{"id":4559,"date":"2013-06-29T06:45:51","date_gmt":"2013-06-29T06:45:51","guid":{"rendered":"http:\/\/www.mometrix.com\/academy\/?page_id=4559"},"modified":"2026-03-26T12:58:53","modified_gmt":"2026-03-26T17:58:53","slug":"volume-and-surface-area-of-a-pyramid","status":"publish","type":"page","link":"https:\/\/www.mometrix.com\/academy\/volume-and-surface-area-of-a-pyramid\/","title":{"rendered":"Finding the Volume and Surface Area of a Pyramid"},"content":{"rendered":"\n\t\t\t<div id=\"mmDeferVideoEncompass_S_84f9IRj7Q\" 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_S_84f9IRj7Q\" data-source-videoID=\"S_84f9IRj7Q\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" alt=\"Finding the Volume and Surface Area of a Pyramid Video\" height=\"464\" width=\"825\" class=\"size-full\" data-matomo-title = \"Finding the Volume and Surface Area of a Pyramid\">\n\t\t\t<\/picture>\n\t\t\t<\/div>\n\t\t\t<style>img#videoThumbnailImage_S_84f9IRj7Q:hover {cursor:pointer;} img#videoThumbnailImage_S_84f9IRj7Q {background-size:contain;background-image:url(\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/07\/updated-finding-the-volume-and-surface-area-of-a-pyramid-64bfec4c06712.webp\");}<\/style>\n\t\t\t<script defer>\n\t\t\t  jQuery(\"img#videoThumbnailImage_S_84f9IRj7Q\").click(function() {\n\t\t\t\tlet videoId = jQuery(this).attr(\"data-source-videoID\");\n\t\t\t\tlet helpTag = '<div id=\"mmDeferVideoYTMessage_S_84f9IRj7Q\" 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\",\"Finding the Volume and Surface Area of a Pyramid\");\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_S_84f9IRj7Q\").html(tag);\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_S_84f9IRj7Q\").prepend(helpTag);\n\t\t\t\tsetTimeout(function(){jQuery(\"div#mmDeferVideoYTMessage_S_84f9IRj7Q\").css(\"display\", \"block\");}, 2000);\n\t\t\t  });\n\t\t\t  \n\t\t\t<\/script>\n\t\t\n<p><script>\nfunction PVS_Function() {\n  var x = document.getElementById(\"PVS\");\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=\"PVS_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=\"PVS\" style=\"display:none;\">\n<ul>\n<li class=\"toc-h2\"><a href=\"#Historical_Pyramids\" class=\"smooth-scroll\">Historical Pyramids<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Pyramids_in_Geometry\" class=\"smooth-scroll\">Pyramids in Geometry<\/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_Pyramid_Practice_Questions\" class=\"smooth-scroll\">Volume and Surface Area of a Pyramid 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>\n<div class=\"spoiler\" id=\"transcript-spoiler\">\n<p>Hello, and welcome to this video about pyramids!  In this video, we will explore different types of pyramids and how to calculate their volume and surface area. Let\u2019s learn about pyramids!<\/p>\n<h2><span id=\"Historical_Pyramids\" class=\"m-toc-anchor\"><\/span>Historical Pyramids<\/h2>\n<p>\nWe cannot talk about pyramids without mentioning the most famous pyramids in the world, located in Egypt, and built as a tomb for Egyptian kings. Although pyramids are not quite common, their shape is so striking that it seems it is used to make a statement, perhaps that is the reason the Egyptian kings used it as their tomb. The Louvre Museum in Paris, which is the largest art museum in the world, is also shaped like a pyramid. <\/p>\n<p><img decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2025\/01\/pyramid-with-drawings-on-it@300-scaled.webp\" alt=\"\" width=\"\" height=\"\" class=\"aligncenter size-full wp-image-215971\"  role=\"img\" style=\"box-shadow: 1.5px 1.5px 3px gray;\"  \/><\/p>\n<h2><span id=\"Pyramids_in_Geometry\" class=\"m-toc-anchor\"><\/span>Pyramids in Geometry<\/h2>\n<p>\nAs you can see from the images of the Egyptian Pyramids, a pyramid is a 3-dimensional figure with triangular sides that meet at the edges and at the top to form an <strong>apex<\/strong> and it has a polygon as its base.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2025\/01\/621932-triangular-base.png\" alt=\"\" width=\"\" height=\"\" class=\"aligncenter size-full wp-image-215971\"  role=\"img\" \/><\/p>\n<p>The polygonal base determines the type of pyramid. Here you see the pyramid with a triangle as its base is a triangular pyramid, the pyramid with a rectangle as its base is a rectangular pyramid, and the pyramid with a pentagon as its base is a pentagonal pyramid. We also have hexagonal pyramids and heptagonal pyramids, and so on.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2025\/01\/621932-bases.png\" alt=\"\" width=\"\" height=\"\" class=\"aligncenter size-full wp-image-215971\"  role=\"img\" \/><\/p>\n<p>Let\u2019s take a moment and recall what the volume and surface area of a 3-dimensional figure means.<\/p>\n<h3><span id=\"Volume_and_Surface_Area_Formula\" class=\"m-toc-anchor\"><\/span>Volume and Surface Area Formula<\/h3>\n<p>\nThe <strong>volume<\/strong> of a 3-dimensional figure is the measure of how much it can hold, and it is measured in cubic units.<\/p>\n<p>The <strong>surface area<\/strong> of a 3-dimensional figure is the measure of the total area that the surface of the figure covers and is measured in square units.<\/p>\n<p>Before we can calculate the volume and surface area of a pyramid, we must know the difference between the height and slant height. The <strong>height<\/strong> of a pyramid is the perpendicular length from the apex to the base, and the <strong>slant height<\/strong> is the length from the apex to the midpoint of the bottom edge of one of the triangular faces.<\/p>\n<p>Here are the formulas for the volume and surface area of any pyramid. <\/p>\n<div class=\"examplesentence\">\\(V=\\frac{1}{3}Bh\\)<br \/>\n\\(SA=B+\\frac{1}{2}ps\\)<\/div>\n<p>\n&nbsp;<br \/>\nTo calculate the volume and surface area of any pyramid we need B, which represents the area of the base, and p, which represents the perimeter of the base. It is important to note, since the base of a pyramid can be any polygon, we will be using our prior knowledge of finding the area and perimeter of different polygons to calculate the volume and surface area of a pyramid. <\/p>\n<div class=\"examplesentence\">\\(V=\\frac{1}{3}Bh\\)<br \/>\n\\(SA=B+\\frac{1}{2}ps\\)<\/div>\n<p>\n&nbsp;<\/p>\n<div style=\"font-style:normal; font-size:90%\">\\(B\\) = area of base<br \/>\n\\(h\\) = height of pyramid<br \/>\n\\(p\\) = perimeter of base<br \/>\n\\(s\\) = slant height<\/div>\n<p>\n&nbsp;<br \/>\nLet\u2019s look at an example.<\/p>\n<h3>Calculating Surface Area &#8211; Example 1<\/h3>\n<p>\nA triangular pyramid has an equilateral triangle as its base with side lengths 6 in and a height of 8 in. <\/p>\n<p>To find the <strong>surface area of a pyramid<\/strong>, we use the formula \\(SA=B+\\frac{1}{2}ps\\), where \\(B\\) is the area of the base, \\(p\\) is the perimeter of the base, and \\(s\\) is the slant height. Since the base is a triangle, we will use the formula for the area of a triangle to find \\(B\\).<\/p>\n<p>\\(\\text{Area of a triangle}=\\frac{1}{2}bh\\), where the \\(b\\) and \\(h\\) are the base and height of the triangular base. We will draw a perpendicular line to the base, which is the height of the triangular base and it divides the base of the triangle into 2 equal parts. Since the triangle is now turned into 2 right triangles, we can use the <a class=\"ylist\" href=\"https:\/\/www.mometrix.com\/academy\/pythagorean-theorem\/\">Pythagorean theorem<\/a> to find the height. <\/p>\n<p>So the Pythagorean theorem is:<\/p>\n<div class=\"examplesentence\">\\(c^{2}=a^{2}+b^{2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd then we\u2019re gonna look at our triangle, which says we have:<\/p>\n<div class=\"examplesentence\">\\(6^{2}=h^{2}+3^{2}\\)<br \/>\n\\(36=h^{2}+9\\)<\/div>\n<p>\n&nbsp;<br \/>\nWe\u2019ll subtract 9 from both sides. This gives us:<\/p>\n<div class=\"examplesentence\">\\(27=h^{2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd then we\u2019ll square root both sides. Which gives us:<\/p>\n<div class=\"examplesentence\">\\(h=3\\sqrt{3}\\)<br \/>\nOR<br \/>\n\\(h\\approx 5.2\\)<\/div>\n<p>\n&nbsp;<br \/>\nTherefore, the area of the base can be found by doing:<\/p>\n<div class=\"examplesentence\">\\(A=\\frac{1}{2}bh\\)<br \/>\n\\(A=\\frac{1}{2}(6)(5.2)\\)<br \/>\n\\(A\\approx 15.6 \\text{ in}^{2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nThe perimeter of the base is equal to all the sides added together, so:<\/p>\n<div class=\"examplesentence\">\\(p=6+6+6=18\\text{ inches}\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow to solve for surface area, all we have to do is plug these values in for our variables. So our surface area is equal to:<\/p>\n<div class=\"examplesentence\">\\(SA=B+\\frac{1}{2}ps\\)<br \/>\n\\(SA=(15.6)+\\frac{1}{2}(18)(7)\\)<br \/>\n\\(SA\\approx 78.6 \\text{ in}^{2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd all this added together is approximately equal to 78.6 square inches.<\/p>\n<h3>Calculating Volume &#8211; Example 1<\/h3>\n<p>\nTo find the <b>volume of the triangular pyramid<\/b>, we need the area of the base, \\(B\\), and the height of the pyramid, which is 8 inches. So let\u2019s plug this in. So:<\/p>\n<div class=\"examplesentence\">\\(V=\\frac{1}{3}Bh\\)<br \/>\n\\(V=\\frac{1}{3}(15.6)(8)\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow all we have to do is calculate this out, which is equal to approximately 41.6 cubic inches.<\/p>\n<div class=\"examplesentence\">\\(V=41.6\\text{ in}^{3}\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>Let\u2019s look at another example. <\/p>\n<h3>Calculating Volume &#8211; Example 2<\/h3>\n<p>\nHere is a pyramid with a square base, with side lengths of 5 centimeters. The height of the pyramid is 11 centimeters. What is the volume of the pyramid?<\/p>\n<p>So our volume formula is:<\/p>\n<div class=\"examplesentence\">\\(V=\\frac{1}{3}Bh\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>Before we can calculate the volume of the pyramid, we need to find the area of the base. Since the base is a square, we use the formula for the area of a square, which is \\(s^{2}\\). So to find our area, we\u2019re gonna use \\(s^{2}\\) and our side length is 5.<\/p>\n<div class=\"examplesentence\">\\(A=s^{2}\\)<br \/>\n\\(A=(5)^{2}\\)<br \/>\n\\(A=25\\text{ cm}^{2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow we can plug this value into our formula.<\/p>\n<div class=\"examplesentence\">\\(V=\\frac{1}{3}Bh\\)<br \/>\n\\( V=\\frac{1}{3}(25)(11)\\)<br \/>\n\\(V\\approx 91.67\\text{ cm}^{3}\\)<\/div>\n<p>\n&nbsp;<br \/>\nThen if we plug this into a calculator, we\u2019ll get that it\u2019s approximately equal to 91.67 cubic centimeters.<\/p>\n<p>But when would we ever need this in real life? Well, I\u2019m glad you asked! Take a look at this next example and try it on your own.<\/p>\n<h3>Calculating Surface Area &#8211; Example 2<\/h3>\n<p>\nThe roof of a wooden cottage is shaped like a square-based pyramid. The length of the sides of the square base are 22 feet and the height of the triangular face is 14 feet. Peter wants to paint the roof of his wooden cottage and needs to determine how much paint he needs to buy. We will assume 1 pint of paint covers 400 square feet. How much paint does Peter need to buy? <\/p>\n<p>The roof of the cottage does not include the base of the pyramid. Therefore, we only need to find the area of the 4 triangular faces. This is called the lateral area. So the lateral area is equal to the surface area minus the area of the base. So all we need is:<\/p>\n<div class=\"examplesentence\">\\(LA=\\frac{1}{2}ps\\)<\/div>\n<p>\n&nbsp;<br \/>\nThe perimeter of our square is equal to 4 times the side length.<\/p>\n<div class=\"examplesentence\">\\(LA=\\frac{1}{2}(4\\times 22)(14)\\)<\/div>\n<p>\n&nbsp;<br \/>\nWhich we can plug into a calculator to get:<\/p>\n<div class=\"examplesentence\">\\(LA=616 \\text{ ft}^{2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo we will need 616 square feet to be covered.<\/p>\n<p>If 1 pint of paint covers 400 square feet, we need to divide 616 by 400 to figure out how much paint we need. So:<\/p>\n<div class=\"examplesentence\">\\(616\\div 400=1.54\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo, Peter will need to buy 2 pints of paint to cover the roof of the cottage because 1 pint won&#8217;t be enough and you can&#8217;t get 0.54 of a pint. So 2 pints of paint will cover the whole roof.<\/p>\n<p>I hope this video on finding the volume and surface area of a pyramid was helpful! Thanks for watching, and happy studying!<\/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 volume of a pyramid?<\/p>\n<\/p><\/div>\n<div class=\"a_item\">\n<h4 class=\"letter text_bold\">A<\/h4>\n<p>To find the volume of a pyramid, multiply the area of its base with the height of the pyramid, and divide by 3.<\/p>\n<p>We express this product with the formula \\(V=\\frac{1}{3}\\times B\\times h\\), where \\(B\\) is the area of the base of the pyramid and \\(h\\) is its height.<\/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;\">Why is \\(\\frac{1}{3}\\) in the formula for the volume of a pyramid?<\/p>\n<\/p><\/div>\n<div class=\"a_item\">\n<h4 class=\"letter text_bold\">A<\/h4>\n<p>In a similar way to how we derive the area of a trapezoid by cutting a rectangle in half, we can derive the volume of a pyramid by cutting a prism into six small pyramids. We then look at half of the prism, which is then the same height as the pyramid in question, to find that the volume of our pyramid is one-third of the half-prism, or one-third the volume of a prism of the same height.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-94099\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/09\/rectangular-prism-with-two-rectangular-pyramids-stacked-inside-with-one-pointing-downward-and-one-pointing-upward-bottom-pyramid-is-yellow.png\" alt=\"rectangular prism, with two rectangular pyramids stacked inside with one pointing downward and one pointing upward, bottom pyramid is yellow\" width=\"315\" height=\"421\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/09\/rectangular-prism-with-two-rectangular-pyramids-stacked-inside-with-one-pointing-downward-and-one-pointing-upward-bottom-pyramid-is-yellow.png 718w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/09\/rectangular-prism-with-two-rectangular-pyramids-stacked-inside-with-one-pointing-downward-and-one-pointing-upward-bottom-pyramid-is-yellow-225x300.png 225w\" sizes=\"auto, (max-width: 315px) 100vw, 315px\" \/><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-94108\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/09\/rectangle-cut-into-two-trapezoids-left-trapezoid-is-pink-with-pink-lines-right-trapezoid-is-blue-1.png\" alt=\"rectangle cut into two trapezoids, left trapezoid is pink with pink lines, right trapezoid is blue\" width=\"439\" height=\"285\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/09\/rectangle-cut-into-two-trapezoids-left-trapezoid-is-pink-with-pink-lines-right-trapezoid-is-blue-1.png 891w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/09\/rectangle-cut-into-two-trapezoids-left-trapezoid-is-pink-with-pink-lines-right-trapezoid-is-blue-1-300x195.png 300w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/09\/rectangle-cut-into-two-trapezoids-left-trapezoid-is-pink-with-pink-lines-right-trapezoid-is-blue-1-768x499.png 768w\" sizes=\"auto, (max-width: 439px) 100vw, 439px\" \/><\/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 starting formula for volume of a pyramid?<\/p>\n<\/p><\/div>\n<div class=\"a_item\">\n<h4 class=\"letter text_bold\">A<\/h4>\n<p>The general formula for the volume of a pyramid is \\(V=\\frac{1}{3}\\times B\\times h\\), where \\(B\\) is the area of the base of the pyramid and \\(h\\) is the height of the pyramid.<\/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 square pyramid?<\/p>\n<\/p><\/div>\n<div class=\"a_item\">\n<h4 class=\"letter text_bold\">A<\/h4>\n<p>For pyramids whose bases are square, we can find their volume using the formula \\(V=\\frac{1}{3}\\times s^2\\times h\\), where \\(s\\) is the length of the sides of the square base.<\/p>\n<p>For example, if a square pyramid has a height of 11&nbsp;cm and base side lengths of 6&nbsp;cm, its volume is as follows: <\/p>\n<p style=\"text-align: center\">\\(\\frac{1}{3}\\times6^2\\times11=132\\text{ cm}^3\\)<\/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 triangular pyramid?<\/p>\n<\/p><\/div>\n<div class=\"a_item\">\n<h4 class=\"letter text_bold\">A<\/h4>\n<p>To find the volume of a pyramid whose base is triangular, use the following formula: <\/p>\n<p style=\"text-align: center\">\\(V=\\dfrac{1}{3}\\times(\\dfrac{1}{2}\\times b_{\\text{triangle}}\\times h_{\\text{triangle}})\\times h_{\\text{pyramid}}\\)<\/p>\n<p>The product inside the parentheses is the area of the base, which we multiply by \\(\\frac{1}{3}\\) and the height of the pyramid.<\/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 rectangular pyramid?<\/p>\n<\/p><\/div>\n<div class=\"a_item\">\n<h4 class=\"letter text_bold\">A<\/h4>\n<p>For rectangular pyramids whose base has length \\((l)\\) and width \\((w)\\) the formula for volume is as follows: <\/p>\n<p style=\"text-align: center\">\\(V=\\dfrac{1}{3}\\times l\\times w\\times h\\)<\/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 volume of a regular pentagonal pyramid?<\/p>\n<\/p><\/div>\n<div class=\"a_item\">\n<h4 class=\"letter text_bold\">A<\/h4>\n<p>To determine the volume of a pyramid, we multiply the area of its base with its height and divide by 3. For pyramids whose base is a regular pentagon (that is, all five sides are equal in length and all five angles are equal as well), it is important to first determine the area of that pentagon.<\/p>\n<p>If you have the length \\(s\\) of each side of the pentagon, the area can be determined using the following formula: <\/p>\n<p style=\"text-align: center\">\\(A_{\\text{pentagon}}=\\dfrac{1}{4}\\sqrt{5(5+2\\sqrt{5})}\\times s^2\\)<\/p>\n<p>Those constants can be reduced to approximately 1.72048. <\/p>\n<p style=\"text-align: center\">\\(A_{\\text{pentagon}}\\approx 1.72\\times s^2\\)<\/p>\n<p>This leads us to the formula for the volume of a regular pentagonal pyramid: <\/p>\n<p style=\"text-align: center\">\\(V\\approx \\dfrac{1}{3}\\times(1.72\\times s^2)\\times h\\)<\/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 total surface area of a pyramid?<\/p>\n<\/p><\/div>\n<div class=\"a_item\">\n<h4 class=\"letter text_bold\">A<\/h4>\n<p>To find the total surface area of a pyramid, we need to add up the areas of each side, including the base. For triangular pyramids, there will be three side panels; for square and rectangular pyramids, there will be four; and so on. <\/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 surface area of a triangular pyramid?<\/p>\n<\/p><\/div>\n<div class=\"a_item\">\n<h4 class=\"letter text_bold\">A<\/h4>\n<p>In order to find the total surface area, we will need to add up the areas of the base and the three other sides. <\/p>\n<p style=\"text-align: center\">\\(SA=B+\\dfrac{1}{2}(P\\times l)\\)<\/p>\n<p> In this case, \\(B\\) is the area of the base of the pyramid, \\(P\\) is the perimeter of the base, and \\(l\\) is the slant height of the pyramid.<\/p>\n<p>We can find \\(B\\) by using the formula \\(\\frac{1}{2}bh\\) for area of a triangle. \\(P\\) can easily be determined by adding up the lengths of the sides of the base triangle. Slant height \\(l\\) can be found using Pythagorean theorem, where \\(l\\) is the hypotenuse length.<\/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 total surface area of a square pyramid?<\/p>\n<\/p><\/div>\n<div class=\"a_item\">\n<h4 class=\"letter text_bold\">A<\/h4>\n<p>To find the total surface area of a square pyramid, we sum the area of the base with the area of the four sides. Because the four side panels are identical, we just need to find that area once, and multiply it by four. <\/p>\n<p style=\"text-align: center\">\\(SA=A_{\\text{base}}+(4\\times A_{\\text{side}})\\)<\/p>\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_Pyramid_Practice_Questions\" class=\"m-toc-anchor\"><\/span>Volume and Surface Area of a Pyramid 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 surface area of the following square pyramid. <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\" wp-image-70190 aligncenter\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/03\/Pyramid-with-a-base-of-5cm-and-height-of-8cm.png\" alt=\"\" width=\"308\" height=\"326\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/03\/Pyramid-with-a-base-of-5cm-and-height-of-8cm.png 785w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/03\/Pyramid-with-a-base-of-5cm-and-height-of-8cm-284x300.png 284w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/03\/Pyramid-with-a-base-of-5cm-and-height-of-8cm-768x812.png 768w\" sizes=\"auto, (max-width: 308px) 100vw, 308px\" \/><\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-1-1\">95 cm<sup>2<\/sup><\/div><div class=\"PQ\"  id=\"PQ-1-2\">100 cm<sup>2<\/sup><\/div><div class=\"PQ correct_answer\"  id=\"PQ-1-3\">105 cm<sup>2<\/sup><\/div><div class=\"PQ\"  id=\"PQ-1-4\">110 cm<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>The following formula can be used to calculate the surface area of a square pyramid.<\/p>\n<p style=\"text-align: center\">\\(SA=B+\\frac{1}{2}pl\\)<\/p>\n<p>In this formula, \\(B\\) represents the area of the base, \\(p\\) represents the perimeter of the base, and \\(l\\) represents the slant height.<\/p>\n<p>When the appropriate values are plugged into the formula, it becomes:<\/p>\n<p style=\"text-align: center\">\\(SA=25+\\frac{1}{2}(20)(8)\\)<\/p>\n<p>This simplifies to 105 cm<sup>2<\/sup>, so the surface area of the square pyramid is 105 cm<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 \/>\nCalculate the surface area of the following equilateral triangular pyramid. <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\" wp-image-70196 aligncenter\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/03\/triangular-pyramid-with-a-base-of-14-in-and-heights-of-12-in-and-13-in.png\" alt=\"\" width=\"249\" height=\"373\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/03\/triangular-pyramid-with-a-base-of-14-in-and-heights-of-12-in-and-13-in.png 631w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/03\/triangular-pyramid-with-a-base-of-14-in-and-heights-of-12-in-and-13-in-200x300.png 200w\" sizes=\"auto, (max-width: 249px) 100vw, 249px\" \/><\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ correct_answer\"  id=\"PQ-2-1\">357 in<sup>2<\/sup><\/div><div class=\"PQ\"  id=\"PQ-2-2\">350 in<sup>2<\/sup><\/div><div class=\"PQ\"  id=\"PQ-2-3\">257 in<sup>2<\/sup><\/div><div class=\"PQ\"  id=\"PQ-2-4\">332 in<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>The formula for calculating the surface area of a triangular pyramid is very similar to the formula for a square pyramid. The only difference occurs when determining the area of the base shape.<\/p>\n<p>The surface area formula will still apply, but \\(B\\) will be calculated by multiplying \\(\\frac{1}{2}bh\\). This can be plugged into the formula for \\(B\\), which results in the following: <\/p>\n<p style=\"text-align: center\">\\(SA=(\\dfrac{1}{2}bh)+\\dfrac{1}{2}pl\\)<\/p>\n<p>From here, we can simply plug in all of the appropriate values:<\/p>\n<p style=\"text-align: center\">\\(SA=(\\dfrac{1}{2})(14)(12)+\\dfrac{1}{2}(42)(13)\\)<\/p>\n<p>This simplifies to \\(SA=357\\text{ in}^2\\), so the surface area of the equilateral triangular pyramid is \\(357\\text{ in}^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 volume of the pyramid. <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\" wp-image-70193 aligncenter\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/03\/pyramid-with-bases-of-21-and-30-in-and-a-height-of-27-in.png\" alt=\"\" width=\"308\" height=\"337\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/03\/pyramid-with-bases-of-21-and-30-in-and-a-height-of-27-in.png 764w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/03\/pyramid-with-bases-of-21-and-30-in-and-a-height-of-27-in-274x300.png 274w\" sizes=\"auto, (max-width: 308px) 100vw, 308px\" \/> <\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-3-1\">2,670 in<sup>3<\/sup><\/div><div class=\"PQ\"  id=\"PQ-3-2\">3,670 in<sup>3<\/sup><\/div><div class=\"PQ\"  id=\"PQ-3-3\">4,670 in<sup>3<\/sup><\/div><div class=\"PQ correct_answer\"  id=\"PQ-3-4\">5,670 in<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>The formula \\(V=\\frac{1}{3}Bh\\) can be used to calculate the volume of a pyramid.<\/p>\n<p>In this formula, \\(B\\) represents the area of the base shape and \\(h\\) represents the height of the pyramid (not the slant height).<\/p>\n<p>When the appropriate values are plugged into the formula, it becomes:<\/p>\n<p style=\"text-align: center\">\\(V=\\dfrac{1}{3}(21)(30)(27)\\)<\/p>\n<p>This simplifies to \\(V=5{,}670\\), so the volume of the pyramid is 5,670 in<sup>3<\/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 \/>\nLeo has a triangular glass pyramid that he is hoping to fill with some beach sand. On his vacation to Florida he collects 6.5 cm<sup>3<\/sup>  of sand. The triangle at the base of the glass pyramid has an area of 5 cm<sup>2<\/sup>, and the height of the pyramid is 6 cm. If he collects 6.5 cm<sup>3<\/sup> of sand, will the pyramid be large enough to hold all of the sand? <\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ correct_answer\"  id=\"PQ-4-1\">Yes, the pyramid is large enough<\/div><div class=\"PQ\"  id=\"PQ-4-2\">No,  the pyramid is too small.<\/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 volume of the triangular based pyramid can be calculated using the formula \\(V=\\frac{1}{3}Bh\\), where \\(B\\) is the area of the base triangle, and \\(h\\) is the height of the pyramid.<\/p>\n<p>When \\(B\\) and \\(h\\) are plugged in, it becomes:<\/p>\n<p style=\"text-align: center\">\\(V=\\dfrac{1}{3}(5)(6)\\)<\/p>\n<p>This simplifies to \\(V=10\\), which means the volume of the pyramid is 10 cm<sup>3<\/sup>. This is enough room for the 6.5 cm<sup>3<\/sup> of sand.<\/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 \/>\nSome of Leonardo da Vinci\u2019s first sketches of parachute designs were in the shape of square based pyramids. For a parachute design with rectangular base side lengths of six feet and eight feet and a height of seven feet, how much air would be captured in the pyramid once it is deployed fully?<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ correct_answer\"  id=\"PQ-5-1\">112 ft<sup>3<\/sup><\/div><div class=\"PQ\"  id=\"PQ-5-2\">109 ft<sup>3<\/sup><\/div><div class=\"PQ\"  id=\"PQ-5-3\">115 ft<sup>3<\/sup><\/div><div class=\"PQ\"  id=\"PQ-5-4\">118 ft<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>The formula \\(V=\\frac{1}{3}Bh\\) can be used to calculate the volume of a pyramid. In this case, \\(B\\) represents the area of the base shape and \\(h\\) represents the height of the pyramid.<\/p>\n<p>When the appropriate values are plugged into the formula, it becomes:<\/p>\n<p style=\"text-align: center\">\\(V=\\dfrac{1}{3}(48)(7)\\)<\/p>\n<p>This simplifies to \\(V=112\\), which means the volume of the fully deployed parachute would be 112 ft<sup>3<\/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":186566,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"open","template":"","meta":{"footnotes":""},"class_list":{"0":"post-4559","1":"page","2":"type-page","3":"status-publish","4":"has-post-thumbnail","6":"page_category-volume-and-surface-area","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\/4559","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=4559"}],"version-history":[{"count":7,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/4559\/revisions"}],"predecessor-version":[{"id":242716,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/4559\/revisions\/242716"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media\/186566"}],"wp:attachment":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media?parent=4559"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}