Lesson Solved problems on surface area of cones

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<H2>Solved problems on surface area of cones</H2>

In this lesson you will find typical solved problems on surface area of cones.

The theoretical base for these problems is the lesson &nbsp;<A HREF=http://www.algebra.com/algebra/homework/Surface-area/Surface-area-of-cones.lesson>Surface area of cones</A>&nbsp; under the topic &nbsp;<B>Area and surface area</B>&nbsp; of the section &nbsp;<B>Geometry</B>&nbsp; in this site.

<H3>Problem 1</H3>Find the lateral surface area of a cone if the base radius of the cone is of &nbsp;10 cm&nbsp; and the height of the cone is of &nbsp;5 cm.

Then find the total surface area of the cylinder.

<B>Solution</B>

First, &nbsp;find the slant height of the cone. &nbsp;It is

{{{l}}} = {{{sqrt(r^2 + h^2)}}} = {{{sqrt(10^2 + 5^2)}}} = {{{sqrt(125)}}} = {{{5sqrt(5)}}}.

The lateral surface area of the cone equals &nbsp;{{{pi}}}&nbsp; times the product of the base radius and the slant height

{{{S[lateral]}}} = {{{pi}}}{{{r}}}{{{l}}} = {{{3.14159}}}*{{{10}}}*{{{5sqrt(5)}}} = 351.24 {{{cm^2}}} (approximately).

The area of the base is

{{{S[base]}}} = {{{pi}}}{{{r^2}}} = 3.14159*100 = 314.159 {{{cm^2}}} (approximately).

So, &nbsp;the total surface area of the cone is &nbsp;{{{S[lateral]}}} + {{{S[base]}}} = 351.24 + 314.159 = 665.40 {{{cm^2}}} (approximately).

<B>Answer</B>. &nbsp;The lateral surface area of the cone is &nbsp;351.24 {{{cm^2}}} (approximately).

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;The total surface area of the cone is &nbsp;665.40 {{{cm^2}}} (approximately).

<H3>Problem 2</H3>Find the lateral surface area of a cone if the triangular axial section of the cone &nbsp;(<B>Figure 2</B>)&nbsp; has the area &nbsp;{{{a}}} = 10 {{{cm^2}}}.

<TABLE>
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  <TD>
<B>Solution</B>

The lateral surface area of the cone equals {{{pi}}} times the product of the radius of&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

the cone at the base and the height of the cone

{{{S[lateral]}}} = {{{pi}}}{{{r}}}{{{h}}}.

The area of the triangle at the axial section is

{{{S[section]}}} = {{{1/2}}}.{{{2r}}}.{{{h}}} = {{{r}}}{{{h}}}.

It implies that the lateral surface area of the cylinder is &nbsp;{{{pi}}}&nbsp; times the area

of the triangle at the axial section

{{{S[lateral]}}} = {{{pi*S[section]}}}.

Hence, &nbsp;{{{S[lateral]}}} = {{{pi*a}}} = 3.14159*10 = 31.4159 {{{cm^2}}} (approximately).

</TD>
  <TD>
{{{drawing( 220, 228, -5.5, 5.5, -6.0, 5.4,

ellipse( 0.0, -3.0, 10.0,  6.0),
            line( -4.5, -1.7,  0.0,  2.0),
            line(  4.5, -1.7,  0.0,  2.0),

blue(line( 0.0, 2.0, 0.0, -2.7)),

green(line(-3.2, -5.2,  2.6, -0.6)), 
      green(line(-3.2, -5.2,  0.0,  2.0)), 
      green(line( 2.6, -0.6,  0.0,  2.0)),

locate (-1.1,  -3.4, r),
            locate ( 1.8,  -1.2, r),
            locate ( 0.15, -0.3, h)

)}}}

<B>Figure 2</B>. &nbsp;To the &nbsp;<B>Problem 2</B>

</TD>
  </TR>
</TABLE>

<B>Answer</B>. &nbsp;The lateral surface area of the given cone is &nbsp;31.4159 {{{cm^2}}} (approximately).

<H3>Problem 3</H3>Find the surface area of a combined solid body which comprises of two identical cones joined base to base &nbsp;(<B>Figure 3</B>), &nbsp;if their common base radius is of &nbsp;4 cm&nbsp; and the height

of each cone is of &nbsp;3 cm.

<TABLE>
  <TR>
  <TD> 
<B>Solution</B>

We are given a 3D body comprised of two identical cones whose bases are joined and&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

overposed each to the other &nbsp;(<B>Figure 3</B>).

First, &nbsp;find the slant height of the cone. &nbsp;It is

{{{l}}} = {{{sqrt(r^2 + h^2)}}} = {{{sqrt(4^2 + 3^2)}}} = {{{sqrt(25)}}} = {{{5}}}.

The lateral surface area of the cone equals &nbsp;{{{pi}}}&nbsp; times the product of the base

radius and the slant height

{{{S[lateral]}}} = {{{pi}}}{{{r}}}{{{l}}} = {{{3.14159}}}*{{{4}}}*{{{5}}} = 62.832 {{{cm^2}}} (approximately).

The total surface area of the combined solid body is doubled the lateral area

of the single cone, &nbsp;i.e.

{{{S}}} = 2*62.832 = 125.664 {{{cm^2}}} (approximately).

</TD>
  <TD>
{{{drawing( 210, 249,  -3.5, 3.5, -3.8, 4.5,

ellipse (0.0, 0.0, 6.0, 3.0),

line ( -2.74, 0.5,  0.0,  3.9),
            line (  2.80, 0.5,  0.0,  3.9),

line ( -2.80, -0.5,  0.0, -3.5),
            line (  2.85, -0.5,  0.0, -3.5)
)}}}

<B>Figure 3</B>. To the <B>Problem 3</B>

</TD>
 </TR>
</TABLE>

<B>Answer</B>. &nbsp;The surface area of the combined body is &nbsp;125.664 {{{cm^2}}} (approximately).

<H3>Problem 4</H3>Find the lateral surface area of a body &nbsp;(a truncated cone)&nbsp; obtained from a cone with the base radius of &nbsp;6 cm&nbsp; and the height of &nbsp;8 cm&nbsp; after cutting off the part of the cone by the plane parallel to the base in a way that the cutting plane bisects the height of the original cone &nbsp;(<B>Figure 4</B>).

<TABLE>
  <TR>
  <TD> 
<B>Solution</B>

The strategy solving this problem is to find first the lateral surface area of the&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

entire cone with the base radius of &nbsp;6 cm&nbsp; and the height of &nbsp;8 cm&nbsp; and then

to distract the lateral surface area of the cone with the base radius of &nbsp;3 cm

and the height of &nbsp;4 cm.

The lateral surface area of the original entire cone is &nbsp;&nbsp;{{{pi*r*l}}}, &nbsp;where &nbsp;{{{r}}}= 6 cm

is the cone base radius and &nbsp;{{{l}}}&nbsp; is the cone slant height. &nbsp;The slant height is

{{{l}}} = {{{sqrt(r^2 + h^2)}}} = {{{sqrt(6^2 + 8^2)}}} = {{{sqrt(100)}}} = {{{10}}},

therefore the lateral area of the original cone is &nbsp;3.14159*6*10 =  188.495{{{cm^2}}}.

The small cone has the base radius of &nbsp;3 cm&nbsp; and the slant height of &nbsp;{{{1/2}}}{{{10}}} = 5 cm.

Therefore, &nbsp;the lateral surface area of the small cone is &nbsp;3.14159*3*5 =  47.124{{{cm^2}}}.

</TD>
  <TD> 
{{{drawing( 210, 249,  -3.5, 3.5, -3.8, 4.5,

ellipse (0.0, -2.0, 6.0, 3.0),

ellipse (0.0,  1.0, 3.0, 1.5),

line ( -2.75, -1.5,  -1.375,  1.25),
            line (  2.82, -1.5,   1.410,  1.25),

green(line ( -1.375,  1.25,  0.0,  4.0)),
      green(line (  1.410,  1.25,  0.0,  4.0)),

green(line (0, -2, 0, 4))
)}}}

&nbsp;&nbsp;&nbsp;&nbsp;<B>Figure 4</B>. To the &nbsp;<B>Problem 4</B>

</TD>
 </TR>
</TABLE>

Thus the lateral surface area of the truncated cone under consideration is &nbsp;188.495 - 47.124 = 141.372 {{{cm^2}}} (approximately).

<B>Answer</B>. &nbsp;The lateral surface area of body under consideration is &nbsp;141.372 {{{cm^2}}} (approximately).

My lessons on surface area of cones and other 3D solid bodies in this site are

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  <TD>
<B>Lessons on surface area of prisms</B>

<A HREF=http://www.algebra.com/algebra/homework/Surface-area/Surface-area-of-prisms.lesson>Surface area of prisms</A>

<A HREF=http://www.algebra.com/algebra/homework/word/geometry/Solved-problems-on-surface-area-of-prisms.lesson>Solved problems on surface area of prisms</A>

<A HREF=http://www.algebra.com/algebra/homework/Surface-area/OVERVIEW-of-LESSONS-on-surface-area-of-prisms.lesson>Overview of lessons on surface area of prisms</A>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

</TD>
  <TD>
<B>Lessons on surface area of pyramids</B>

<A HREF=http://www.algebra.com/algebra/homework/Surface-area/Surface-area-of-pyramids.lesson>Surface area of pyramids</A>

<A HREF=http://www.algebra.com/algebra/homework/word/geometry/Solved-problems-on-surface-area-of-pyramids.lesson>Solved problems on surface area of pyramids</A>

<A HREF=http://www.algebra.com/algebra/homework/Surface-area/OVERVIEW-of-LESSONS-on-surface-area-of-pyramids.lesson>Overview of lessons on surface area of pyramids</A>

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 </TR>
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  <TR>
  <TD>
<B>Lessons on surface area of cylinders</B>

<A HREF=http://www.algebra.com/algebra/homework/Surface-area/Surface-area-of-cylinders.lesson>Surface area of cylinders</A>

<A HREF=http://www.algebra.com/algebra/homework/word/geometry/Solved-problems-on-surface-area-of-cylinders.lesson>Solved problems on surface area of cylinders</A>

<A HREF=http://www.algebra.com/algebra/homework/Surface-area/OVERVIEW-of-LESSONS-on-surface-area-of-cylinders.lesson>Overview of lessons on surface area of cylinders</A>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

</TD>
  <TD>
<B>Lessons on surface area of cones</B>

<A HREF=http://www.algebra.com/algebra/homework/Surface-area/Surface-area-of-cones.lesson>Surface area of cones</A>

Solved problems on surface area of cones

<A HREF=http://www.algebra.com/algebra/homework/Surface-area/OVERVIEW-of-LESSONS-on-surface-area-of-cones.lesson>Overview of lessons on surface area of cones</A>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

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  <TD>
<B>Lessons on surface area of spheres</B>

<A HREF=http://www.algebra.com/algebra/homework/Surface-area/Surface-area-of-spheres.lesson>Surface area of spheres</A>

<A HREF=http://www.algebra.com/algebra/homework/word/geometry/Solved-problems-on-surface-area-of-spheres.lesson>Solved problems on surface area of spheres</A>

<A HREF=http://www.algebra.com/algebra/homework/Surface-area/OVERVIEW-of-LESSONS-on-surface-area-of-spheres.lesson>Overview of lessons on surface area of spheres</A>

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</Table>

To navigate over all topics/lessons of the Online Geometry Textbook use this file/link &nbsp;<A HREF=https://www.algebra.com/algebra/homework/Triangles/GEOMETRY-your-online-textbook.lesson>GEOMETRY - YOUR ONLINE TEXTBOOK</A>.