{"id":4565,"date":"2013-06-29T06:46:55","date_gmt":"2013-06-29T06:46:55","guid":{"rendered":"http:\/\/www.mometrix.com\/academy\/?page_id=4565"},"modified":"2026-03-26T11:57:05","modified_gmt":"2026-03-26T16:57:05","slug":"volume-and-surface-area-of-a-right-circular-cone","status":"publish","type":"page","link":"https:\/\/www.mometrix.com\/academy\/volume-and-surface-area-of-a-right-circular-cone\/","title":{"rendered":"Volume and Surface Area of a Right&nbsp;Circular&nbsp;Cone"},"content":{"rendered":"\n\t\t\t<div id=\"mmDeferVideoEncompass_PfXaB4Xwqek\" 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_PfXaB4Xwqek\" data-source-videoID=\"PfXaB4Xwqek\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" alt=\"Volume and Surface Area of a Right&nbsp;Circular&nbsp;Cone Video\" height=\"464\" width=\"825\" class=\"size-full\" data-matomo-title = \"Volume and Surface Area of a Right&nbsp;Circular&nbsp;Cone\">\n\t\t\t<\/picture>\n\t\t\t<\/div>\n\t\t\t<style>img#videoThumbnailImage_PfXaB4Xwqek:hover {cursor:pointer;} img#videoThumbnailImage_PfXaB4Xwqek {background-size:contain;background-image:url(\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/02\/701-thumb-final-1.webp\");}<\/style>\n\t\t\t<script defer>\n\t\t\t  jQuery(\"img#videoThumbnailImage_PfXaB4Xwqek\").click(function() {\n\t\t\t\tlet videoId = jQuery(this).attr(\"data-source-videoID\");\n\t\t\t\tlet helpTag = '<div id=\"mmDeferVideoYTMessage_PfXaB4Xwqek\" 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\",\"Volume and Surface Area of a Right&nbsp;Circular&nbsp;Cone\");\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_PfXaB4Xwqek\").html(tag);\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_PfXaB4Xwqek\").prepend(helpTag);\n\t\t\t\tsetTimeout(function(){jQuery(\"div#mmDeferVideoYTMessage_PfXaB4Xwqek\").css(\"display\", \"block\");}, 2000);\n\t\t\t  });\n\t\t\t  \n\t\t\t<\/script>\n\t\t\n<p><script>\nfunction z23_Function() {\n  var x = document.getElementById(\"z23\");\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=\"z23_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=\"z23\" style=\"display:none;\">\n<ul>\n<li class=\"toc-h2\"><a href=\"#Volume_and_Surface_Area_%E2%80%93_Review\" class=\"smooth-scroll\">Volume and Surface Area \u2013 Review<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Volume_and_Surface_Area_Formulas\" class=\"smooth-scroll\">Volume and Surface Area Formulas<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Finding_Volume_and_Surface_Area\" class=\"smooth-scroll\">Finding Volume and Surface Area<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Frequently_Asked_Questions\" class=\"smooth-scroll\">Frequently Asked Questions<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Volume_and_Surface_Area_of_a_Right_Circular_Cone_Practice_Questions\" class=\"smooth-scroll\">Volume and Surface Area of a Right Circular Cone 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=\"FAQs\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"FAQs\">FAQs<\/label><input id=\"PQs\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQs\">Practice<\/label><a href=\"https:\/\/www.mometrix.com\/academy\/cone-volume-calculator\/\" target=\"none\" style=\"margin: 0 auto;\"><span class=\"accordion_calculator_button\">Calculator<\/span><\/a><\/p>\n<div class=\"spoiler\" id=\"transcript-spoiler\">\n<p>Hey guys! Welcome to today\u2019s video where we\u2019re going to talk about the volume and surface area of a cone. <\/p>\n<p>We know that a cone is actually a lot like a pyramid. While a pyramid has a square base that connects to a pointy tip at the opposite end, a cone\u2019s base is, instead, a circle.<\/p>\n<h2>Volume and Surface Area &#8211; Review<\/h2>\n<p>\nBefore diving into the details, let\u2019s make sure you\u2019re comfortable with the concepts of volume and surface area. These are two key features that all 3-dimensional shapes have. <\/p>\n<p><strong>Volume<\/strong> is the space inside of a 3D object, and <strong>surface area<\/strong> is, well, just that! It\u2019s the total area of the surface of a shape. <\/p>\n<p>Think of volume as the amount of liquid that you could fill an object with, and think of surface area as how much paper you could wrap over that object. Every cube, sphere, cylinder, cone (of course), and so on has a volume and a surface area; and the formulas used for finding these measurements is different for each shape.<\/p>\n<h2><span id=\"Volume_and_Surface_Area_Formulas\" class=\"m-toc-anchor\"><\/span>Volume and Surface Area Formulas<\/h2>\n<p>\n<img decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2025\/02\/Surface-Area-Cone@72.webp\" alt=\"\" width=\"\" height=\"\" class=\"aligncenter size-full wp-image-215971\"  role=\"img\" style=\"box-shadow: 1.5px 1.5px 3px gray;\"  \/><br \/>\nIn the case of a cone, our volume formula looks like this:<\/p>\n<div class=\"examplesentence\">\\( V=\\frac{1}{3}\\pi r^{2}h\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd our surface area formula looks like this:<\/p>\n<div class=\"examplesentence\">\\(SA=\\pi r^{2}+\\pi rl\\)<\/div>\n<p>\n&nbsp;<\/p>\n<h3><span id=\"Variables\" class=\"m-toc-anchor\"><\/span>Variables<\/h3>\n<p>\nThe letters <span style=\"font-style:normal; font-size:90%\">\\(r\\)<\/span>, <span style=\"font-style:normal; font-size:90%\">\\(h\\)<\/span>, and <span style=\"font-style:normal; font-size:90%\">\\(l\\)<\/span> represent different measurements on the cone. But which measurements do these letters (or, as we call them in the &#8220;math world,&#8221; <strong>variables<\/strong>) represent?<\/p>\n<p>The <span style=\"font-style:normal; font-size:90%\">\\(r\\)<\/span> represents the radius of the circular base of the cone.<\/p>\n<p>The <span style=\"font-style:normal; font-size:90%\">\\(h\\)<\/span> represents the height of the cone. More specifically, it\u2019s the length of an imaginary line that runs from the center of the circular base to the very tip of the cone.<\/p>\n<p>Finally, the <span style=\"font-style:normal; font-size:90%\">\\(l\\)<\/span> represents the slant height. Think of this as a straight line that runs from the tip of the cone to the edge of its base.<\/p>\n<p>So, to solve the volume and surface area equations, we\u2019d simply plug a cone\u2019s measurements into the respective variables. <\/p>\n<h2><span id=\"Finding_Volume_and_Surface_Area\" class=\"m-toc-anchor\"><\/span>Finding Volume and Surface Area<\/h2>\n<h3><span id=\"Example_1\" class=\"m-toc-anchor\"><\/span>Example #1<\/h3>\n<p>\nLet\u2019s try a quick example.<\/p>\n<p>Say we have a cone whose base radius measures <span style=\"font-style:normal; font-size:90%\">\\(3\\)<\/span> units, height measures <span style=\"font-style:normal; font-size:90%\">\\(4\\)<\/span> units, and slant height measures <span style=\"font-style:normal; font-size:90%\">\\(5\\)<\/span> units. First, let\u2019s find its volume.<\/p>\n<p>So, <span style=\"font-style:normal; font-size:90%\">\\(V=\\frac{1}{3}\\pi r^{2}h\\)<\/span>. So we\u2019re gonna rewrite this formula. We know our radius is equal to <span style=\"font-style:normal; font-size:90%\">\\(3\\)<\/span>, so we&#8217;re gonna substitute <span style=\"font-style:normal; font-size:90%\">\\(3\\)<\/span> for our <span style=\"font-style:normal; font-size:90%\">\\(r\\)<\/span>. And our height is equal to <span style=\"font-style:normal; font-size:90%\">\\(4\\)<\/span>. So when we multiply this out, we get:<\/p>\n<div class=\"examplesentence\">\\(V=\\frac{1}{3}\\pi (9)(4)\\)<\/div>\n<p>\n&nbsp;<br \/>\nWhich is:<\/p>\n<div class=\"examplesentence\">\\(V=\\frac{1}{3}\\pi (36)\\)<\/div>\n<p>\n&nbsp;<br \/>\nWhich is equal to <span style=\"font-style:normal; font-size:90%\">\\(12\\pi\\)<\/span>. So the volume of this cone is <span style=\"font-style:normal; font-size:90%\">\\(12\\pi\\)<\/span> (or approximately <span style=\"font-style:normal; font-size:90%\">\\(37.7\\)<\/span>) cubic units.<\/p>\n<p>Notice that volume is measured in units cubed. Think centimeters cubed, inches cubed, feet cubed (or \u201ccubic feet\u201d), etc.<\/p>\n<p>Okay, now let\u2019s find the surface area. So:<\/p>\n<div class=\"examplesentence\">\\(SA=\\pi r^{2}+\\pi rl\\)<\/div>\n<p>\n&nbsp;<br \/>\nThen we\u2019ll substitute in the numbers for our variables. So:<\/p>\n<div class=\"examplesentence\">\\(SA=\\pi (3)^{2}+\\pi (3)(5)\\)<\/div>\n<\/blockquote>\n<p>\n&nbsp;<br \/>\nSo:<\/p>\n<div class=\"examplesentence\">\\(SA=\\pi(9)+\\pi (15)\\)<\/div>\n<\/blockquote>\n<p>\n&nbsp;<br \/>\nWhich is equal to <span style=\"font-style:normal; font-size:90%\">\\(24\\pi\\)<\/span>. In the end, we\u2019d say that the surface area is <span style=\"font-style:normal; font-size:90%\">\\(24\\pi\\)<\/span> (or approximately <span style=\"font-style:normal; font-size:90%\">\\(75.4\\)<\/span>) square units. Notice the way that\u2019s measured! Since we are talking about area, we use square units.<\/p>\n<h3><span id=\"Example_2\" class=\"m-toc-anchor\"><\/span>Example #2<\/h3>\n<p>\nIt\u2019s pretty straightforward to plug in values for <span style=\"font-style:normal; font-size:90%\">\\(r\\)<\/span>, <span style=\"font-style:normal; font-size:90%\">\\(h\\)<\/span>, and <span style=\"font-style:normal; font-size:90%\">\\(l\\)<\/span>, right? Well, what if we wanted to figure out what size ice cream cone we would need in order to fit <span style=\"font-style:normal; font-size:90%\">\\(30\\)<\/span> cubic inches of soft serve into it? <\/p>\n<p>Because we can only have one unknown variable in an equation right now, and because we already know the volume of this cone (<span style=\"font-style:normal; font-size:90%\">\\(30\\text{ in}^{3}\\)<\/span>), we\u2019ll need to decide on a fixed value of either the cone\u2019s height or the base\u2019s radius. <\/p>\n<p>Let\u2019s say that we want the cone to be exactly <span style=\"font-style:normal; font-size:90%\">\\(10\\)<\/span> inches in height (so <span style=\"font-style:normal; font-size:90%\">\\(h=10\\text{ in}\\)<\/span>). Now all that\u2019s left is to figure out what <span style=\"font-style:normal; font-size:90%\">\\(r\\)<\/span> equals! In other words, we need to find out how wide the opening of this cone gets.<\/p>\n<p>First, we set up the volume equation using all the information we already have. So:<\/p>\n<div class=\"examplesentence\">\\(V=\\frac{1}{3}\\pi r^{2}h\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd our volume is equal to <span style=\"font-style:normal; font-size:90%\">\\(30\\)<\/span> cubic inches. <span style=\"font-style:normal; font-size:90%\">\\(\\frac{1}{3}\\pi \\)<\/span> stays the same. We\u2019re looking for <span style=\"font-style:normal; font-size:90%\">\\(r\\)<\/span>, so we\u2019re gonna leave that as <span style=\"font-style:normal; font-size:90%\">\\(r^{2}\\)<\/span>, and <span style=\"font-style:normal; font-size:90%\">\\(h=10\\text{ in}\\)<\/span>.<\/p>\n<div class=\"examplesentence\">\\( 30=\\frac{1}{3}\\pi r^{2}(10)\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo, if we rearrange a little, we\u2019ll get this:<\/p>\n<div class=\"examplesentence\">\\(30=\\frac{10}{3}\\pi r^{2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow, we\u2019re trying to get <span style=\"font-style:normal; font-size:90%\">\\(r\\)<\/span> all by itself on one side of the equals sign, so to do this, we\u2019re gonna multiply both sides by <span style=\"font-style:normal; font-size:90%\">\\(\\frac{3}{10\\pi }\\)<\/span>. When we do this, multiply by <span style=\"font-style:normal; font-size:90%\">\\(\\frac{3}{10\\pi }\\)<\/span> on the left, <span style=\"font-style:normal; font-size:90%\">\\(\\frac{3}{10\\pi }\\)<\/span> on the right:<\/p>\n<div class=\"examplesentence\">\\(\\frac{3}{10\\pi }\\cdot 30=\\frac{10}{3}\\pi r^{2}\\cdot \\frac{3}{10\\pi }\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd that gives us:<\/p>\n<div class=\"examplesentence\">\\(\\frac{90}{10\\pi }=r^{2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nThen we can simplify this further to get:<\/p>\n<div class=\"examplesentence\">\\(\\frac{9}{\\pi }=r^{2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nThen, all we have to do is take the square root of both sides. So when we do this, we get:<\/p>\n<div class=\"examplesentence\">\\(r=\\frac{3}{\\sqrt{\\pi }}\\approx 1.7\\)<\/div>\n<p>\n&nbsp;<br \/>\nIn fact, since the linear dimensions of the cone are measured in inches, we know that <span style=\"font-style:normal; font-size:90%\">\\(r\\)<\/span> is approximately <span style=\"font-style:normal; font-size:90%\">\\(1.7\\text{ inches}\\)<\/span>.<\/p>\n<p>Okay, cool! So what if we want to figure out the surface area of that cone we were just talking about? We now know the three important measurements of this cone: its volume, its height, and its base\u2019s radius. Remember that in our surface area equation, we need both the radius and the slant height of the cone in order to find the surface area. Believe it or not, we can calculate the slant height by using the <span style=\"font-style:normal; font-size:90%\">\\(h\\)<\/span>-value that we were given and using the <span style=\"font-style:normal; font-size:90%\">\\(r\\)<\/span>-value that we just found. We just need to use the Pythagorean Theorem.<\/p>\n<p>Let\u2019s look at this particular cross-section of our cone:<\/p>\n<p>Since this is a right triangle, we can use <span style=\"font-style:normal; font-size:90%\">\\(r^{2}+h^{2}=l^{2}\\)<\/span> to find the hypotenuse, which is the slant height. Let\u2019s work that out. <\/p>\n<p>Remember, we just saw that  <span style=\"font-style:normal; font-size:90%\">\\(r^{2}=\\frac{9}{\\pi }\\)<\/span>. So we can substitute <span style=\"font-style:normal; font-size:90%\">\\(\\frac{9}{\\pi }\\)<\/span> in for <span style=\"font-style:normal; font-size:90%\">\\(r^{2}\\)<\/span>. Plus <span style=\"font-style:normal; font-size:90%\">\\(h^{2}\\)<\/span>, <span style=\"font-style:normal; font-size:90%\">\\(h\\)<\/span> is <span style=\"font-style:normal; font-size:90%\">\\(10\\)<\/span>, <span style=\"font-style:normal; font-size:90%\">\\(10^{2}=100\\)<\/span>, so plus <span style=\"font-style:normal; font-size:90%\">\\(100\\)<\/span>. Equals <span style=\"font-style:normal; font-size:90%\">\\(l^{2}\\)<\/span>, which we\u2019re looking for <span style=\"font-style:normal; font-size:90%\">\\(l\\)<\/span>.<\/p>\n<div class=\"examplesentence\">\\(\\frac{9}{\\pi }+100=l^{2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow we\u2019re gonna want to simplify our addition on the left side. So we\u2019re gonna keep <span style=\"font-style:normal; font-size:90%\">\\(\\frac{9}{\\pi }\\)<\/span> the same. And then we&#8217;re gonna need a common denominator for both of these parts, so we&#8217;re gonna do that by multiplying <span style=\"font-style:normal; font-size:90%\">\\(100\\times \\frac{\\pi }{\\pi }\\)<\/span>. When we do that, we get <span style=\"font-style:normal; font-size:90%\">\\( \\frac{100\\pi }{\\pi }\\)<\/span>. Equals <span style=\"font-style:normal; font-size:90%\">\\(l^{2}\\)<\/span>.<\/p>\n<div class=\"examplesentence\">\\(\\frac{9}{\\pi }+\\frac{100\\pi }{\\pi }=l^{2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow we have the same denominator in both our fractions, so we can add them like normal.<\/p>\n<div class=\"examplesentence\">\\(\\frac{9+100\\pi }{\\pi }=l^{2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow all that we have to do is take the square root of both sides. When we do this, we get that:<\/p>\n<div class=\"examplesentence\">\\(l\\approx 10.1\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo our slant height is approximately <span style=\"font-style:normal; font-size:90%\">\\(10.1\\)<\/span> inches.<\/p>\n<p>Awesome! Now, let\u2019s use this to find the surface area.<\/p>\n<div class=\"examplesentence\">\\(SA=\\pi r^{2}+\\pi rl\\)<\/div>\n<p>\n&nbsp;<br \/>\n<span style=\"font-style:normal; font-size:90%\">\\(r\\)<\/span> is approximately <span style=\"font-style:normal; font-size:90%\">\\(1.7\\)<\/span>.<\/p>\n<div class=\"examplesentence\">\\(SA=\\pi (1.7)^{2}+\\pi (1.7)(10.1)\\)<\/div>\n<p>\n&nbsp;<br \/>\nThen, if you plug this into a calculator, you&#8217;ll get that:<\/p>\n<div class=\"examplesentence\">\\(SA\\approx 63\\text{ in}^{2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd that\u2019s it! By just knowing the cone\u2019s volume and height, we were able to find its radius, its slant height, and its surface area.<\/p>\n<h3><span id=\"Example_3\" class=\"m-toc-anchor\"><\/span>Example #3<\/h3>\n<p>\nI want to work on one more example together, but first, let me tell you about how ancient mathematicians came up with these volume and surface area formulas. Why do we divide <span style=\"font-style:normal; font-size:90%\">\\(\\pi r^{2}h\\)<\/span> by <span style=\"font-style:normal; font-size:90%\">\\(3\\)<\/span> in the volume equation? And why are we only concerned with the <span style=\"font-style:normal; font-size:90%\">\\(r\\)<\/span>&#8211; and <span style=\"font-style:normal; font-size:90%\">\\(l\\)<\/span>-values when we find surface area?<\/p>\n<p>To answer that first question, there\u2019s one fact you need to know: a cone has a volume that\u2019s <span style=\"font-style:normal; font-size:90%\">\\(\\frac{1}{3}\\)<\/span> the size of a cylinder\u2019s volume, as long as they have a same-sized circular base. Take a look:<\/p>\n<p>The <span style=\"font-style:normal; font-size:90%\">\\(r\\)<\/span>&#8211; and <span style=\"font-style:normal; font-size:90%\">\\(h\\)<\/span>-values of these two objects are the same, and we know that the volume equation of a cylinder is <span style=\"font-style:normal; font-size:90%\">\\(V=\\pi r^{2}h\\)<\/span>. Hence, since this cylinder could hold <span style=\"font-style:normal; font-size:90%\">\\(3\\)<\/span> times the amount of stuff inside of it, we have that the volume of the cone is equal to <span style=\"font-style:normal; font-size:90%\">\\(\\frac{\\pi r^{2}h}{3}\\)<\/span>.<\/p>\n<p>As for the cone\u2019s surface area, we need to consider the two different \u201csides\u201d of this <span style=\"font-style:normal; font-size:90%\">\\(3D\\)<\/span> shape. We have a circle sitting on the bottom, and we have this curved surface connecting the base to the endpoint. So, we need to add the area of the circle (remember, that\u2019s <span style=\"font-style:normal; font-size:90%\">\\(\\pi r^{2}\\)<\/span>) to the area of the curved piece (which is <span style=\"font-style:normal; font-size:90%\">\\(\\pi rl\\)<\/span>).<\/p>\n<p>With all of this in mind, let\u2019s look at our final example.<\/p>\n<p>Say we have a huge cone whose base\u2019s diameter measures <span style=\"font-style:normal; font-size:90%\">\\(10\\)<\/span> feet, height measures <span style=\"font-style:normal; font-size:90%\">\\(12\\)<\/span> feet, and would need a <span style=\"font-style:normal; font-size:90%\">\\(13\\)<\/span>-foot-long rope to connect a straight line from the tip to the edge of the base. We know how to find this cone\u2019s surface area and volume.<\/p>\n<p>First, let\u2019s work on the volume! Again, we have that <span style=\"font-style:normal; font-size:90%\">\\(V=\\frac{1}{3}\\pi r^{2}h\\)<\/span>. But, remember, we were given the diameter of the circular base, not its radius. We couldn\u2019t just plug the <span style=\"font-style:normal; font-size:90%\">\\(10\\)<\/span> into the <span style=\"font-style:normal; font-size:90%\">\\(r\\)<\/span>\u2019s place in that equation. Instead, we use what we know about the relationship between diameter and radius: a circle\u2019s radius is <span style=\"font-style:normal; font-size:90%\">\\(\\frac{1}{2}\\)<\/span> of its diameter. So this cone\u2019s radius is actually <span style=\"font-style:normal; font-size:90%\">\\(5\\)<\/span> feet; and now finding the volume is pretty straightforward.<\/p>\n<p>So if we plug in the variables that we know, we get:<\/p>\n<div class=\"examplesentence\">\\(V=\\frac{1}{3}\\pi (5)^{2}(12)\\)<br \/>\n\\(V=\\frac{1}{3}\\pi (25)(12)\\)<br \/>\n\\(V=\\frac{1}{3}\\pi (300)\\)<br \/>\n\\(V=100\\pi \\)<\/div>\n<p>\n&nbsp;<br \/>\nSo the volume is <span style=\"font-style:normal; font-size:90%\">\\(100\\pi \\text{ ft}^{3}\\)<\/span>.<\/p>\n<p>Finally, let\u2019s find the surface area. <\/p>\n<div class=\"examplesentence\">\\(SA=\\pi r^{2}+\\pi rl\\)<\/div>\n<p>\n&nbsp;<br \/>\nWhen we plug in our known variables, we get:<\/p>\n<div class=\"examplesentence\">\\(SA= \\pi (5)^{2}+\\pi (5)(13)\\)<br \/>\n\\(SA=25\\pi +65\\pi \\)<\/div>\n<p>\n&nbsp;<br \/>\nWhich is equal to <span style=\"font-style:normal; font-size:90%\">\\( 90\\pi \\text{ ft}^{2}\\)<\/span>. So we see that we\u2019d need <span style=\"font-style:normal; font-size:90%\">\\( 90\\pi \\text{ ft}^{2}\\)<\/span> of wrapping paper to cover this thing!<\/p>\n<p>Well, I hope that this has given you a greater understanding of cones and has made you confident in your ability to find any cone\u2019s volume and surface area. Thanks for watching, and happy studying!<\/p>\n<p>For more help, check out our <a class=\"ylist\" target=\"_blank\" rel=\"noopener noreferrer\" href=\"https:\/\/www.mometrix.com\/academy\/cone-volume-calculator\/\">volume of a cone calculator<\/a>!<\/p>\n<\/div>\n<div class=\"spoiler\" id=\"FAQs-spoiler\">\n<h2 style=\"text-align:center\"><span id=\"Frequently_Asked_Questions\" class=\"m-toc-anchor\"><\/span>Frequently Asked Questions<\/h2>\n<div class=\"faq-list\">\n<div class=\"qa_wrap\">\n<div class=\"q_item text_bold\">\n<h4 class=\"letter\">Q<\/h4>\n<p style=\"line-height: unset;\">How do you find the surface area and volume of a cone?<\/p>\n<\/p><\/div>\n<div class=\"a_item\">\n<h4 class=\"letter text_bold\">A<\/h4>\n<p>The surface area of a cone is found by using the formula \\(\\pi r^2+\\pi rl\\), where \\(r\\) represents the radius of the circular base, \\(h\\) represents the height of the cone, \\(l\\) represents the slant height, and \\(\\pi\\) can be approximated as 3.14. If the slant height is not given, it can be found using the formula \\(l=\\sqrt{r^2+h^2}\\).<\/p>\n<p>The volume of a cone is found by using the formula \\(\\frac{1}{3}\\pi r^2h\\). The volume of a cone is simply \\(\\frac{1}{3}\\) the volume of a cylinder. This is why the formulas for the volume of cylinders and cones are so similar.<\/p>\n<\/p><\/div>\n<\/p><\/div>\n<div class=\"qa_wrap\">\n<div class=\"q_item text_bold\">\n<h4 class=\"letter\">Q<\/h4>\n<p style=\"line-height: unset;\">What is the formula for finding the surface area of a cone?<\/p>\n<\/p><\/div>\n<div class=\"a_item\">\n<h4 class=\"letter text_bold\">A<\/h4>\n<p>The formula for finding the surface area of a cone is \\(\\pi r^2+\\pi rl\\). In this formula, \\(r\\) represents the radius of the circular base, \\(h\\) represents the height of the cone, \\(l\\) represents the slant height, and \\(\\pi\\) can be approximated as \\(3.14\\). If the slant height is not given, it can be found using the formula \\(l=\\sqrt{r^2+h^2}\\). <\/p>\n<\/p><\/div>\n<\/p><\/div>\n<div class=\"qa_wrap\">\n<div class=\"q_item text_bold\">\n<h4 class=\"letter\">Q<\/h4>\n<p style=\"line-height: unset;\">What is the volume of a cone?<\/p>\n<\/p><\/div>\n<div class=\"a_item\">\n<h4 class=\"letter text_bold\">A<\/h4>\n<p>The volume of a cone is the amount of space taken up by a cone. The volume of a cone is \\(\\frac{1}{3}\\) the volume of a cylinder. A cone\u2019s volume can be calculated by using the formula \\(\\frac{1}{3}\\pi r^2h\\). <\/p>\n<\/p><\/div>\n<\/p><\/div>\n<div class=\"qa_wrap\">\n<div class=\"q_item text_bold\">\n<h4 class=\"letter\">Q<\/h4>\n<p style=\"line-height: unset;\">How do you find the surface area of a cone without the base?<\/p>\n<\/p><\/div>\n<div class=\"a_item\">\n<h4 class=\"letter text_bold\">A<\/h4>\n<p>A cone consists of a circular base, and a lateral face that is wrapped around the base forming a point. The surface area of the circular portion of a cone is simply \\(\\pi r^2\\).<\/p>\n<p>The lateral face wrapping around this base is calculated as \\(\\pi rl\\), or if you aren\u2019t given the slant height, you can use the formula \\(\\pi r\\sqrt{(r^2+h^2)}\\). The formula of an entire cone is the result of adding these two pieces together (\\(\\pi r^2+\u03c0rl\\)).<\/p>\n<\/p><\/div>\n<\/p><\/div>\n<div class=\"qa_wrap\">\n<div class=\"q_item text_bold\">\n<h4 class=\"letter\">Q<\/h4>\n<p style=\"line-height: unset;\">What is the slant height of a cone?<\/p>\n<\/p><\/div>\n<div class=\"a_item\">\n<h4 class=\"letter text_bold\">A<\/h4>\n<p>The slant height of a cone should not be confused with the height of a cone. Slant height is the distance from the top of a cone, down the side to the edge of the circular base.<\/p>\n<p>Slant height is calculated as \\(\\sqrt{(r^2+h^2)}\\), where \\(r\\) represents the radius of the circular base, and \\(h\\) represents the height, or altitude, of the cone.<\/p>\n<\/p><\/div>\n<\/p><\/div>\n<div class=\"qa_wrap\">\n<div class=\"q_item text_bold\">\n<h4 class=\"letter\">Q<\/h4>\n<p style=\"line-height: unset;\">What is the difference between height and slant height? <\/p>\n<\/p><\/div>\n<div class=\"a_item\">\n<h4 class=\"letter text_bold\">A<\/h4>\n<p>The \u201cheight\u201d of a cone, and the \u201cslant height\u201d of a cone are not the same thing.<\/p>\n<p>The <strong>height<\/strong> of a cone is considered the vertical height or altitude of the cone. This is the perpendicular distance from the top of the cone down to the center of the circular base.<\/p>\n<p>The <strong>slant height<\/strong> of a cone is the distance from the top of the cone, down the side of the cone to the edge of the circular base. <\/p>\n<\/p><\/div>\n<\/p><\/div>\n<div class=\"qa_wrap\">\n<div class=\"q_item text_bold\">\n<h4 class=\"letter\">Q<\/h4>\n<p style=\"line-height: unset;\">How do you find the vertical height of a cone given the radius and slant height?<\/p>\n<\/p><\/div>\n<div class=\"a_item\">\n<h4 class=\"letter text_bold\">A<\/h4>\n<p>The vertical height, radius, and slant height of a cone form three lines that create a right triangle. This means that the Pythagorean theorem can be used to determine a missing value if at least two values are known. <\/p>\n<div class=\"lightbulb-example-2\"><span class=\"lightbulb-icon\">\ud83d\udca1<\/span><span class=\"faq-example-question\">Example: What is the vertical height of a cone with a radius of <span style=\"white-space:nowrap\">9 cm<\/span> and a slant height of <span style=\"white-space:nowrap\">15 cm<\/span>?<\/span><\/p>\n<hr style=\"padding: 0; margin-top: -0.2em; margin-bottom: 1.2em\">If a cone has a radius of <span style=\"white-space:nowrap\">9 cm<\/span> and a slant height of <span style=\"white-space:nowrap\">15 cm<\/span>, then the formula \\(a^2+b^2=c^2\\) becomes: <\/p>\n<p style=\"text-align: center; line-height: 35px\"> \\((\\text{vertical height})^2+9^2=15^2\\)<br \/>\\((\\text{vertical height})^2+81=225\\)<\/p>\n<p style=\"margin-bottom: 0em\">This becomes <span style=\"font-size: 95%\">\\(\\text{vertical height}\\)\\(\\:=12 \\mathrm{\\:cm}\\).<\/p>\n<\/div>\n<\/p><\/div>\n<\/p><\/div>\n<div class=\"qa_wrap\">\n<div class=\"q_item text_bold\">\n<h4 class=\"letter\">Q<\/h4>\n<p style=\"line-height: unset;\">How do you find the radius of a cone with slant height and height? <\/p>\n<\/p><\/div>\n<div class=\"a_item\">\n<h4 class=\"letter text_bold\">A<\/h4>\n<p>The Pythagorean theorem can be used to determine the radius of a cone\u2019s base if the slant height and the height are provided. The Pythagorean theorem states that \\(a^2+b^2=c^2\\) for right triangles. Conveniently, the radius, slant height, and height of a cone forms a right triangle. <\/p>\n<div class=\"lightbulb-example-2\"><span class=\"lightbulb-icon\">\ud83d\udca1<\/span><span class=\"faq-example-question\">Example: What is the radius of a cone with a slant height of <span style=\"white-space:nowrap\">5 m<\/span> and a height of <span style=\"white-space:nowrap\">4 m<\/span>?<\/span><\/p>\n<hr style=\"padding: 0; margin-top: -0.2em; margin-bottom: 1.2em\">We can plug these values into the formula and solve for the radius.<\/p>\n<p style=\"margin-bottom: 0em\">The formula \\(a^2+b^2=c^2\\) becomes \\(r^2+4^2=5^2\\), which simplifies to \\(r=3 \\mathrm{\\:m}\\). <\/p>\n<\/div>\n<\/p><\/div>\n<\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"spoiler\" id=\"PQs-spoiler\">\n<h2 style=\"text-align:center\"><span id=\"Volume_and_Surface_Area_of_a_Right_Circular_Cone_Practice_Questions\" class=\"m-toc-anchor\"><\/span>Volume and Surface Area of a Right Circular Cone 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 \/>\nWhat is the volume to the nearest whole number of a cone that has a height of 14 inches and a radius of 6 inches?<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-1-1\">535 in<sup>3<\/sup><\/div><div class=\"PQ\"  id=\"PQ-1-2\">582 in<sup>3<\/sup><\/div><div class=\"PQ correct_answer\"  id=\"PQ-1-3\">528 in<sup>3<\/sup><\/div><div class=\"PQ\"  id=\"PQ-1-4\">498 in<sup>3<\/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>Start with the volume formula for a cone:<\/p>\n<p style=\"text-align: center\">\\(V=\\frac{1}{3}\\pi r^2h\\)<\/p>\n<p>Plug in 6 for \\(r\\) and 14 for \\(h\\).<\/p>\n<p style=\"text-align: center; line-height: 45px;\">\n\\(V=\\frac{1}{3}\\pi (6)^2(14)\\)<br \/>\n\\(V=\\frac{1}{3}\\pi (36)(14)\\)<br \/>\n\\(V=\\frac{1}{3}\\pi (504)\\)<br \/>\n\\(V=168\\pi \\)<br \/>\n\\(V=168(3.14159)=527.8\\)<br \/>\n\\(V=528\\text{ in}^3\\)<\/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 \/>\nWhat is the volume in terms of pi of a cone that has a height of 12&nbsp;cm and a radius of 9&nbsp;cm?<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ correct_answer\"  id=\"PQ-2-1\">324\\(\\pi\\) cm<sup>3<\/sup><\/div><div class=\"PQ\"  id=\"PQ-2-2\">346\\(\\pi\\) cm<sup>3<\/sup><\/div><div class=\"PQ\"  id=\"PQ-2-3\">445\\(\\pi\\) cm<sup>3<\/sup><\/div><div class=\"PQ\"  id=\"PQ-2-4\">389\\(\\pi\\) cm<sup>3<\/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>Start with the volume formula for a cone and plug in 9 for \\(r\\) and 12 for \\(h\\).<\/p>\n<p style=\"text-align: center; line-height: 45px\">\\(V=\\frac{1}{3}\\pi (9)^2(12)\\)<br \/>\n\\(V=\\frac{1}{3}\\pi (81)(12)\\)<br \/>\n\\(V=\\frac{1}{3}\\pi (972)\\)<br \/>\n\\(V=324\\pi \\text{ cm}^3\\)<\/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 \/>\nFind the volume of a cone that has a radius of 6 meters and a height of 11 meters. Express your answer in terms of pi.<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-3-1\">227\\(\\pi\\) m<sup>3<\/sup><\/div><div class=\"PQ\"  id=\"PQ-3-2\">432\\(\\pi\\) m<sup>3<\/sup><\/div><div class=\"PQ\"  id=\"PQ-3-3\">145\\(\\pi\\) m<sup>3<\/sup><\/div><div class=\"PQ correct_answer\"  id=\"PQ-3-4\">132\\(\\pi\\) m<sup>3<\/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>Start with the volume formula for a cone and plug in 6 for \\(r\\) and 11 for \\(h\\).<\/p>\n<p style=\"text-align: center; line-height: 45px\">\\(V=\\frac{1}{3}\\pi (6)^2(11)\\)<br \/>\n\\(V=\\frac{1}{3}\\pi (36)(11)\\)<br \/>\n\\(V=\\frac{1}{3}\\pi (396)\\)<br \/>\n\\(V=132\\pi \\text{ m}^3\\)<\/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 \/>\nFind the volume of a cone that has a diameter of 18 meters and a height of 30 meters. Express your answer to the nearest whole number.<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-4-1\">2,676 m<sup>3<\/sup><\/div><div class=\"PQ\"  id=\"PQ-4-2\">2,304 m<sup>3<\/sup><\/div><div class=\"PQ\"  id=\"PQ-4-3\">2,499 m<sup>3<\/sup><\/div><div class=\"PQ correct_answer\"  id=\"PQ-4-4\">2,545 m<sup>3<\/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>If the diameter is 18 meters, then the diameter is 9 meters. Start with the volume formula for a cylinder and plug in 9 for \\(r\\) and 30 for \\(h\\).<\/p>\n<p style=\"text-align: center; line-height: 45px\">\\(V=\\frac{1}{3}\\pi (9)^2(30)\\)<br \/>\n\\(V=\\frac{1}{3}\\pi (81)(30)\\)<br \/>\n\\(V=\\frac{1}{3}\\pi (2,430)\\)<br \/>\n\\(V=810\\pi \\)<br \/>\n\\(V=810(3.14159)=2,544.69\\)<br \/>\n\\(V=2,545\\text{ m}^3\\)<\/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 \/>\nFind the volume of a cone that has a radius of 6 yards and a height of 10 yards. Express your answer to the nearest whole number.<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-5-1\">488 yd<sup>3<\/sup><\/div><div class=\"PQ correct_answer\"  id=\"PQ-5-2\">377 yd<sup>3<\/sup><\/div><div class=\"PQ\"  id=\"PQ-5-3\">366 yd<sup>3<\/sup><\/div><div class=\"PQ\"  id=\"PQ-5-4\">411 yd<sup>3<\/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>Start with the volume formula for a cylinder and plug in 6 for \\(r\\) and 10 for \\(h\\).<\/p>\n<p style=\"text-align: center; line-height: 45px\">\\(V=\\frac{1}{3}\\pi (6)^2(10)\\)<br \/>\n\\(V=\\frac{1}{3}\\pi (36)(10)\\)<br \/>\n\\(V=\\frac{1}{3}\\pi (360)\\)<br \/>\n\\(V=120\\pi \\)<br \/>\n\\(V=120(3.14159)=376.99\\)<br \/>\n\\(V=377\\text{ yd}^3\\)<\/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":162224,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"open","template":"","meta":{"footnotes":""},"class_list":{"0":"post-4565","1":"page","2":"type-page","3":"status-publish","4":"has-post-thumbnail","6":"page_category-finding-volume-in-geometry","7":"page_category-math-advertising-group","8":"page_category-volume-and-surface-area","9":"page_type-video","10":"content_type-practice-questions","11":"subject_matter-math"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/4565","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=4565"}],"version-history":[{"count":6,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/4565\/revisions"}],"predecessor-version":[{"id":289735,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/4565\/revisions\/289735"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media\/162224"}],"wp:attachment":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media?parent=4565"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}