Lesson Solved problems on surface area of spheres

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

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

The theoretical base for these problems is the lesson &nbsp;<A HREF=http://www.algebra.com/algebra/homework/Surface-area/Surface-area-of-spheres.lesson>Surface area of spheres</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 surface area of a sphere if its radius is of &nbsp;5 cm.

<B>Solution</B>

The surface area of the sphere is

{{{4pi*r^2}}} = {{{4*3.14159*5^2}}} = {{{3.14159*100}}} = 314.159 {{{cm^2}}} (approximately).

<B>Answer</B>. &nbsp;The surface area of the sphere is &nbsp;314.159 {{{cm^2}}} (approximately).

<H3>Problem 2</H3>Find the surface area of a composite body comprised of a right circular cylinder and a hemisphere attached center-to-center to one of the cylinder bases &nbsp;(<B>Figure 1</B>)&nbsp; if both the cylinder diameter and the hemisphere diameter are of &nbsp;10 cm, &nbsp;and the cylinder height is of &nbsp;20 cm.

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

The full surface area of the composite body under consideration is the sum of&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

the lateral surface area of the cylinder &nbsp;{{{2pi*r*h}}}, &nbsp;the area of the base of the

cylinder &nbsp;{{{pi*r^2}}}&nbsp; and the area of the hemisphere &nbsp;{{{2pi*r^2}}}.

So, &nbsp;the total surface area of the composite body is equal to

{{{S}}} = {{{2pi*r*h}}} + {{{pi*r^2}}} + {{{2pi*r^2}}} = {{{2pi*r*h}}} + {{{3pi*r^2}}} = {{{pi}}}*({{{2r*h + 3r^2}}}) =

= 3.14159*(2*5*20 + 3*5^2) = 3.14159*275 = 863.937 {{{cm^2}}} (approximately).

</TD>
  <TD>
{{{drawing( 210, 195,  -3.5, 3.5, -1.0, 5.5,

ellipse( 0.5, 3.5, 3.0, 1.0),
            ellipse( 0.5, 0.0, 3.0, 1.0),

line( -1, 3.5, -1, 0.0),
              line(  2, 3.5,  2, 0.0),

arc ( 0.5, 3.5, 3.00, 3.00, 180, 360),
              arc ( 0.5, 3.5, 3.06, 3.06, 180, 360)
)}}}

<B>Figure 1</B>. &nbsp;To the &nbsp;<B>Problem 2</B>

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

<B>Answer</B>. &nbsp;The surface area of the composite body under consideration is &nbsp;863.937 {{{cm^2}}} (approximately).

<H3>Problem 3</H3>Find the surface area of a composite body comprised of a cone and a hemisphere attached center-to-center to the cone base &nbsp;(<B>Figure 2</B>)&nbsp; if both the cone base diameter and the hemisphere diameter are of &nbsp;10 cm&nbsp; and the cone height is of &nbsp;10 cm.

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

the lateral surface area of the cone &nbsp;{{{pi*r*slant_height}}} &nbsp;and the area of the

hemisphere &nbsp;{{{2pi*r^2}}}.

So, &nbsp;the total surface area of the composite body is equal to

{{{S}}} = {{{pi*r*slant_height}}} + {{{2pi*r^2}}} = {{{pi*r*sqrt(r^2 + h^2)}}} + {{{2pi*r^2}}} = {{{pi*r}}}*({{{sqrt(r^2+h^2) + 2r}}}) =

= {{{3.14159*5}}}*({{{sqrt(5^2 + 10^2) + 2*5}}}) = {{{3.14159*5}}}*{{{(sqrt(125) + 10)}}} = {{{3.14159*5}}}*{{{(5sqrt(5) + 10)}}} =

= 332.7 {{{cm^2}}} (approximately).

</TD>
  <TD>
{{{drawing( 195, 195,  -3.0, 3.5, -1.0, 5.5,

ellipse( 0.5, 3.5, 3.0, 1.0),

line( -1.0, 3.5,  0.5, 0.0),
              line(  2.0, 3.5,  0.5, 0.0),

arc ( 0.5, 3.5, 3.00, 3.00, 180, 360),
              arc ( 0.5, 3.5, 3.06, 3.06, 180, 360)
)}}}

<B>Figure 2</B>. To the <B>Problem 3</B>

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

<B>Answer</B>. &nbsp;The surface area of the composite body under consideration is &nbsp;332.7 {{{cm^2}}} (approximately).

<H3>Problem 4</H3>Find the surface area of a composite body comprised of a cube and a hemisphere attached center-to-center to one of the cube faces &nbsp;(<B>Figure 3</B>)&nbsp; if both the cube edge measure and the hemisphere diameter are of &nbsp;10 cm.

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

the surface area of the six cube faces &nbsp;{{{6a^2}}} &nbsp;minus the area of the base of the

hemisphere &nbsp;{{{pi*r^2}}}&nbsp; plus the area of the hemisphere &nbsp;{{{2pi*r^2}}}, &nbsp;where {{{r}}} = {{{10/2}}} = 5 cm.

So, &nbsp;the total surface area of the composite body is equal to

{{{S}}} = {{{6a^2}}} - {{{pi*(a/2)^2}}} + {{{2pi*(a/2)^2}}} = {{{6a^2}}} + {{{pi*(a/2)^2}}} = {{{6*10^2}}} + {{{3.14159*5^2}}} =

= 678.54 {{{cm^2}}} (approximately).

</TD>
  <TD>
{{{drawing( 180, 180,  -2.5, 3.5, -0.5, 5.5,

line (  0.0, 0.0,  3.0,  0.0),
            line (  0.0, 0.0,  0.0,  3.0),
            line (  0.0, 0.0, -2.0,  0.8),

line (  0.0, 3.0,  3.0,  3.0),
            line (  3.0, 3.0,  3.0,  0.0),

line (  0.0, 3.0, -2.0,  3.8),
            line ( -2.0, 3.8,  1.0,  3.8),
            line ( -2.0, 3.8, -2.0,  0.8),
      green(line (  1.0, 3.8,  1.0,  0.8)),

line (  1.0, 3.8,  3.0,  3.0),
      green(line ( -2.0, 0.8,  1.0,  0.8)),
      green(line (  1.0, 0.8,  3.0,  0.0)),

locate( 1.2,  0.1, a),
            locate(-1.5,  0.6, a),
            locate(-0.4,  2.1, a),

ellipse( 0.52, 3.43, 2.54, 0.77),
            arc    ( 0.52, 3.43, 2.54, 2.54, 175, 354)
)}}}

<B>Figure 3</B>. To the <B>Problem 4</B>

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

<B>Answer</B>. &nbsp;The surface area of the composite body under consideration is &nbsp;678.54 {{{cm^2}}} (approximately).

<H3>Problem 5</H3>Find the surface area of the sphere inscribed in a cone if the base diameter of the cone is of &nbsp;80 cm&nbsp; and the height of the cone is of &nbsp;75 cm &nbsp;(<B>Figure 4a</B>).

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

<B>Figure 4a</B>&nbsp; shows &nbsp;<B>3D</B>&nbsp; view of the cone with the inscribed sphere. &nbsp;<B>Figure 4b</B>&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;

shows the axial section of this cone and the inscribed sphere as the isosceles

triangle with the inscribed circle.

The radius of the inscribed sphere in the cone in the &nbsp;<B>Figure 4a</B>&nbsp; is the same

as the radius of the inscribed circle in the triangle in the &nbsp;<B>Figure 4b</B>. &nbsp;So,

instead of determining the radius of the sphere we will find the radius of the

inscribed circle. &nbsp;For it, &nbsp;use the formula &nbsp;{{{r}}} = {{{2S/P}}}, &nbsp;where &nbsp;{{{r}}}&nbsp; is the radius

of the inscribed circle in a triangle, &nbsp;{{{S}}}&nbsp; is the area of the triangle and &nbsp;{{{P}}}&nbsp; is

the perimeter of the triangle. &nbsp;The proof of this formula is in the lesson

<A HREF=http://www.algebra.com/algebra/homework/Surface-area/Proof-of-the-formula-for-the-area-of-a-triangle-via-the-radius-of-the-inscribed-circle.lesson>Proof of the formula for the area of a triangle via the radius of the inscribed

circle</A>&nbsp; under the topic &nbsp;<B>Area and surface area</B>&nbsp; of the section &nbsp;<B>Geometry</B>

in this site.

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

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

green(ellipse (0.0,  0.8, 3.2, 1.6)),

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

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

green(line (0, -2, 0, 4)),
            circle(  0.0, -0.12, 1.88, 1.88)
)}}}

<B>Figure 4a</B>. To the &nbsp;<B>Problem 5</B>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

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

line (-3.0, -2.0, 3.0, -2.0),

line ( -3.0, -2.0,   0.0, 4.0),
            line (  3.0, -2.0,   0.0, 4.0),

green(line (0, -2, 0, 4)),
            circle(  0.0, -0.12, 1.88, 1.88),

line ( 0.0, -0.2, -1.6,  0.8),
            line ( 0.0, -0.2,  1.6,  0.8),
            line ( 0.0, -0.2,  0.0, -2.0)
)}}}

&nbsp;&nbsp;<B>Figure 4b</B>. To the solution 
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;of the &nbsp;<B>Problem 5</B>

</TD>
 </TR>
</TABLE>For our isosceles triangle, &nbsp;we have {{{l}}} = {{{sqrt((80/2)^2 + 75^2)}}} = 85 cm&nbsp; for its lateral side length, &nbsp;{{{P}}} = {{{85 + 85 + 80}}} = 250 cm &nbsp;for the perimeter &nbsp;and &nbsp;{{{S}}} = {{{1/2}}}{{{80*75}}} = 3000 {{{cm^2}}}&nbsp; for the area. &nbsp;Therefore, &nbsp;the radius of the inscribed circle is &nbsp;{{{r}}} = {{{2*3000/250}}} = {{{24}}} {{{cm}}}&nbsp; in accordance with the formula above.

Hence, &nbsp;the area of the sphere inscribed in the cone is &nbsp;{{{4pi*r^2}}} = {{{4*3.14159*24^2}}} = 7238.223 {{{cm^2}}}.

<B>Answer</B>. &nbsp;The surface area of the sphere inscribed in the cone is &nbsp;7238.223 {{{cm^2}}} (approximately).

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

<TABLE>
  <TR>
  <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>

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

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

<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;

</TD>
  <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>

Solved problems on surface area of spheres

<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>

</TD>
 </TR>
</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>.