{"id":90241,"date":"2021-08-25T11:29:52","date_gmt":"2021-08-25T16:29:52","guid":{"rendered":"https:\/\/www.mometrix.com\/academy\/?page_id=90241"},"modified":"2026-03-25T10:49:29","modified_gmt":"2026-03-25T15:49:29","slug":"area-of-a-rectangle-with-a-semicircle","status":"publish","type":"page","link":"https:\/\/www.mometrix.com\/academy\/area-of-a-rectangle-with-a-semicircle\/","title":{"rendered":"Area of a Rectangle with a Semicircle"},"content":{"rendered":"\n\t\t\t<div id=\"mmDeferVideoEncompass_87BRAxIcxW0\" style=\"position: relative;\">\n\t\t\t<picture>\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.webp\" type=\"image\/webp\">\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" type=\"image\/jpeg\"> \n\t\t\t\t<img fetchpriority=\"high\" decoding=\"async\" loading=\"eager\" id=\"videoThumbnailImage_87BRAxIcxW0\" data-source-videoID=\"87BRAxIcxW0\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" alt=\"Area of a Rectangle with a Semicircle Video\" height=\"1080\" width=\"1920\" class=\"size-full\" data-matomo-title = \"Area of a Rectangle with a Semicircle\">\n\t\t\t<\/picture>\n\t\t\t<\/div>\n\t\t\t<style>img#videoThumbnailImage_87BRAxIcxW0:hover {cursor:pointer;} img#videoThumbnailImage_87BRAxIcxW0 {background-size:contain;background-image:url(\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2025\/01\/1871-thumb-final.webp\");}<\/style>\n\t\t\t<script defer>\n\t\t\t  jQuery(\"img#videoThumbnailImage_87BRAxIcxW0\").click(function() {\n\t\t\t\tlet videoId = jQuery(this).attr(\"data-source-videoID\");\n\t\t\t\tlet helpTag = '<div id=\"mmDeferVideoYTMessage_87BRAxIcxW0\" style=\"display: none;position: absolute;top: -24px;width: 100%;text-align: center;\"><span style=\"font-style: italic;font-size: small;border-top: 1px solid #fc0;\">Having trouble? <a href=\"https:\/\/www.youtube.com\/watch?v='+videoId+'\" target=\"_blank\">Click here to watch on YouTube.<\/a><\/span><\/div>';\n\t\t\t\tlet tag = document.createElement(\"iframe\");\n\t\t\t\ttag.id = \"yt\" + videoId;\n\t\t\t\ttag.src = \"https:\/\/www.youtube-nocookie.com\/embed\/\" + videoId + \"?autoplay=1&controls=1&wmode=opaque&rel=0&egm=0&iv_load_policy=3&hd=0&enablejsapi=1\";\n\t\t\t\ttag.frameborder = 0;\n\t\t\t\ttag.allow = \"autoplay; fullscreen\";\n\t\t\t\ttag.width = this.width;\n\t\t\t\ttag.height = this.height;\n\t\t\t\ttag.setAttribute(\"data-matomo-title\",\"Area of a Rectangle with a Semicircle\");\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_87BRAxIcxW0\").html(tag);\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_87BRAxIcxW0\").prepend(helpTag);\n\t\t\t\tsetTimeout(function(){jQuery(\"div#mmDeferVideoYTMessage_87BRAxIcxW0\").css(\"display\", \"block\");}, 2000);\n\t\t\t  });\n\t\t\t  \n\t\t\t<\/script>\n\t\t\n<p><script>\nfunction frt_Function() {\n  var x = document.getElementById(\"frt\");\n  if (x.style.display === \"none\") {\n    x.style.display = \"block\";\n  } else {\n    x.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<div class=\"moc-toc hide-on-desktop hide-on-tablet\">\n<div><button onclick=\"frt_Function()\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2024\/12\/toc2.svg\" width=\"16\" height=\"16\" alt=\"show or hide table of contents\"><\/button><\/p>\n<p>On this page<\/p>\n<\/div>\n<nav id=\"frt\" style=\"display:none;\">\n<ul>\n<li class=\"toc-h2\"><a href=\"#Example_1\" class=\"smooth-scroll\">Example #1<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Example_2\" class=\"smooth-scroll\">Example #2<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Example_3\" class=\"smooth-scroll\">Example #3<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Area_of_Composite_Shape_Practice_Questions\" class=\"smooth-scroll\">Area of Composite Shape Practice Questions<\/a><\/li>\n<\/ul>\n<\/nav>\n<\/div>\n<div class=\"accordion\"><input id=\"transcript\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"transcript\">Transcript<\/label><input id=\"PQs\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQs\">Practice<\/label>\n<div class=\"spoiler\" id=\"transcript-spoiler\">\n<p>Hi, and welcome to this video about <strong>finding the area of composite shapes<\/strong> made of rectangles and semicircles. <\/p>\n<p>Many geometric shapes have their very own formulas for finding their areas. But other shapes don\u2019t have common names or formulas. Today we\u2019ll find the area for a couple of those shapes. We\u2019ll even create a formula for one of them just for fun and then not use it. <\/p>\n<h2><span id=\"Example_1\" class=\"m-toc-anchor\"><\/span>Example #1<\/h2>\n<p>\nHere\u2019s our shape: <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-90397\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/08\/Area-of-a-Rectangle-and-Semicircle-1-2.png\" alt=\"Image of a 'tombstone' shape with a length of 20m, and a height of 50m.\" width=\"600\" height=\"600\" \/><\/p>\n<p>It\u2019s a classic \u201ctombstone\u201d shape. It\u2019s a composite shape, which means it\u2019s made up of two or more simpler shapes. This one is made up of a rectangle and a semicircle, which is half a circle, as we can see here: <\/p>\n<p>To find the area of this shape, we need to find the area of the rectangle at the bottom and the area of the semicircle at the top and then simply add the two together to get our total area. But first we need to review the formulas used to find the areas of rectangles and circles. <\/p>\n<div class=\"examplesentence\">\\(\\text{Area of a Circle}=\u03c0r^{2}\\)<\/div>\n<p>\n&nbsp;<\/p>\n<div class=\"examplesentence\">\\(\\text{Area of a Rectangle}=lw\\)<\/div>\n<p>\n&nbsp;<br \/>\nIn this case, we only have half of a circle, so we need to modify our circle formula a bit. Dividing by 2 will make it the area of a semicircle: <\/p>\n<div class=\"examplesentence\">\\(\\text{Area of a Semicircle}=\\frac{\u03c0r^{2}}{2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow we can create the formula for the area of our \u201ctombstone\u201d shape: <\/p>\n<div class=\"examplesentence\">\\(\\text{Area of a &#8220;Tombstone&#8221; Shape}\\)\\(=lw+\\frac{\u03c0r^{2}}{2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nIt\u2019s just the area of the rectangle plus the area of the semicircle. In truth, we don\u2019t really need the formula. Logic dictates that we\u2019re going to add the pieces of this composite shape to get our total area anyway. <\/p>\n<h2><span id=\"Example_2\" class=\"m-toc-anchor\"><\/span>Example #2<\/h2>\n<p>\nLet\u2019s try a problem:<\/p>\n<div class=\"transcriptcallout\">Find the area of this composite shape made of a semicircle and rectangle.<\/div>\n<p>\n&nbsp;<br \/>\nWe can see that our shape is \\(50\\) meters tall and \\(20\\) meters wide. This is the typical information given for this type of problem. What makes this tricky is that we know the width of the shape, which is the width of the rectangle, but we don\u2019t know the radius of the circle or the length of the rectangle. Uh-oh! <\/p>\n<p>Looking at our diagram, though, we see that we could draw another radius that is horizontal. <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-90403\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/08\/Area-of-a-Rectangle-and-Semicircle-2-1.png\" alt=\"Image of a 'tombstone' shape with a length of 20m, and a height of 50m. Two orange lines showing the radius. \" width=\"600\" height=\"600\" \/><\/p>\n<p>Now we can see that this radius is half the width of our shape! And since we know that the width of our shape is \\(20\\) meters, we can find the radius by dividing that by \\(2\\)! So our radius is \\(10\\) meters! <\/p>\n<p>Now we can not only find the area of the semicircle, but we can also find the length of the rectangle! Since the height of our object is \\(50\\) meters and the height of the semicircle part, our radius, is \\(10\\) meters, that means the length of our rectangle is \\(50\\) meters-\\(10\\) <em>meters<\/em>, which is \\(40\\) <em>meters<\/em>! Now we have all the information we need: <\/p>\n<div class=\"examplesentence\">\\(r=10\\text{ m}\\)<br \/>\n\\(l=40\\text{ m}\\)<br \/>\n\\(w=20\\text{ m}\\)<\/div>\n<p>\n&nbsp;<br \/>\nLet\u2019s find the area of our semicircle first: <\/p>\n<div class=\"examplesentence\">\\(\\text{Area of a Semicircle}=\\frac{\u03c0r}{22}\\)<br \/>\n\\(A=\\frac{\u03c0(10 m)^2}{2}\\)<br \/>\n\\(A=\\frac{100\u03c0 m^2}{2}\\)<br \/>\n\\(A=50\u03c0\\text{ m}^2\\)<\/div>\n<p>\n&nbsp;<br \/>\nThe area of our semicircle is \\(50\\) pi meters squared. That\u2019s the exact area. We could use \\(3.14\\) as an approximation for pi and multiply by \\(50\\) to get an approximate area that could be a bit more useful for a real world problem, but first, let\u2019s find the area of our rectangle. <\/p>\n<div class=\"examplesentence\">\\(\\text{Area of a Rectangle}=lw\\)<br \/>\n\\(A=(40\\text{m})(20\\text{m})\\)<br \/>\n\\(A=800\\text{m}^2\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow that we know the area of both objects that make up our composite shape, we can add them together to get our total area. <\/p>\n<div class=\"examplesentence\">\\(\\text{Exact Area of }\\)\\(\\text{&#8220;Tombstone&#8221; Shape}\\)\\(=(800+50\u03c0)m^{2}\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>To get a more useful approximate value, we can multiply \\(50\\) by an approximation of pi using the key on our calculator to get an approximate area of \\(157.08\\) meters squared. <\/p>\n<div class=\"examplesentence\">\\(\\text{Approximate Area of }\\)\\(\\text{&#8220;Tombstone&#8221; Shape}\\approx{800} <br \/>\nm^2\\)<br \/>\n\\(+157.08 m^2\\approx{957.08} m^2\\)<\/div>\n<p>\n&nbsp;<br \/>\nNotice that we used the \u201capproximately equal to\u201d symbol \\(( \\approx)\\) since our semicircle area is now an approximation. <\/p>\n<h2><span id=\"Example_3\" class=\"m-toc-anchor\"><\/span>Example #3<\/h2>\n<p>\nFor our next problem, let\u2019s try to find the area for another related shape, the stadium: <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/08\/Area-of-a-Rectangle-and-Semicircle-3-1.png\" alt=\"\" width=\"600\" height=\"337\" class=\"alignnone size-full wp-image-90382\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/08\/Area-of-a-Rectangle-and-Semicircle-3-1.png 600w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/08\/Area-of-a-Rectangle-and-Semicircle-3-1-300x169.png 300w\" sizes=\"auto, (max-width: 600px) 100vw, 600px\" \/><\/p>\n<p>It\u2019s not hard to see where it gets its name! It\u2019s shaped like a track and field stadium. For a real world problem, let\u2019s calculate the area of an outdoor track and field stadium. <\/p>\n<p>Just like with our \u201ctombstone\u201d-shaped problem, problems of this type only give two dimensions of our shape \u2013 the total width and the total length. <\/p>\n<p>But that\u2019s all the information that we need! As we saw in the last problem, our width is also the diameter of our circle:<\/p>\n<p>That means that we can find the radius by simply dividing our diameter by \\(2\\). Here the width, or the diameter, is \\(73\\) meters, so our radius is \\(36.5\\) meters. That\u2019s one piece of the puzzle completed. Now we just need to find the length of the rectangle. <\/p>\n<p>The total length of our stadium infield is \\(157.4\\) meters, but we need to subtract the length of each semicircle from that. So we subtract the radius of the left semicircle and the right semicircle: <\/p>\n<div class=\"examplesentence\">\\(\\text{Rectangle Length}\\)\\(=157.4\\text{ meters}-36.5\\text{ meters}\\)<br \/>\n\\(-36.5\\text{ meters}=84.4\\text{ meters}\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow we have all of the information we need to find our total area. <\/p>\n<div class=\"examplesentence\">\\(r=36.5\\text{ m}\\)<br \/>\n\\(l=84.4\\text{ m}\\)<br \/>\n\\(w=73\\text{ m}\\)<\/div>\n<p>\n&nbsp;<br \/>\nBefore we begin calculating, let\u2019s take another look at our stadium. Notice that there are two semicircles. If we removed the rectangle between them and pushed them together, we\u2019d have a complete circle! So to find the area of both semicircle ends at once, we can just use the formula for the area of a whole circle: <\/p>\n<div class=\"examplesentence\">\\(\\text{Area of a Circle}=\u03c0r^{2}\\)<br \/>\n\\(A=\u03c0(36.5\\text{ m})^{2}\\)<br \/>\n\\(A=1,332.25\u03c0\\text{ m}^2\\)<\/div>\n<p>\n&nbsp;<br \/>\nThe area of our two semicircles together is \\(1,332.25\u03c0\\text{ m}^2\\). Remember, that\u2019s the exact area. We\u2019ll find the approximate total area after we find our rectangle\u2019s area. <\/p>\n<div class=\"examplesentence\">\\(\\text{Area of a Rectangle}=lw\\)<br \/>\n\\(A=(84.4\\text{ m})(73\\text{ m})^2\\)<br \/>\n\\(A=6,161.2\\text{ m}^2\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow that we know the area of the parts that make up our stadium shape we simply add them together:<\/p>\n<div class=\"examplesentence\">\\(\\text{Exact Area of Stadium Infield}\\)\\(=6,161.2\\text{ m}^2\\)<br \/>\n\\(+1,332.25\u03c0\\text{ m}^2\\)<\/div>\n<p>\n&nbsp;<br \/>\nWhile exact, this answer wouldn\u2019t be decipherable by most people. So it makes sense to write it as an approximation. We find the approximate area of our two semicircles by multiplying \\(1,332.25\\) by \u03c0 on our calculators to find \\(4,185.39\\text{ m}^2\\). <\/p>\n<div class=\"examplesentence\">\\(\\text{Area of Stadium}\\approx 6,161.2\\text{ m}^2\\)<br \/>\n\\(+4,185.39\\text{ m}^2\\approx 10,346.6\\text{ m}^2\\)<\/div>\n<p>\n&nbsp;<br \/>\nOnce again, we use the \u201capproximately equal to\u201d symbol because our semicircle areas are now an approximation. <\/p>\n<p>That\u2019s it for finding the area of composite shapes made of circles and rectangles. Thanks for watching, and happy studying!<\/p>\n<ul class=\"citelist\">\n<li><a href=\"https:\/\/www.mathsisfun.com\/geometry\/rectangle.html\"target=\"_blank\">\u201cRectangle.\u201d n.d.<\/a><\/li>\n<li><a href=\"https:\/\/www.mathsisfun.com\/geometry\/circle-area.html\"target=\"_blank\">https:\/\/www.mathsisfun.com\/geometry\/circle-area.html<\/a><\/li>\n<li><a href=\"https:\/\/www.dimensions.com\/element\/track-and-field-400m-running-track\"target=\"_blank\">\u201c400m Running Track Dimensions &#038; Drawings | Dimensions.com.\u201d n.d.<\/a><\/li>\n<\/ul>\n<\/div>\n<div class=\"spoiler\" id=\"PQs-spoiler\">\n<h2 style=\"text-align:center\"><span id=\"Area_of_Composite_Shape_Practice_Questions\" class=\"m-toc-anchor\"><\/span>Area of Composite Shape Practice Questions<\/h2>\n\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #1:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nCalculate the area of the compound figure. Use 3.14 for \\(\\pi\\).<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2026\/02\/Area-of-Rectangle-with-Semicircle-Example-1.svg\" alt=\"Diagram of an oval-shaped area 55 feet wide, with a marked point 30 feet from one side along a dashed line.\" width=\"404.745\" height=\"481.8\" class=\"aligncenter size-full wp-image-287092\"  role=\"img\" \/><\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ correct_answer\"  id=\"PQ-1-1\">6,126 ft<sup>2<\/sup><\/div><div class=\"PQ\"  id=\"PQ-1-2\">6,122 ft<sup>2<\/sup><\/div><div class=\"PQ\"  id=\"PQ-1-3\">6,926 ft<sup>2<\/sup><\/div><div class=\"PQ\"  id=\"PQ-1-4\">6,387 ft<sup>2<\/sup><\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-1\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-1\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-1-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>Calculate the area of the rectangle:<\/p>\n<p style=\"text-align: center; line-height: 35px\">\n\\(A=l\\times w\\)<br \/>\n\\(A=55\\text{ ft}\\times60\\text{ ft}\\)<br \/>\n\\(A=3{,}300\\text{ ft}^2\\)<\/p>\n<p>Calculate the area of the two semicircles (one circle):<\/p>\n<p style=\"text-align: center; line-height: 35px\">\n\\(A=\\pi r^2\\)<br \/>\n\\(A=(3.14)(30)^2\\)<br \/>\n\\(A=2{,}826\\text{ ft}^2\\)<\/p>\n<p>Add the two areas together to get 6,126 ft<sup>2<\/sup>.<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-1-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-1-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #2:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nThe shape below is made up of a semicircle and a square. The radius of the semicircle is 5 centimeters. What is the area of the entire figure? Use 3.14 for \\(\\pi\\).<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2026\/02\/Area-of-Rectangle-with-Semicircle-Example-2.svg\" alt=\"A semicircle sits atop a rectangle; a horizontal line across the semicircle&#039;s diameter is marked with a blue dot, labeled as 5 cm from the left end.\" width=\"244.8\" height=\"328.1\" class=\"aligncenter size-full wp-image-287095\"  role=\"img\" \/><\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-2-1\">134 cm<sup>2<\/sup><\/div><div class=\"PQ correct_answer\"  id=\"PQ-2-2\">139.25 cm<sup>2<\/sup><\/div><div class=\"PQ\"  id=\"PQ-2-3\">219.25 cm<sup>2<\/sup><\/div><div class=\"PQ\"  id=\"PQ-2-4\">130 cm<sup>2<\/sup><\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-2\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-2\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-2-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>Add the area of the square to the area of the semicircle. <\/p>\n<p>Each side of the square has length \\(2\\times5\\text{ cm}=10\\text{ cm}\\). The area of the square is \\(10\\text{ cm} \\times\\ 10\\text{ cm}\\), which equals \\(100\\text{ cm}^2\\).<\/p>\n<p>The area of the semicircle is \\(\\frac{\\pi r^2}{2}\\), which becomes \\(\\frac{(3.14){(5)}^2}{2}\\), which simplifies to \\(39.25\\text{ cm}^2\\). <\/p>\n<p>Add the two areas:<\/p>\n<p style=\"text-align: center\">\\(100\\text{ cm}^2+39.25\\text{ cm}^2=139.25\\text{ cm}^2\\)<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-2-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-2-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #3:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nCalculate the surface area of the image below. Round your answer to the nearest whole number. Use 3.14 for \\(\\pi\\).<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2026\/02\/Area-of-Rectangle-with-Semicircle-Example-3.svg\" alt=\"A rectangle measuring 9 cm by 4.5 cm with a semicircular cutout 5 cm wide along the bottom edge.\" width=\"370.7\" height=\"199.1\" class=\"aligncenter size-full wp-image-287098\"  role=\"img\" \/><\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-3-1\">32 cm<sup>2<\/sup><\/div><div class=\"PQ\"  id=\"PQ-3-2\">31.5 cm<sup>2<\/sup><\/div><div class=\"PQ correct_answer\"  id=\"PQ-3-3\">31 cm<sup>2<\/sup><\/div><div class=\"PQ\"  id=\"PQ-3-4\">33 cm<sup>2<\/sup><\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-3\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-3\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-3-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>Subtract the area of the semicircle from the area of the rectangle. The area of the rectangle is found using the formula \\(A=l\\times w\\).<\/p>\n<p>In this case, \\(A=9\\times\\ 4.5\\), which simplifies to \\(A=40.5\\). The area of the rectangle is \\(40.5\\text{ cm}^2\\).<\/p>\n<p>Now find the area of the semicircle using the formula \\(A=\\frac{\\pi r^2}{2}\\). In this case, \\(A=\\frac{(3.14)({2.5)}^2}{2}\\), which simplifies to \\(A=9.8125\\). <\/p>\n<p>Now subtract this from the rectangle\u2019s area. <\/p>\n<p style=\"text-align: center;\">\\(40.5-9.8125=30.6875\\)<\/p>\n<p>Rounded to the nearest whole number, the shape\u2019s area is 31 cm<sup>2<\/sup>. <\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-3-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-3-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #4:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nA landscaper is designing a backyard with a pool for a client. The pool will be surrounded by grass. How much grass will be planted? Use 3.14 for \\(\\pi\\).<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2026\/02\/Area-of-Rectangle-with-Semicircle-Example-4.svg\" alt=\"A rectangular white rectangular object with black text.\" width=\"398.695\" height=\"166.375\" class=\"aligncenter size-full wp-image-287089\"  role=\"img\" \/><\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-4-1\">179.86 ft<sup>2<\/sup><\/div><div class=\"PQ correct_answer\"  id=\"PQ-4-2\">289.86 ft<sup>2<\/sup><\/div><div class=\"PQ\"  id=\"PQ-4-3\">209.86 ft<sup>2<\/sup><\/div><div class=\"PQ\"  id=\"PQ-4-4\">299.86 ft<sup>2<\/sup><\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-4\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-4\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-4-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>The two semicircles on either end make a circle. Find the area of the circle. Add this to the large green rectangle, then subtract the blue rectangle. <\/p>\n<p>The area of the circle (two semicircles) is calculated as \\(\\pi r^2\\). This becomes \\((3.14)(7)^2\\), which simplifies to 153.86 ft<sup>2<\/sup>.<\/p>\n<p>Now calculate the area of the large green rectangle.<\/p>\n<p style=\"text-align: center\">\\(24\\text{ ft} \\times 14\\text{ ft}=336\\text{ ft}^2\\)<\/p>\n<p>Add this to the green circle.  <\/p>\n<p style=\"text-align: center;\">\\(153.86\\text{ ft}^2+336\\text{ ft}^2=489.86\\text{ ft}^2\\)<\/p>\n<p>Now subtract the blue rectangle from this. The blue rectangle (swimming pool) measures \\(20\\text{ ft} \\times\\ 10\\text{ ft}=200\\text{ ft}^2\\).<\/p>\n<p style=\"text-align: center;\">\\(489.86\\text{ ft}^2-200\\text{ ft}^2=289.86\\text{ ft}^2\\)<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-4-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-4-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #5:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nAntonio wants to use chalk to draw a basketball court on his driveway similar to the image. He wants the rectangle to be 18 feet by 6 feet. The semicircle at the top of the rectangle will have a radius of 3 feet. What is the surface area of the yellow region of the court? Use 3.14 for \\(\\pi\\), and round your answer to the nearest whole number.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2025\/10\/Basketball-court.svg\" alt=\"A basketball hoop and backboard on a court, with the key area and the free throw lane highlighted in yellow.\" width=\"577\" height=\"407\" class=\"aligncenter size-full wp-image-274306\"  role=\"img\" \/><\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-5-1\">118 ft<sup>2<\/sup><\/div><div class=\"PQ\"  id=\"PQ-5-2\">130 ft<sup>2<\/sup><\/div><div class=\"PQ\"  id=\"PQ-5-3\">126 ft<sup>2<\/sup><\/div><div class=\"PQ correct_answer\"  id=\"PQ-5-4\">122 ft<sup>2<\/sup><\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-5\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-5\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-5-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>Add the area of the rectangle to the area of the semicircle. The area of the rectangle is found using the formula \\(A=l\\times w\\). The area of the semicircle is found using the formula \\(A=\\frac{\\pi r^2}{2}\\).<\/p>\n<p>Rectangle: \\(A=l\\times w\\) becomes \\(A=18\\times6\\), which simplifies to \\(A=108\\text{ ft}^2\\).<br \/>\nSemicircle: \\(A=\\frac{\\pi r^2}{2}\\) becomes \\(A=\\frac{(3.14)(3)^2}{2}\\), which simplifies to \\(A=14.13\\text{ ft}^2\\).<\/p>\n<p>Now add the two areas. <\/p>\n<p><\/p>\n<p style=\"text-align: center;\">\\(108\\text{ ft}^2+14.13\\text{ ft}^2=122.13\\text{ ft}^2\\)<\/p>\n<p>When 122.13 ft<sup>2<\/sup> is rounded to the nearest whole number, we get 122 ft<sup>2<\/sup>.<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-5-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-5-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div><\/div>\n<\/div>\n\n<div class=\"home-buttons\">\n<p><a href=\"https:\/\/www.mometrix.com\/academy\/geometry\/\">Return to Geometry Videos<\/a><\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Return to Geometry Videos<\/p>\n","protected":false},"author":1,"featured_media":240589,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":{"0":"post-90241","1":"page","2":"type-page","3":"status-publish","4":"has-post-thumbnail","6":"page_category-area-and-perimeter-videos","7":"page_type-video","8":"content_type-practice-questions","9":"subject_matter-math"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/90241","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/comments?post=90241"}],"version-history":[{"count":9,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/90241\/revisions"}],"predecessor-version":[{"id":240583,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/90241\/revisions\/240583"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media\/240589"}],"wp:attachment":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media?parent=90241"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}