<\/strong><\/h2>\nVOLUME OF RIGHT CIRCULAR CONE<\/strong><\/h2>\n<\/strong><\/span><\/h3>\nMEANING OF RIGHT CIRCULAR CONE<\/strong>
<\/span><\/h3>\nA cone is a 3-dimensional figure which has two ends, at one end there is a circular plane(base) and the other end is pointed. When the axis of a cone meets the vertex and joins the midpoint of the circular plane(base) perpendicularly then it is known as the right circular cone.<\/span><\/p>\n![\"\"](\"https:\/\/eistudymaterial.s3.amazonaws.com\/rightcircularcone1-300x254.jpg\")
<\/span><\/p>\n\u00a0<\/span><\/p>\nPROPERTIES OF RIGHT CIRCULAR CONE<\/strong>\u00a0 <\/span><\/h3>\n1. A right circular cone is formed when a right-angled triangle is rotated about its perpendicular and the perpendicular becomes the axis of the cone.<\/span><\/p>\n\n- The axis of the right circular cone is perpendicular to its circular base. Due to this, the axis of the cone overlaps its height.<\/span><\/li>\n
- When the vertex and any two points of the circular plane(base) are joined, an isosceles triangle is formed.<\/span><\/li>\n<\/ol>\n
<\/span><\/p>\n<\/span><\/p>\nVOLUME AND FORMULA<\/strong><\/h3>\nThe volume of a right circular cone is the maximum amount of space it can contain. In simple words, it is the capacity of the cone.\u00a0<\/span><\/p>\nVolume of a cone = \u2153 rd volume of a cylinder.<\/span><\/p>\n<\/span>Formula:<\/strong><\/p>\nVolume =\\frac{1}{3} \\pi r^{2}h<\/span> cubic units<\/span><\/p>\nwhere r = radius of the circular base & <\/span>h = height of the cone.<\/span><\/p>\n<\/span><\/p>\n![\"\"](\"https:\/\/eistudymaterial.s3.amazonaws.com\/rightcircularcone2-300x237.jpg\")
<\/p>\n
As we know, the axis of the cone is perpendicular to the base,\u00a0<\/span>therefore, by using Pythagoras\u2019s theorem we get,<\/span><\/p>\nl^{2}=r^{2}+h^{2}<\/span><\/span><\/p>\n
l=\\sqrt{{r}^{2}+h^{2}}<\/span><\/span><\/p>\nThis is the formula for finding the slant height of the cone. If we know any of the two parameters, we can easily find the third parameter using this formula.
<\/span><\/p>\nDERIVATION OF VOLUME OF CONE<\/strong><\/h2>\n<\/strong><\/p>\nSTEP 1<\/strong>
Take a cylindrical vessel and a conical flask of the same radius and height.<\/p>\nSTEP 2<\/strong>
Fill the conical flask fully with water and pour this water into the cylindrical vessel. We will observe that the cylindrical vessel is not filled.<\/p>\nSTEP 3<\/strong>
On repeating this process we will see that there is again some volume left in the cylinder.<\/p>\nSTEP 4<\/strong>
When the same is done the third time we will observe that the cylindrical vessel is filled.<\/p>\n <\/p>\n
![\"\"](\"https:\/\/eistudymaterial.s3.amazonaws.com\/rightcircularcone3-300x104.jpg\")
<\/p>\n
Hence, the volume of a cone = 1\u20443 volume of a\u00a0 cylinder.<\/span><\/span><\/p>\nVolume of cone =\\frac{1}{3} \\pi r^{2} h<\/span> <\/span>cubic units<\/span><\/p>\nwhere r = radius of the circular <\/span>base & h = height of the cone.<\/span><\/p>\n <\/p>\n
ILLUSTRATION<\/strong><\/strong><\/strong><\/h2>\n<\/strong><\/p>\nQ1.<\/strong> If the height of the cone is 14 cm and the diameter is 18 cm. Find the volume of the cone.<\/span><\/p>\nSolution:<\/strong><\/p>\nradius(r) = Diameter\/2<\/span><\/p>\n\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 = 18\/2 cm<\/span><\/p>\n\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 = 9 cm<\/span><\/p>\nVolume of cone<\/span><\/p>\n=\\frac{1}{3}\\pi r^{2} h<\/span><\/span><\/p>\n<\/span><\/p>\n=\\frac{1}{3}\\times \\pi \\times 9^{2}\\times 14 <\/span>\n <\/p>\n=\\frac{1}{3}\\times \\frac{22}{7} \\times 9^{2}\\times 14 <\/span>\n <\/p>\n=1188 \\mathrm{~cm}^{3}<\/span>\n <\/p>\n
Q2.<\/strong> The slant height of the cone is 13 cm and the diameter is 10 cm. Find the volume of the<\/span> cone.<\/span><\/p>\n![\"\"](\"https:\/\/eistudymaterial.s3.amazonaws.com\/rightcircularcone4-300x258.jpg\")
<\/p>\n
Solution:<\/strong><\/p>\n radius(r) = Diameter\/2<\/span><\/p>\n\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 = 10\/2 cm<\/span><\/p>\n\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 = 5 cm<\/span><\/p>\nWe know that<\/span><\/p>\n\\mathrm{l}^{2} =\\mathrm{r}^{2}+\\mathrm{h}^{2} <\/span> <\/span>( \\mathrm{l}<\/span>= slant height)<\/span><\/p>\n13^{2} =5^{2}+\\mathrm{h}^{2} <\/span>
\\Rightarrow \\mathrm{~h}^{2} =169-25 <\/span>
\\Rightarrow \\mathrm{~h} =\\sqrt{144} <\/span>
\\therefore \\mathrm{~h} =12 \\mathrm{~cm}<\/span>
<\/span><\/p>\nVolume of cone<\/span><\/p>\n=\\frac{1}{3}\\pi r^{2}\u00a0 h <\/span>
=\\frac{1}{3}\\times 3.14 \\times 5^{2}\\times 12<\/span>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(\u03c0 = 3.14)
=314 \\mathrm{~cm}^{3}<\/span><\/span><\/p>\n <\/p>\n
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\nSurface Area and Volume<\/a><\/div>\nUnits of Volume<\/a><\/div>\n
<\/strong><\/span><\/h3>\nMEANING OF RIGHT CIRCULAR CONE<\/strong>
<\/span><\/h3>\nA cone is a 3-dimensional figure which has two ends, at one end there is a circular plane(base) and the other end is pointed. When the axis of a cone meets the vertex and joins the midpoint of the circular plane(base) perpendicularly then it is known as the right circular cone.<\/span><\/p>\n<\/span><\/p>\n\u00a0<\/span><\/p>\nPROPERTIES OF RIGHT CIRCULAR CONE<\/strong>\u00a0 <\/span><\/h3>\n1. A right circular cone is formed when a right-angled triangle is rotated about its perpendicular and the perpendicular becomes the axis of the cone.<\/span><\/p>\n\n- The axis of the right circular cone is perpendicular to its circular base. Due to this, the axis of the cone overlaps its height.<\/span><\/li>\n
- When the vertex and any two points of the circular plane(base) are joined, an isosceles triangle is formed.<\/span><\/li>\n<\/ol>\n
<\/span><\/p>\n<\/span><\/p>\nVOLUME AND FORMULA<\/strong><\/h3>\nThe volume of a right circular cone is the maximum amount of space it can contain. In simple words, it is the capacity of the cone.\u00a0<\/span><\/p>\nVolume of a cone = \u2153 rd volume of a cylinder.<\/span><\/p>\n<\/span>Formula:<\/strong><\/p>\nVolume =\\frac{1}{3} \\pi r^{2}h<\/span> cubic units<\/span><\/p>\nwhere r = radius of the circular base & <\/span>h = height of the cone.<\/span><\/p>\n<\/span><\/p>\n<\/p>\n
As we know, the axis of the cone is perpendicular to the base,\u00a0<\/span>therefore, by using Pythagoras\u2019s theorem we get,<\/span><\/p>\nl^{2}=r^{2}+h^{2}<\/span><\/span><\/p>\n
l=\\sqrt{{r}^{2}+h^{2}}<\/span><\/span><\/p>\nThis is the formula for finding the slant height of the cone. If we know any of the two parameters, we can easily find the third parameter using this formula.
<\/span><\/p>\nDERIVATION OF VOLUME OF CONE<\/strong><\/h2>\n<\/strong><\/p>\nSTEP 1<\/strong>
Take a cylindrical vessel and a conical flask of the same radius and height.<\/p>\nSTEP 2<\/strong>
Fill the conical flask fully with water and pour this water into the cylindrical vessel. We will observe that the cylindrical vessel is not filled.<\/p>\nSTEP 3<\/strong>
On repeating this process we will see that there is again some volume left in the cylinder.<\/p>\nSTEP 4<\/strong>
When the same is done the third time we will observe that the cylindrical vessel is filled.<\/p>\n <\/p>\n
<\/p>\n
Hence, the volume of a cone = 1\u20443 volume of a\u00a0 cylinder.<\/span><\/span><\/p>\nVolume of cone =\\frac{1}{3} \\pi r^{2} h<\/span> <\/span>cubic units<\/span><\/p>\nwhere r = radius of the circular <\/span>base & h = height of the cone.<\/span><\/p>\n <\/p>\n
ILLUSTRATION<\/strong><\/strong><\/strong><\/h2>\n<\/strong><\/p>\nQ1.<\/strong> If the height of the cone is 14 cm and the diameter is 18 cm. Find the volume of the cone.<\/span><\/p>\nSolution:<\/strong><\/p>\nradius(r) = Diameter\/2<\/span><\/p>\n\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 = 18\/2 cm<\/span><\/p>\n\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 = 9 cm<\/span><\/p>\nVolume of cone<\/span><\/p>\n=\\frac{1}{3}\\pi r^{2} h<\/span><\/span><\/p>\n<\/span><\/p>\n=\\frac{1}{3}\\times \\pi \\times 9^{2}\\times 14 <\/span>\n <\/p>\n=\\frac{1}{3}\\times \\frac{22}{7} \\times 9^{2}\\times 14 <\/span>\n <\/p>\n=1188 \\mathrm{~cm}^{3}<\/span>\n <\/p>\n
Q2.<\/strong> The slant height of the cone is 13 cm and the diameter is 10 cm. Find the volume of the<\/span> cone.<\/span><\/p>\n<\/p>\n
Solution:<\/strong><\/p>\n radius(r) = Diameter\/2<\/span><\/p>\n\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 = 10\/2 cm<\/span><\/p>\n\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 = 5 cm<\/span><\/p>\nWe know that<\/span><\/p>\n\\mathrm{l}^{2} =\\mathrm{r}^{2}+\\mathrm{h}^{2} <\/span> <\/span>( \\mathrm{l}<\/span>= slant height)<\/span><\/p>\n13^{2} =5^{2}+\\mathrm{h}^{2} <\/span>
\\Rightarrow \\mathrm{~h}^{2} =169-25 <\/span>
\\Rightarrow \\mathrm{~h} =\\sqrt{144} <\/span>
\\therefore \\mathrm{~h} =12 \\mathrm{~cm}<\/span>
<\/span><\/p>\nVolume of cone<\/span><\/p>\n=\\frac{1}{3}\\pi r^{2}\u00a0 h <\/span>
=\\frac{1}{3}\\times 3.14 \\times 5^{2}\\times 12<\/span>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(\u03c0 = 3.14)
=314 \\mathrm{~cm}^{3}<\/span><\/span><\/p>\n <\/p>\n
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\nSurface Area and Volume<\/a><\/div>\nUnits of Volume<\/a><\/div>\n
<\/span><\/h3>\n
A cone is a 3-dimensional figure which has two ends, at one end there is a circular plane(base) and the other end is pointed. When the axis of a cone meets the vertex and joins the midpoint of the circular plane(base) perpendicularly then it is known as the right circular cone.<\/span><\/p>\n \u00a0<\/span><\/p>\n 1. A right circular cone is formed when a right-angled triangle is rotated about its perpendicular and the perpendicular becomes the axis of the cone.<\/span><\/p>\n <\/span><\/p>\n <\/span><\/p>\n The volume of a right circular cone is the maximum amount of space it can contain. In simple words, it is the capacity of the cone.\u00a0<\/span><\/p>\n Volume of a cone = \u2153 rd volume of a cylinder.<\/span><\/p>\n <\/span>Formula:<\/strong><\/p>\n Volume =\\frac{1}{3} \\pi r^{2}h<\/span> cubic units<\/span><\/p>\n where r = radius of the circular base & <\/span>h = height of the cone.<\/span><\/p>\n <\/span><\/p>\n As we know, the axis of the cone is perpendicular to the base,\u00a0<\/span>therefore, by using Pythagoras\u2019s theorem we get,<\/span><\/p>\n l^{2}=r^{2}+h^{2}<\/span><\/span><\/p>\n This is the formula for finding the slant height of the cone. If we know any of the two parameters, we can easily find the third parameter using this formula. <\/strong><\/p>\n STEP 1<\/strong> STEP 2<\/strong> STEP 3<\/strong> STEP 4<\/strong> <\/p>\n Hence, the volume of a cone = 1\u20443 volume of a\u00a0 cylinder.<\/span><\/span><\/p>\n Volume of cone =\\frac{1}{3} \\pi r^{2} h<\/span> <\/span>cubic units<\/span><\/p>\n where r = radius of the circular <\/span>base & h = height of the cone.<\/span><\/p>\n <\/p>\n <\/strong><\/p>\n Q1.<\/strong> If the height of the cone is 14 cm and the diameter is 18 cm. Find the volume of the cone.<\/span><\/p>\n Solution:<\/strong><\/p>\n radius(r) = Diameter\/2<\/span><\/p>\n \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 = 18\/2 cm<\/span><\/p>\n \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 = 9 cm<\/span><\/p>\n Volume of cone<\/span><\/p>\n =\\frac{1}{3}\\pi r^{2} h<\/span><\/span><\/p>\n <\/span><\/p>\n=\\frac{1}{3}\\times \\pi \\times 9^{2}\\times 14 <\/span>\n <\/p>\n=\\frac{1}{3}\\times \\frac{22}{7} \\times 9^{2}\\times 14 <\/span>\n <\/p>\n=1188 \\mathrm{~cm}^{3}<\/span>\n <\/p>\n Q2.<\/strong> The slant height of the cone is 13 cm and the diameter is 10 cm. Find the volume of the<\/span> cone.<\/span><\/p>\n Solution:<\/strong><\/p>\n radius(r) = Diameter\/2<\/span><\/p>\n \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 = 10\/2 cm<\/span><\/p>\n \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 = 5 cm<\/span><\/p>\n We know that<\/span><\/p>\n \\mathrm{l}^{2} =\\mathrm{r}^{2}+\\mathrm{h}^{2} <\/span> <\/span>( \\mathrm{l}<\/span>= slant height)<\/span><\/p>\n 13^{2} =5^{2}+\\mathrm{h}^{2} <\/span> Volume of cone<\/span><\/p>\n =\\frac{1}{3}\\pi r^{2}\u00a0 h <\/span> <\/p>\n [\/et_pb_text][\/et_pb_column][et_pb_column type=”2_5″ module_id=”stickysideR” admin_label=”Column R” _builder_version=”4.10.4″ _module_preset=”default” background_color=”#fdefe0″ custom_padding=”25px|25px|25px|25px|true|true” sticky_position=”top” sticky_offset_top=”-280px” sticky_limit_top=”row” sticky_limit_bottom=”row” sticky_position_tablet=”none” sticky_position_phone=”none” sticky_position_last_edited=”on|desktop” sticky_limit_bottom_tablet=”” sticky_limit_bottom_phone=”” sticky_limit_bottom_last_edited=”on|phone” border_radii=”on|15px|15px|15px|15px” box_shadow_style=”preset3″ global_colors_info=”{}”][et_pb_image src=”https:\/\/eistudymaterial.s3.amazonaws.com\/1080×1080.png” alt=”Free Trial banner” title_text=”Mindspark Free Trial Banner” url=”https:\/\/mindspark.in\/free-trial” align=”center” module_class=”adsimg” _builder_version=”4.11.1″ _module_preset=”default” custom_padding=”||||false|false” border_radii=”on|10px|10px|10px|10px” global_colors_info=”{}” transform_styles__hover_enabled=”on|hover” transform_scale__hover_enabled=”on|hover” transform_translate__hover_enabled=”on|desktop” transform_rotate__hover_enabled=”on|desktop” transform_skew__hover_enabled=”on|desktop” transform_origin__hover_enabled=”on|desktop” transform_scale__hover=”102%|102%”][\/et_pb_image][et_pb_text admin_label=”Explore Other Topics [\/et_pb_text][et_pb_text _builder_version=”4.10.7″ _module_preset=”default” text_line_height=”2.2em” link_font_size=”16px” custom_margin=”||0px||false|false” custom_padding=”10px|15px|10px|28px|true|false” locked=”off” global_colors_info=”{}”]<\/p>\n [\/et_pb_text][et_pb_text admin_label=”Related Concepts [\/et_pb_text][et_pb_text _builder_version=”4.13.1″ _module_preset=”default” text_line_height=”2.2em” link_font_size=”16px” custom_margin=”||0px||false|false” custom_padding=”10px|15px|10px|28px|true|false” hover_enabled=”0″ locked=”off” global_colors_info=”{}” sticky_enabled=”0″]<\/p>\n<\/span><\/p>\nPROPERTIES OF RIGHT CIRCULAR CONE<\/strong>\u00a0 <\/span><\/h3>\n
\n
VOLUME AND FORMULA<\/strong><\/h3>\n
<\/p>\n
l=\\sqrt{{r}^{2}+h^{2}}<\/span><\/span><\/p>\n
<\/span><\/p>\nDERIVATION OF VOLUME OF CONE<\/strong><\/h2>\n
Take a cylindrical vessel and a conical flask of the same radius and height.<\/p>\n
Fill the conical flask fully with water and pour this water into the cylindrical vessel. We will observe that the cylindrical vessel is not filled.<\/p>\n
On repeating this process we will see that there is again some volume left in the cylinder.<\/p>\n
When the same is done the third time we will observe that the cylindrical vessel is filled.<\/p>\n<\/p>\nILLUSTRATION<\/strong><\/strong><\/strong><\/h2>\n
<\/p>\n
\\Rightarrow \\mathrm{~h}^{2} =169-25 <\/span>
\\Rightarrow \\mathrm{~h} =\\sqrt{144} <\/span>
\\therefore \\mathrm{~h} =12 \\mathrm{~cm}<\/span>
<\/span><\/p>\n
=\\frac{1}{3}\\times 3.14 \\times 5^{2}\\times 12<\/span>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(\u03c0 = 3.14)
=314 \\mathrm{~cm}^{3}<\/span><\/span><\/p>\n
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![\"\"](\"https:\/\/eistudymaterial.s3.amazonaws.com\/rightcircularcone5-300x122.jpg\")
<\/p>\n