distance of 20 feet while rotating 30 degrees. What is the area of the sector of grass the sprinkler waters?<\/p>\n<figure style=\"width: 487px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003454\/CNX_Precalc_Figure_05_01_0272.jpg\" alt=\"Illustration of a 30 degree ange with a terminal and initial side with length of 20 feet.\" width=\"487\" height=\"169\" \/><figcaption class=\"wp-caption-text\">The sprinkler sprays 20 ft within an arc of 30\u00b0.<\/figcaption><\/figure>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q785904\">Show Solution<\/button><\/p>\n<div id=\"q785904\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, we need to convert the angle measure into radians. Because 30 degrees is one of our special angles, we already know the equivalent radian measure, but we can also convert:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}30\\text{ degrees}&=30\\cdot \\frac{\\pi }{180} \\\\ &=\\frac{\\pi }{6}\\text{ radians} \\end{align}[\/latex]<\/p>\n<p>The area of the sector is then<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\text{Area} &= \\frac{1}{2}\\left(\\frac{\\pi }{6}\\right){\\left(20\\right)}^{2} \\\\ &\\approx 104.72 \\end{align}[\/latex]<\/p>\n<p>So the area is about [latex]104.72{\\text{ ft}}^{2}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\">In central pivot irrigation, a large irrigation pipe on wheels rotates around a center point. A farmer has a central pivot system with a radius of 400 meters. If water restrictions only allow her to water 150 thousand square meters a day, what angle should she set the system to cover? Write the answer in radian measure to two decimal places.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q495928\">Show Solution<\/button><\/p>\n<div id=\"q495928\" class=\"hidden-answer\" style=\"display: none\">1.88<\/div>\n<\/div>\n<\/section>\n<\/section>\n<\/section>\n","protected":false},"author":13,"menu_order":15,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":178,"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\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/1795"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/users\/13"}],"version-history":[{"count":4,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/1795\/revisions"}],"predecessor-version":[{"id":2418,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/1795\/revisions\/2418"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/parts\/178"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/1795\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=1795"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=1795"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=1795"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=1795"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}