{"id":3849,"date":"2021-11-15T13:08:51","date_gmt":"2021-11-15T13:08:51","guid":{"rendered":"https:\/\/stgwebsite.mindspark.in\/wordpress\/?page_id=3849"},"modified":"2022-01-03T06:53:34","modified_gmt":"2022-01-03T06:53:34","slug":"lateral-surface-area-with-examples-and-faqs","status":"publish","type":"page","link":"https:\/\/stgwebsite.mindspark.in\/studymaterial\/math-concepts\/lateral-surface-area-with-examples-and-faqs\/","title":{"rendered":"Lateral Surface Area with Examples and FAQs"},"content":{"rendered":"

[et_pb_section fb_built=”1″ admin_label=”Section” module_class=”mainsec” _builder_version=”4.10.4″ _module_preset=”default” background_color=”#e0f2fd” z_index=”1″ custom_padding=”5px||5px||true|false” locked=”off” collapsed=”off” global_colors_info=”{}”][et_pb_row column_structure=”3_5,2_5″ custom_padding_last_edited=”on|phone” _builder_version=”4.10.8″ _module_preset=”default” background_color=”#FFFFFF” width=”100%” max_width=”1310px” custom_padding=”|51px|40px|51px|false|true” custom_padding_tablet=”” custom_padding_phone=”|40px|30px|40px|false|true” border_radii=”on|10px|10px|10px|10px” global_colors_info=”{}”][et_pb_column type=”3_5″ admin_label=”Column L” _builder_version=”4.9.10″ _module_preset=”default” global_colors_info=”{}”][et_pb_text admin_label=”Acute Angles
\n” _builder_version=”4.11.3″ _module_preset=”default” header_font=”|700|||||||” header_text_align=”left” header_font_size=”50px” header_line_height=”1.18em” custom_padding=”|0px||4px|false|false” header_font_size_tablet=”” header_font_size_phone=”35px” header_font_size_last_edited=”on|phone” global_colors_info=”{}”]<\/p>\n

Lateral Surface Area with Examples and FAQs<\/b><\/h1>\n

[\/et_pb_text][et_pb_text admin_label=”Text” _builder_version=”4.11.3″ _module_preset=”default” text_font_size=”16px” header_2_font=”|600|||||||” header_2_text_color=”#a01414″ header_3_font=”|600|||||||” custom_padding=”15px|15px|1px|4px|false|false” global_colors_info=”{}”]<\/p>\n

LATERAL SURFACE AREA<\/b><\/h2>\n

The lateral surface area is the area of any solid which gives the area occupied by its sides, it excludes the top and bottom bases. It is also termed as the curved surface area for cylindrical, spherical and conical objects, which is the area of the curved surface or the sides of the three-dimensional figures. The methods to find out the lateral surface area of solid figures like cube, cuboid, cone, sphere, hemisphere, cylinder, etc., are all different but can be easily determined if the basic dimensions of important parts are known<\/span>.\u00a0\u00a0\u00a0<\/span><\/p>\n

<\/span><\/p>\n

<\/span><\/p>\n

<\/span><\/p>\n

FORMULA FOR LATERAL SURFACE AREA OF DIFFERENT SOLIDS<\/b><\/h2>\n

<\/b><\/p>\n

1.Cube<\/strong><\/p>\n

<\/h2>\n

[\/et_pb_text][et_pb_image src=”https:\/\/eistudymaterial.s3.amazonaws.com\/CUBE.png” title_text=”CUBE” align=”center” _builder_version=”4.11.3″ _module_preset=”default” min_height=”226px” custom_margin=”-35px||||false|false” global_colors_info=”{}”][\/et_pb_image][et_pb_text content_tablet=”” content_phone=”<\/p>\n

Cube<\/strong><\/p>\n

The Lateral Surface area of a cube with side length %22a%22 = %91latex%934 a^{2}%91\/latex%93<\/strong><\/p>\n

2. Cuboid<\/strong><\/h5>\n

%22%22<\/strong><\/p>\n

The Length of a cuboid is ‘I’ its breadth is ‘b’ and it’s height is ‘h’<\/strong><\/p>\n

The lateral surface area of the cuboid %91latex%93=2(1+b) h%91\/latex%93<\/strong><\/p>\n

<\/strong><\/p>\n

3. Cylinder<\/strong><\/p>\n

%22%22<\/strong><\/p>\n

Cylinder with height h and base radius r<\/strong><\/p>\n

The radius of the base cylinder of the cylinder is ‘r’ and its the height is ‘h’<\/strong><\/p>\n

The lateral surface area of cylinder %91latex%93=2 pi mathrm{rh}%91\/latex%93<\/strong><\/p>\n

<\/strong><\/p>\n

4. Cone<\/strong><\/p>\n

<\/strong><\/p>\n

%22%22<\/strong><\/p>\n

The radius of the base of the cone is \u2018r\u2019, the cone\u2019s height is \u2018h\u2019<\/span><\/p>\n

The slant height %91latex%931=sqrt{h^{2}+r^{2}}%91\/latex%93<\/span><\/p>\n

The Lateral Surface Of Cone %91latex%93=pi mathrm{rl}%91\/latex%93<\/span><\/p>\n

<\/span><\/p>\n

<\/span><\/p>\n

5. Sphere<\/strong><\/span><\/p>\n

%22%22<\/span><\/p>\n

The Radius of Sphere is ‘r’<\/span><\/p>\n

The Lateral Surface Of Sphere %91latex%93=4 pi mathrm{r}^{2}%91\/latex%93<\/span><\/p>\n

<\/span><\/p>\n

6. Hemisphere<\/strong><\/h5>\n

%22%22<\/strong><\/p>\n

The Radius of hemisphere is ‘r’<\/strong><\/p>\n

The Lateral Surface Of Hemisphere %91latex%93=2 pi mathrm{r}^{2}%91\/latex%93<\/span><\/strong><\/p>\n

<\/span><\/strong><\/h2>\n

<\/span><\/strong><\/p>\n

Examples <\/span><\/strong><\/h2>\n

<\/span><\/strong><\/p>\n

Example 1:<\/strong> The diameter of the base and height of a cylinder is 14 units and 20 units respectively. Calculate the lateral surface area.
use%91latex%93pi=frac{22}{7}%91\/latex%93<\/span><\/strong><\/p>\n

Solution: <\/strong><\/p>\n

Diameter of the base of the cylinder = 14 units<\/span><\/p>\n

\u21d2<\/span> radius, r =14\/2 units =7 units<\/span><\/p>\n

Height of cylinder, h = 20 units<\/span><\/p>\n

Lateral area of the cylinder <\/span><\/p>\n

%91latex%93=2 pi mathrm{rh}%91\/latex%93<\/span><\/p>\n

%91latex%93=2 times frac{22}{7} times 7 times 20%91\/latex%93<\/span><\/p>\n

%91latex%93=2 times 22 times 20 text { square units }%91\/latex%93<\/span><\/span><\/p>\n

%91latex%93=880 text { square units }%91\/latex%93<\/span><\/p>\n

Hence, the lateral surface area of the given cylinder is 880 square units. <\/span><\/p>\n

<\/span><\/p>\n

Example 2 : <\/strong><\/p>\n

The radius of a cone measures 5 cm and its height is 12 cm. Determine the LSA of the cone. %91use value of \u03c0 =3.14%93<\/span><\/p>\n

Solution:<\/strong><\/p>\n

Radius of the cone r = 5cm<\/p>\n

Height of the cone h = 12cm<\/p>\n

Slant height %91latex%93 l=sqrt{r^{2}+h^{2}}%91\/latex%93<\/p>\n

<\/span>%91latex%93therefore mathrm{l}=sqrt{5^{2}+12^{2}}=sqrt{25+144}=sqrt{169}=13 mathrm{~cm}%91\/latex%93<\/p>\n

Hence Lateral Surface area of cone <\/p>\n

%91latex%93=pi mathrm{rl}%91\/latex%93<\/p>\n

\n

%91latex%93=3.14 times 5 times 13 mathrm{~cm}^{2}%91\/latex%93<\/p>\n

\n

= 204.1 %91latex%93{~cm}^{2}%91\/latex%93<\/p>\n

Therefore, the LSA of the cone is 204.1 %91latex%93{~cm}^{2}%91\/latex%93<\/span><\/p>\n<\/p>\n

Example 3<\/strong>:<\/strong> Find the lateral area of a cuboid whose length is 12 units, breadth is 8 units and height is 5 units.<\/span><\/p>\n

Solution:<\/strong><\/p>\n

Length of cuboid, l = 12 units.<\/span><\/p>\n

Breadth of cuboid, b = 8 units.<\/span><\/p>\n

Height of cuboid, h = 5 units.<\/span><\/p>\n

Therefore Lateral <\/span><\/p>\n

” content_last_edited=”on|phone” admin_label=”Text” _builder_version=”4.11.3″ _module_preset=”default” text_font_size=”16px” header_2_font=”|600|||||||” header_2_text_color=”#a01414″ header_3_font=”|600|||||||” header_3_text_color=”#898989″ custom_margin=”-40px||||false|false” custom_padding=”0px|15px|1px|4px|false|false” global_colors_info=”{}”]<\/p>\n

The Lateral Surface area of a cube with side length “a” = 4 a^{2}<\/span><\/p>\n

2. Cuboid<\/strong><\/h5>\n

\"\"<\/strong><\/p>\n

The Length of a cuboid is ‘l’ its breadth is ‘b’ and its height is ‘h’<\/p>\n

The lateral surface area of the cuboid =2(l+b) h<\/span><\/p>\n

<\/strong><\/p>\n

3. Cylinder<\/strong><\/p>\n

\"\"<\/strong><\/p>\n

The radius of the base cylinder of the cylinder is ‘r’ and its height is ‘h’<\/p>\n

The lateral surface area of cylinder =2 \\pi \\mathrm{rh}<\/span><\/p>\n

<\/strong><\/p>\n

4. Cone<\/strong><\/p>\n

<\/strong><\/p>\n

\"\"<\/strong><\/p>\n

The radius of the base of the cone is \u2018r\u2019, the cone\u2019s height is \u2018h\u2019<\/span><\/p>\n

The slant height l=\\sqrt{h^{2}+r^{2}}<\/span><\/span><\/p>\n

The Lateral Surface Area of Cone =\\pi \\mathrm{rl}<\/span><\/span><\/p>\n

<\/span><\/p>\n

<\/span><\/p>\n

5. Sphere<\/strong><\/span><\/p>\n

\"\"<\/span><\/p>\n

The Radius of Sphere is ‘r’<\/span><\/p>\n

The Lateral Surface Area of Sphere =4 \\pi \\mathrm{r}^{2}<\/span><\/span><\/p>\n

<\/span><\/p>\n

6. Hemisphere<\/strong><\/h5>\n

\"\"<\/strong><\/p>\n

The Radius of the hemisphere is ‘r’<\/p>\n

The Lateral Surface Area of\u00a0 Hemisphere =2 \\pi \\mathrm{r}^{2}<\/span><\/span><\/strong><\/p>\n

<\/span><\/strong><\/h2>\n

<\/span><\/strong><\/p>\n

Examples\u00a0<\/span><\/strong><\/h2>\n

<\/span><\/strong><\/p>\n

Example 1:<\/strong> The diameter of the base and height of a cylinder is 14 units and 20 units respectively. Calculate the lateral surface area.
use\\pi=\\frac{22}{7}<\/span><\/span><\/strong><\/p>\n

Solution:\u00a0\u00a0\u00a0\u00a0\u00a0<\/strong><\/p>\n

Diameter of the base of the cylinder = 14 units<\/span><\/p>\n

\u21d2<\/span> radius, r =14\/2 units =7 units<\/span><\/p>\n

Height of cylinder, h = 20 units<\/span><\/p>\n

Lateral area of the cylinder\u00a0<\/span><\/p>\n

=2 \\pi \\mathrm{rh}<\/span><\/span><\/p>\n

=2 \\times \\frac{22}{7} \\times 7 \\times 20<\/span><\/span><\/p>\n

=2 \\times 22 \\times 20 \\text { square units }<\/span><\/span><\/span><\/p>\n

=880 \\text { square units }<\/span><\/span><\/p>\n

Hence, the lateral surface area of the given cylinder is 880 square units.\u00a0<\/span><\/p>\n

<\/span><\/p>\n

Example 2 :\u00a0<\/strong><\/p>\n

\u00a0The radius of a cone measures 5 cm and its height is 12 cm. Determine the LSA of the cone. [use value of \u03c0 =3.14]<\/span><\/p>\n

Solution:<\/strong><\/p>\n

Radius of the cone\u00a0 r = 5cm<\/p>\n

Height of the cone\u00a0 h = 12cm<\/p>\n

Slant height l=\\sqrt{r^{2}+h^{2}}<\/span><\/p>\n

<\/span>\\therefore \\mathrm{l}=\\sqrt{5^{2}+12^{2}}=\\sqrt{25+144}=\\sqrt{169}=13 \\mathrm{~cm}<\/span><\/p>\n

Hence Lateral\u00a0 Surface area of cone =\\pi \\mathrm{rl}<\/span><\/p>\n

\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0=3.14 \\times 5 \\times 13 \\mathrm{~cm}^{2}<\/span><\/p>\n

\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0= 204.1 {~cm}^{2}<\/span><\/p>\n

Therefore, the LSA of the cone is 204.1 {~cm}^{2}<\/span><\/span><\/p>\n

 <\/p>\n

Example 3<\/strong>:<\/strong> Find the lateral area of a cuboid whose length is 12 units, breadth is 8 units and height is 5 units.<\/span><\/p>\n

Solution:<\/strong><\/p>\n

Length of cuboid, l = 12 units.<\/span><\/p>\n

Breadth of cuboid, b = 8 units.<\/span><\/p>\n

Height of cuboid, h = 5 units.<\/span><\/p>\n

Therefore\u00a0 Lateral\u00a0 Area of a cuboid<\/span><\/p>\n

=2 \\times(1+b) \\times h<\/span><\/span><\/p>\n

=2 \\times(12+8) \\times 5<\/span> Square units<\/span><\/p>\n

=2 \\times 20 \\times 5<\/span> Square units<\/span><\/p>\n

Therefore, the lateral area of the given cuboid is 200 square units.\u00a0<\/span><\/p>\n

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Explore Other Topics<\/h1>\n

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Geometry<\/a><\/div>\n
Trigonometry<\/a><\/div>\n
Operations<\/a><\/div>\n
Numbers<\/a><\/div>\n<\/div>\n

[\/et_pb_text][et_pb_text admin_label=”Related Concepts
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Related Concepts<\/h1>\n

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\n
Mensuration Formula<\/a><\/div>\n
Surface Area and Volume<\/a><\/div>\n
Surface Area of Cone<\/a><\/div>\n<\/div>\n

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Frequently Asked Questions\u00a0<\/span><\/span><\/h1>\n
    <\/ol>\n

    1. How is the Lateral Surface Area of a Cone determined?<\/strong><\/h3>\n

    Ans: <\/strong>The LSA of a cone = \u03c0rl.
    <\/strong><\/span><\/p>\n

    Where r is the radius, l is its slant height and \u03c0 (pi) is a constant term with a value of 3.14 or \\frac{22}{7}<\/span><\/span>.<\/span>\u00a0<\/span><\/p>\n

    <\/b><\/span><\/span><\/h3>\n

    2. What Is the Lateral Surface Area of a Cylinder?<\/strong><\/h3>\n

    Ans: <\/strong>The LSA of a cylinder = 2\u03c0rh.
    <\/strong>Where r is the radius, h is its height and \u03c0 (pi) is a constant term with a value of 3.14 or \\frac{22}{7}<\/span><\/span>.<\/span>
    <\/strong><\/p>\n

    <\/span><\/p>\n

    3. What are the LSA of a Cube and a Cuboid?<\/strong><\/h3>\n

    Ans:<\/strong> The LSA of a Cube =4 \\mathrm{p}^{2}<\/span>
    <\/span><\/p>\n

    Where \u2018p\u2019 is the length of the edge of the square.<\/span><\/p>\n

    The LSA of a Cuboid = 2 \u2715(L + B ) \u2715 H.<\/span><\/p>\n

    Where \u2018L\u2019 is the length, \u2018B\u2019 is the breadth and \u2018H\u2019 is the height of the cuboid.<\/span><\/p>\n

    <\/span><\/p>\n

    4. What are the LSA of a sphere and a hemisphere?<\/strong>
    <\/span><\/h2>\n

    Ans: <\/strong>The LSA of a sphere =4 \\pi \\mathrm{r}^{2}<\/span><\/span><\/p>\n

    Where r is the radius of the sphere and \u03c0 (pi) is a constant term with a value of 3.14 or \\frac{22}{7}<\/span>.<\/span><\/p>\n

    Hemisphere is half of the sphere hence the LSA will also be half that of the sphere.<\/span><\/p>\n

    The LSA of a hemisphere =2 \\pi \\mathrm{r}^{2}<\/span><\/span><\/p>\n

    Where r is the radius of the hemisphere and \u03c0 (pi) is a constant term with a value of 3.14 or \\frac{22}{7}<\/span>.<\/span><\/p>\n

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