{"id":2904,"date":"2023-04-20T10:00:17","date_gmt":"2023-04-20T04:30:17","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=2904"},"modified":"2023-04-21T09:27:12","modified_gmt":"2023-04-21T03:57:12","slug":"area-right-circular-cone","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/area-right-circular-cone\/","title":{"rendered":"What is the Area of a Right Circular Cone"},"content":{"rendered":"<h2><strong>What is the Area of a Right Circular Cone<\/strong><\/h2>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-12352\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/09\/Surface-Area-of-Cones-1.jpg\" alt=\"\" width=\"1500\" height=\"1125\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/09\/Surface-Area-of-Cones-1.jpg 1500w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/09\/Surface-Area-of-Cones-1-300x225.jpg 300w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/09\/Surface-Area-of-Cones-1-768x576.jpg 768w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/09\/Surface-Area-of-Cones-1-1024x768.jpg 1024w\" sizes=\"(max-width: 1500px) 100vw, 1500px\" \/><\/p>\n<p>If r, h and \u2113 denote respectively the radius of base, height and slant height of a right circular cone, then-<\/p>\n<ol>\n<li>\u2113<sup> 2<\/sup>\u00a0= r<sup>2<\/sup>\u00a0+ h<sup>2<\/sup><\/li>\n<li>Area of base = \u03c0r<sup>2<\/sup><\/li>\n<li>Curved (lateral) surface area = \u03c0r\u2113<\/li>\n<li>Total surface area = \u03c0r (\u2113 + r)<\/li>\n<li>Volume = \\(\\frac{1}{3}\\)\u00a0\u03c0r<sup>2<\/sup>h<\/li>\n<\/ol>\n<p>Read more about\u00a0<a href=\"https:\/\/www.aplustopper.com\/radius-right-circular-cylinder\/\">Radius of a Right Circular Cylinder<\/a><\/p>\n<p><iframe loading=\"lazy\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/rd8tbD2eekM?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/p>\n<h2><strong>Right Circular Cone Example Problems with Solutions<\/strong><\/h2>\n<p><strong>Example 1: \u00a0 \u00a0\u00a0<\/strong>The height of a cone is 48 cm and the radius of its base is 36 cm. Find the curved surface<br \/>\narea and the total surface area of the cone.<br \/>\n(Take \u03c0 = 3.14).<br \/>\n<strong>Solution: \u00a0 \u00a0\u00a0<\/strong>Given : h = 48 cm and r = 36 cm.<br \/>\n\u2234\u00a0\u00a0\u00a0 \u2113<sup>2<\/sup>\u00a0= h<sup>2<\/sup>\u00a0+ r<sup>2<\/sup><br \/>\n\u21d2 \u00a0 \u2113<sup>2<\/sup>\u00a0= 48<sup>2<\/sup>\u00a0+ 36<sup>2<\/sup>\u00a0= 2304 + 1296 = 3600<br \/>\n\u21d2 \u00a0 \u2113 = \\(\\sqrt {3600} \\,\\,cm\\)\u00a0= 60 cm<br \/>\n\u2234 \u00a0\u00a0 The curved surface area = \u03c0r\u2113<br \/>\n= 3.14 \u00d7 36 \u00d7 60 cm<sup>2<\/sup>\u00a0= 6782.4 cm<sup>2<\/sup><br \/>\nAnd, the total surface area of the cone<br \/>\n= \u03c0r\u2113 \u00a0+ \u03c0r<sup>2<\/sup>\u00a0= \u03c0r (\u2113 + r)<br \/>\n= 3.14 \u00d7 36 \u00d7 (60 + 36)cm<sup>2<\/sup><br \/>\n= 10851.84 cm<sup>2<\/sup><\/p>\n<p><strong>Example 2: \u00a0 \u00a0\u00a0<\/strong>Curved surface area of a cone is 2200 cm<sup>2<\/sup>.<br \/>\nIt its slant height is 50 cm, find :<br \/>\n(i)\u00a0\u00a0 radius of the base.<br \/>\n(ii)\u00a0 total surface area.<br \/>\n(iii) height of the cone.<br \/>\n<strong>Solution: \u00a0 \u00a0\u00a0<\/strong>(i)\u00a0\u00a0 Given : \u03c0r\u2113 = 2200 cm<sup>2<\/sup>\u00a0and \u2113 = 50 cm<br \/>\n\u21d2\u00a0\u00a0\\(\\frac{{22}}{7} \\times r \\times 50\\) = 2200<br \/>\ni.e., r = \\(\\frac{{2200 \\times 7}}{{22 \\times 50}}cm\\) = 14 cm<br \/>\n(ii)\u00a0 Total surface area = \u03c0r\u2113 (\u2113 + r)<br \/>\n= \\(\\frac{{22}}{7}cm \\times 14\\)\u00a0\u00d7 (50 + 14) cm<sup>2<\/sup><br \/>\n= 2816 cm<sup>2<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Ans.<br \/>\n(iii) \u00a0\u2113<sup>2<\/sup>\u00a0 = h<sup>2<\/sup>\u00a0+ r<sup>2<\/sup>\u00a0\u00a0 \u21d2 \u00a0 h<sup>2<\/sup>\u00a0= \u2113<sup>2<\/sup>\u00a0\u2013 r<sup>2<\/sup><br \/>\n= 50<sup>2<\/sup>\u00a0\u2013 14<sup>2<\/sup>\u00a0= 2500 \u2013 196 = 2304<br \/>\n\u2234\u00a0\u00a0\u00a0 h = \\(\\sqrt {2304}\\) cm = 48 cm<\/p>\n<p><iframe loading=\"lazy\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/LWy9zamzYhI?start=53&#038;feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/p>\n<p><strong>Example 3: \u00a0 \u00a0\u00a0<\/strong>A conical tent is 10 m high and radius of its base is 24 m. Find<br \/>\n(i) \u00a0 slant height of the \u00a0tent.<br \/>\n(ii) cost of the canvas required to make the tent, if the cost of 1 m<sup>2<\/sup>\u00a0canvas is Rs. 70.<br \/>\n<strong>Solution: \u00a0 \u00a0\u00a0<\/strong>(i)\u00a0\u00a0 Given : h = 10 m and r = 24 m<br \/>\n\u2234\u00a0\u00a0\u00a0 \u2113<sup>2<\/sup>\u00a0= h<sup>2<\/sup>\u00a0+ r<sup>2<\/sup><br \/>\n\u21d2 \u00a0 \u2113<sup>2<\/sup>\u00a0= 10<sup>2<\/sup>\u00a0+ 24<sup>2<\/sup>\u00a0\u00a0= 100 + 576 = 676<br \/>\n\u21d2 \u00a0 \u2113 = \\(\\sqrt {676}\\) m = 26 m<br \/>\n(ii) Area of canvas required<br \/>\n= Curved surface area of the tent<br \/>\n= \u03c0r\u2113 \u00a0= \\(\\frac{{22}}{7} \\times 24 \\times 26m\\)\u00a0=\u00a0\\(\\frac{{13728}}{7}{m^2}\\)<br \/>\n\u2235\u00a0\u00a0\u00a0 Cost of 1 m<sup>2<\/sup>\u00a0canvas is Rs. 70<br \/>\n\u2234 \u00a0\u00a0 Total cost of canvas required<br \/>\n= \\(\\frac{{13728}}{7} \\times Rs.70\\)<br \/>\n= Rs. 1,37,280<\/p>\n<p><strong>Example 4: \u00a0 \u00a0\u00a0<\/strong>What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm (Use \u03c0 = 3.14).<br \/>\n<strong>Solution: \u00a0 \u00a0\u00a0<\/strong>For the tent : h = 8 m and r = 6 m<br \/>\n\u2113<sup>2<\/sup>\u00a0 = h<sup>2<\/sup>\u00a0+ r<sup>2<\/sup>\u00a0 \u21d2\u00a0\u00a0 \u2113<sup>2<\/sup>\u00a0= 8<sup>2<\/sup>\u00a0+ 6<sup>2<\/sup><br \/>\n= 64 + 36 = 100 and \u00a0\u00a0\u2113 = \\(\\sqrt {100} \\,\\,m\\) = 10m<br \/>\nCurved surface area of the tent = \u03c0r\u2113<br \/>\n= 3.14 \u00d7 6 \u00d7 10 m<sup>2<\/sup>\u00a0\u00a0= 188.4 m<sup>2<\/sup><br \/>\n\u21d2 \u00a0Area of tarpaulin used = 188.4 m<sup>2<\/sup><br \/>\n\u21d2 \u00a0Length of tarpaulin \u00d7 its width = 188.4 m<sup>2<\/sup><br \/>\n\u21d2 \u00a0Length of tarpaulin \u00d7 3 m = 188.4 m<sup>2<\/sup><br \/>\n\u21d2 \u00a0Length of tarpaulin =\u00a0\\(\\frac{{188.4}}{3}m = 62.8m\\)<br \/>\n\u2235\u00a0\u00a0\u00a0 Extra length of tarpaulin required<br \/>\n= 20 cm = 0.2 m<br \/>\n\u2234 \u00a0\u00a0 Total length of tarpaulin required<br \/>\n= 62.8 m + 0.2 m\u00a0 = 63 m<\/p>\n<p><strong>Example 5: \u00a0 \u00a0\u00a0<\/strong>A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is Rs. 12 per m<sup>2<\/sup>, what will be the cost of painting all these cones?<br \/>\n(Use \u03c0 = 3.14 and take\u00a0\\(\\sqrt {1.04}\\)\u00a0= 1.02)<br \/>\n<strong>Solution: \u00a0 \u00a0\u00a0<\/strong>For each cone :<br \/>\nr = \\(\\frac{{40}}{2}\\) cm = 20 cm = 0.2 m and h = 1 m<br \/>\n\u2234\u00a0\u00a0\u00a0 \u2113<sup>2<\/sup>\u00a0= h<sup>2<\/sup>\u00a0+ r<sup>2<\/sup>\u00a0\u21d2 \u00a0\u00a0\u2113<sup>2<\/sup>\u00a0= (1)<sup>2<\/sup>\u00a0+ (0.2)<sup>2<\/sup><br \/>\n= 1 + 0.04 = 1.04<br \/>\n\u21d2\u00a0\u00a0 \u2113 =\u00a0\\(\\sqrt {1.04}\\)m = 1.02 m<br \/>\nC.S.A of each cone = \u03c0r\u2113<br \/>\n= 3.14 \u00d7 0.2 \u00d7 1.02 m<sup>2<\/sup> = 0.64056 m<sup>2<\/sup><br \/>\n\u21d2\u00a0\u00a0 C.S.A of 50 cones = 50 \u00d7 0.64056 m<sup>2<\/sup><br \/>\n= 30.028 m<sup>2<\/sup>\u00a0= Area to be painted<br \/>\n\u2235 The cost of painting is Rs. 12 per m<sup>2<\/sup><br \/>\n\u2234 The total cost of painting outer sides of 50 cones<br \/>\n= 30.028 \u00d7 Rs. 12 = Rs. 384.34<\/p>\n<p><strong>Example 6: \u00a0 \u00a0\u00a0<\/strong>The radius and the slant height of a cone are in the ratio 3 : 5. If its curved surface area is 2310 cm2, find its height.<br \/>\n<strong>Solution: \u00a0 \u00a0\u00a0<\/strong>Given : r : \u2113 = 3 : 5<br \/>\n\u21d2\u00a0\u00a0 if r = 3x cm, \u2113 = 5x cm<br \/>\nC.S.A. = \u03c0r\u2113 \u00a0\u00a0\u21d2 \u00a0\\(\\frac{22}{7}\\)\u00a0\u00d7 3x \u00d7 5x = 2310<br \/>\n\u21d2\u00a0\u00a0 x<sup>2<\/sup>\u00a0= \\(\\frac{{2310 \\times 7}}{{22 \\times 3 \\times 5}} = 49\\)\u00a0\u21d2 \u00a0\u00a0\u00a0x = 7<br \/>\n\u2234\u00a0\u00a0\u00a0 r = 3x = 3 \u00d7 7 cm = 21 cm<br \/>\nand \u2113 = 5x = 5 \u00d7 7 cm = 35 cm,<br \/>\n\u2113<sup>2<\/sup>\u00a0= h<sup>2<\/sup>\u00a0+ r<sup>2<\/sup>\u00a0\u21d2 \u00a0h<sup>2<\/sup>\u00a0= 35<sup>2<\/sup>\u00a0\u201321<sup>2<\/sup><br \/>\n= 1225 \u2013 441 = 784<br \/>\n\u2234\u00a0\u00a0\u00a0 Height (h) = \\(\\sqrt {784} cm\\)\u00a0= 28 cm<\/p>\n<p><strong>Example 7: \u00a0 \u00a0\u00a0<\/strong>A circus tent is in the shape of a cylinder, upto a height of 8 m, surmounted by a cone of the same radius 28 m. If the total height of the tent is 13 m, find:<br \/>\n(i)\u00a0\u00a0 total inner curved surface area of the tent.<br \/>\n(ii)\u00a0 cost of painting its inner surface at the rate of Rs. 3.50 per m<sup>2<\/sup>.<br \/>\n<strong>Solution: \u00a0 \u00a0\u00a0<\/strong>According to the given statement, the rough sketch of the circus tent will be as shown:<br \/>\n(i)\u00a0\u00a0 For the cylindrical portion :<br \/>\nr = 28 and h = 8 m<br \/>\n\u2234\u00a0\u00a0\u00a0 Curved surface area = 2\u03c0rh<br \/>\n= 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 28 \u00d7 8 m<sup>2<\/sup>\u00a0= 1408 m<sup>2<\/sup><br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-120374 aligncenter\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/04\/What-is-the-Area-of-a-Right-Circular-Cone-1.png\" alt=\"What is the Area of a Right Circular Cone 1\" width=\"196\" height=\"209\" \/> For conical portion :<br \/>\nr = 28 m and h = 13 m \u2013 8 m = 5 m<br \/>\n\u2234 \u00a0\u00a0 \u2113<sup>2<\/sup>\u00a0= h<sup>2<\/sup>\u00a0+ r<sup>2<\/sup>\u00a0\u21d2 \u00a0\u00a0\u2113<sup>2<\/sup>\u00a0= 5<sup>2<\/sup>\u00a0+ 28<sup>2<\/sup>\u00a0= 809<br \/>\n\u21d2\u00a0\u00a0 \u2113 = \\(\\sqrt {809}\\)m = 28.4 m<br \/>\n\u2234\u00a0\u00a0\u00a0 Curved surface area = \u03c0r\u2113<br \/>\n= \u00a0\\(\\frac{22}{7}\\) \u00d7 28 \u00d7 28.4 m<sup>2<\/sup>\u00a0\u00a0= 2499.2 m<sup>2<\/sup><br \/>\n\u2234\u00a0\u00a0\u00a0 Total inner curved surface area of the tent.<br \/>\n= \u00a0\u00a0 C.S.A. of cylindrical portion + C.S.A. of the conical portion<br \/>\n1408 m<sup>2<\/sup>\u00a0+ 2499.2 m<sup>2<\/sup>\u00a0= 3907.2 m<sup>2<\/sup><br \/>\n(ii)\u00a0 Cost of painting the inner surface<br \/>\n= 3907.2 \u00d7 Rs. 3.50 = Rs. 13675.20<\/p>\n<p><strong>Example 8: \u00a0 \u00a0\u00a0<\/strong>The height of a cone is 30 cm and its volume is 3140 cm<sup>3<\/sup>. Taking \u03c0 = 3.14, find :<br \/>\n(i) radius of the base.<br \/>\n(ii) area of the base.<br \/>\n<strong>Solution: \u00a0 \u00a0\u00a0<\/strong>(i)\u00a0\u00a0 Given : h = 30 cm and volume = 3140 cm<sup>3<\/sup><br \/>\nVolume =\u00a0\\(\\frac{1}{3}\\pi {r^2}h\\)<br \/>\n\u21d2 \u00a0 3140 = \\(\\frac{1}{3} \\times 3.14\\)\u00a0\u00d7 r<sup>2<\/sup>\u00a0\u00d7 30<br \/>\n\u21d2 \u00a0 r<sup>2<\/sup> = \\(\\frac{{3140 \\times 3}}{{3.14 \\times 30}} = 100\\) and r = 10cm<br \/>\n(ii)\u00a0 Area of the base = \u03c0r<sup>2<\/sup><br \/>\n= 3.14 \u00d7 10<sup>2<\/sup>\u00a0cm<sup>2<\/sup>\u00a0= 314 cm<sup>2<\/sup><br \/>\nAlternative Method :<br \/>\n\\(\\frac{1}{3}\\) \u00d7\u00a0area of base \u00d7 height = volume<br \/>\n\u21d2\u00a0 \u00a0\\(\\frac{1}{3}\\) \u00d7 area of base \u00d7 30 = 3140<br \/>\n\u21d2\u00a0 Area of base =\u00a0\\(\\frac{{3140 \\times 3}}{{30}}c{m^2}\\)<br \/>\n= 314 cm<sup>2<\/sup><\/p>\n<p><strong>Example 9: \u00a0 \u00a0\u00a0<\/strong>A right triangle ABC has sides 5 cm, 12 cm and 13 cm. Find the :<br \/>\n(i)\u00a0\u00a0 volume of solid obtained by revolving\u00a0\u2206 ABC about the side 12 cm.<br \/>\n(ii)\u00a0 volume of solid obtained by revolving\u00a0\u2206 ABC about side 5 cm.<br \/>\n(iii) difference between the volumes of the solids obtained in step (i) and step (ii).<br \/>\n<strong>Solution: \u00a0 \u00a0\u00a0<\/strong>\u2235\u00a0\u00a0\u00a0 5<sup>2<\/sup>\u00a0+ 12<sup>2<\/sup>\u00a0= 13<sup>2<\/sup>\u00a0\u21d2 \u00a0Angle opp. to 13 cm is right angle<br \/>\n(i)\u00a0\u00a0 When the \u2206 is revolved about the side of<br \/>\n12 cm, for the cone formed :<br \/>\nh = 12 cm and r = 5.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-120377 aligncenter\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/04\/What-is-the-Area-of-a-Right-Circular-Cone-2.png\" alt=\"What is the Area of a Right Circular Cone 2\" width=\"143\" height=\"144\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/04\/What-is-the-Area-of-a-Right-Circular-Cone-2.png 143w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/04\/What-is-the-Area-of-a-Right-Circular-Cone-2-100x100.png 100w\" sizes=\"(max-width: 143px) 100vw, 143px\" \/> \u2234 \u00a0\u00a0 Volume of solid obtained =\u00a0\\(\\frac{1}{3}\\pi {r^2}h\\)<br \/>\n=\u00a0\\(\\frac{1}{3}\\) \u00d7 \\(\\frac{22}{7}\\)\u00a0\u00d7 5 \u00d7 5 \u00d7 12 cm<sup>3<\/sup><br \/>\n= 314.29 cm<sup>3<\/sup><br \/>\n(ii)\u00a0 When the \u2206 is revolved about the side of 5 cm, for the cone formed :<br \/>\nh = 5 cm, and r = 12 cm.<br \/>\n\u2234 \u00a0\u00a0 Volume of solid obtained =\u00a0\\(\\frac{1}{3}\\pi {r^2}h\\)<br \/>\n= \\(\\frac{1}{3}\\) \u00d7 \\(\\frac{22}{7}\\)\u00a0\u00d7 12\u00a0\u00d7 12 \u00d7 5 cm<sup>3<\/sup>\u00a0= 754.29 cm<sup>3<\/sup><br \/>\n(iii) Required difference<br \/>\n= 754.29 cm<sup>3<\/sup>\u00a0\u2013 314.29 cm<sup>3<\/sup>\u00a0= 440 cm<sup>3<\/sup><\/p>\n<p><strong>Example 10: \u00a0 \u00a0\u00a0<\/strong>The volume of a right circular cone is 9856 cm<sup>3<\/sup>. If the diameter of the base is 28 cm, find:<br \/>\n(i) height of the cone.<br \/>\n(ii) slant height of the cone<br \/>\n(iii) curved surface area of the cone.<br \/>\n<strong>Solution: \u00a0 \u00a0\u00a0<\/strong>Given : volume of cone = 9856 cm<sup>3<\/sup><br \/>\nand radius (r) = \\(\\frac{28}{2}\\) cm = 14cm<br \/>\n(i)\u00a0\u00a0 Volume =\u00a0\\(\\frac{1}{3}\\)\u03c0r<sup>2<\/sup>h \u00a0 \u21d2\u00a0 9856 = \\(\\frac{1}{3}\\) \u00d7 \\(\\frac{22}{7}\\) \u00d7 14 \u00d7 14 \u00d7 h<br \/>\n\u21d2 \u00a0 \u00a0\u00a0h = \\(\\frac{{9856 \\times 3 \\times 7}}{{22 \\times 14 \\times 14}}cm\\)\u00a0= 48 cm<br \/>\n(ii)\u00a0 \u00a0\u2113<sup>2<\/sup>\u00a0 = h<sup>2<\/sup>\u00a0+ r<sup>2\u00a0\u00a0<\/sup>\u21d2\u00a0\u00a0 \u2113<sup>2<\/sup>\u00a0= 48<sup>2<\/sup>\u00a0+ 14<sup>2<\/sup><br \/>\n= 2304 + 196 = 2500<br \/>\n\u21d2\u00a0\u00a0 \u2113 = \\(\\sqrt {2500}\\)cm = 50 cm<br \/>\n\u2234 \u00a0\u00a0 Slant height = 50 cm<br \/>\n(iii) Curved surface area = \u03c0r\u2113<br \/>\n= \\(\\frac{22}{7}\\)\u00a0\u00d7 14 \u00d7 50 cm<sup>2<\/sup>\u00a0= 2200 cm<sup>2<\/sup><\/p>\n<p><strong>Example 11: \u00a0 \u00a0\u00a0<\/strong>A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.<br \/>\n<strong>Solution: \u00a0 \u00a0\u00a0<\/strong>For the conical heap :<br \/>\nRadius (r) = \\(\\frac{{10.5}}{2}m\\) = 5.25 m<br \/>\nand height (h) = 3m.<br \/>\n\u2234\u00a0\u00a0\u00a0 Volume =\u00a0\\(\\frac{1}{3}\\)\u03c0r<sup>2<\/sup>h<br \/>\n= \\(\\frac{1}{3} \\times \\frac{{22}}{7} \\times 5.25\\) \u00d7 3 m<sup>3<\/sup>\u00a0= 86.625 m<sup>3<\/sup><br \/>\nNow, \u2113<sup>2<\/sup>\u00a0= h<sup>2<\/sup>\u00a0+ r<sup>2<\/sup>\u00a0 \u21d2\u00a0 \u2113<sup>2<\/sup>\u00a0= (3)<sup>2<\/sup>\u00a0+ (5.25)<sup>2<\/sup><br \/>\n= 9 + 27.5625 = 36.5625<br \/>\n\u21d2\u00a0\u00a0 \u2113 = \\(\\sqrt {36.5625}\\)m = 6.047 m<br \/>\n\u2234\u00a0\u00a0\u00a0 Area of canvas required<br \/>\n= curved surface area of the conical heap<br \/>\n= \u03c0r\u2113 = \\(\\frac{{22}}{7} \\times 5.25 \\times 6.047{m^2}\\)\u00a0= 99.7755 m<sup>2<\/sup><\/p>\n<p><strong>Example 12: \u00a0 \u00a0\u00a0<\/strong>A cylinder and a cone have same base area. But the volume of cylinder is twice the volume of cone. Find the ratio between their heights.<br \/>\n<strong>Solution: \u00a0 \u00a0\u00a0<\/strong>Since, the base areas of the cylinder and the cone are the same.<br \/>\n\u21d2\u00a0\u00a0 their radius are equal (same).<br \/>\nLet the radius of their base be r and their heights be h<sub>1<\/sub>\u00a0and h<sub>2<\/sub>\u00a0respectively.<br \/>\nClearly, volume of the cylinder = \\(\\frac{1}{3}\\)\u03c0r<sup>2<\/sup>h<sub>1<\/sub><br \/>\nand, volume of the cone = \\(\\frac{1}{3}\\)\u03c0r<sup>2<\/sup>h<sub>2<\/sub><br \/>\nGiven :<br \/>\nVolume of cylinder = 2 \u00d7 \\(\\frac{1}{3}\\) volume of cone<br \/>\n\u21d2 \u00a0 \u03c0r<sup>2<\/sup>h<sub>1<\/sub>\u00a0= 2 \u00d7 \u03c0r<sup>2<\/sup>h<sub>2<\/sub><br \/>\n\u21d2\u00a0\u00a0 h<sub>1<\/sub>\u00a0=\u00a0\\(\\frac{2}{3}\\)h<sub>2<\/sub><br \/>\n\u21d2\u00a0\\(\\frac{{{h_1}}}{{{h_2}}} = \\frac{2}{3}\\)<br \/>\ni.e., h<sub>1<\/sub>\u00a0: h<sub>2<\/sub>\u00a0= 2 : 3<\/p>\n","protected":false},"excerpt":{"rendered":"<p>What is the Area of a Right Circular Cone If r, h and \u2113 denote respectively the radius of base, height and slant height of a right circular cone, then- \u2113 2\u00a0= r2\u00a0+ h2 Area of base = \u03c0r2 Curved (lateral) surface area = \u03c0r\u2113 Total surface area = \u03c0r (\u2113 + r) Volume = [&hellip;]<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_genesis_hide_title":false,"_genesis_hide_breadcrumbs":false,"_genesis_hide_singular_image":false,"_genesis_hide_footer_widgets":false,"_genesis_custom_body_class":"","_genesis_custom_post_class":"","_genesis_layout":"","footnotes":""},"categories":[5],"tags":[1258,1259,1260,1251],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v17.8 (Yoast SEO v17.8) - 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