Lesson Solved problems on volume of pyramids

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<H2>Solved problems on volume of pyramids</H2>

In this lesson you will find typical solved problems on volume of pyramids.

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

<H3>Problem 1</H3>Find the volume of a triangular pyramid &nbsp;ABCD&nbsp; if its edges issued from the vertex &nbsp;A&nbsp; are of &nbsp;8 cm, &nbsp;6 cm &nbsp;and &nbsp;6 cm&nbsp; long and each of these three edges is perpendicular

to the two others &nbsp;(Figure 1).

<TABLE>
 <TR>
 <TD> 
Solution

First, &nbsp;let us find the area of the triangle &nbsp;{{{DELTA}}}ABC&nbsp; at the base of the pyramid&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;

(Figure 1).

It is a right-angled triangle, &nbsp;and its area is half of the product of its legs:

{{{S[base]}}} = {{{S[ABC]}}} = {{{1/2}}}{{{abs(AB)}}}.{{{abs(AC)}}} = {{{1/2}}}{{{6}}}.{{{6}}} = 18 {{{cm^2}}}.

The edge &nbsp;AD&nbsp; is the height of the given pyramid. &nbsp;So, &nbsp;the volume of the pyramid is

{{{V}}} = {{{1/3}}}{{{S[base]}}}.{{{h}}} = {{{1/3}}}.{{{18}}}.{{{8}}} = 48 {{{cm^3}}}.

Answer. &nbsp;The volume of the given pyramid is &nbsp;48 {{{cm^3}}}.

</TD>
 <TD> 
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line ( -3.0,  1.5,  4.5,  0.0),
      green(line ( -3.0,  1.5,  2.0,  2.0)),
      green(line (  2.0,  2.0,  4.5,  0.0)),

line ( -3.0,  1.5,  2.0,  8.0),
            line (  4.5,  0.0,  2.0,  8.0),

green(line (  2.0,  2.0,  2.0,  8.0)),

locate ( 2.1, 4.7, 8),
            locate (-0.2, 2.4, 6),
            locate ( 3.0, 1.8, 6),

locate ( 2.1, 2.5, A),
            locate (-3.4, 1.7, B),
            locate ( 4.6, 0.2, C),
            locate ( 1.9, 8.6, D)
)}}}

&nbsp;&nbsp;Figure 1. To the &nbsp;Problem 1&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

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

<H3>Problem 2</H3>Find the volume of a rectangular pyramid &nbsp;ABCDE&nbsp; if its base &nbsp;ABCD&nbsp; is a square with the side measure of &nbsp;6 cm&nbsp; and the lateral edge &nbsp;AE&nbsp; is perpendicular to the base plane and has the measure of &nbsp;8 cm &nbsp;(Figure 2).

<TABLE>
 <TR>
 <TD> 
Solution

First, &nbsp;let us find the area of the square &nbsp;ABCD&nbsp; at the base of the pyramid&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;

(Figure 2).

This area is squared length of the base edge:

{{{S[base]}}} = {{{S[ABCD]}}} = {{{abs(AB)}}}.{{{abs(AC)}}} = {{{6}}}.{{{6}}} = 36 {{{cm^2}}}.

The edge &nbsp;AD&nbsp; is the height of the given pyramid. &nbsp;So, &nbsp;the volume of the pyramid is

{{{V}}} = {{{1/3}}}{{{S[base]}}}.{{{h}}} = {{{1/3}}}.{{{36}}}.{{{8}}} = 96 {{{cm^3}}}.

Answer. &nbsp;The volume of the given pyramid is &nbsp;96 {{{cm^3}}}.

</TD>
 <TD> 
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      green(line (  2.0,  2.0,  4.5,  0.0)), 
            line ( -0.5, -0.5,  4.5,  0.0),

line ( -3.0,  1.5,  2.0,  8.0),
            line (  4.5,  0.0,  2.0,  8.0),

green(line (  2.0,  2.0,  2.0,  8.0)),

line ( -0.5, -0.5,  2.0,  8.0),

locate ( 2.1,  4.7,  8),
            locate (-0.4,  2.4,  6),
            locate ( 3.0,  1.8,  6),
            locate ( 1.8, -0.25, 6),
            locate (-2.3,  0.84,  6),

locate ( 2.1,  2.5, A),
            locate (-3.4,  1.7, B),
            locate (-0.6, -0.5, C),
            locate ( 4.6,  0.2, D),
            locate ( 1.9,  8.6, E),

arc( 2.0, 2.0, 1.0, 1.0, 172, 270),
            arc( 2.0, 2.0, 1.4, 1.4, 270,  32),

arc(-3.0, 1.5, 1.0, 1.0, 310,  40),
            arc( 4.5, 0.0, 1.2, 1.2, 172, 250)
)}}}

&nbsp;&nbsp;Figure 2. To the &nbsp;Problem 2

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

<H3>Problem 3</H3>Find the volume of a regular pyramid with the square base &nbsp;(Figure 3a)&nbsp; if the lateral edge of the pyramid has the same measure of &nbsp;10 cm&nbsp; as the the base edge has.

Also find the angle between the lateral edge and the base of the pyramid.

<TABLE>
 <TR>
 <TD>
Solution

First, &nbsp;let us find the area of the base of the given pyramid.

Since the base is a square with the side measure of &nbsp;10 cm, &nbsp;the area of the base is&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

{{{S[base]}}} = {{{10^2}}} = 100 {{{cm^2}}}.

Next, &nbsp;let us find the height of the pyramid.

For it, &nbsp;let us consider the triangle &nbsp;{{{DELTA}}}AOP&nbsp; (Figure 3b), &nbsp;where the point &nbsp;A&nbsp; is one

of the base vertices of the pyramid, &nbsp;the point &nbsp;O&nbsp; is the center of the square base

of the pyramid, &nbsp;and the point &nbsp;P&nbsp; is the pyramid vertex.

It is a right-angled triangle &nbsp;(the segment &nbsp;OP&nbsp; is the height of the pyramid). &nbsp;Its

leg &nbsp;AO&nbsp; is half of the diagonal of the square base, &nbsp;and its measure is {{{10sqrt(2)/2}}} = {{{5sqrt(2)}}}.

Therefore, &nbsp;the measure of the height &nbsp;OP&nbsp; is

|OP| = {{{sqrt(abs(AP)^2 - abs(AO)^2)}}} = {{{sqrt(10^2 - (5sqrt(2))^2)}}} = {{{sqrt(100 - 50)}}} = {{{sqrt(50)}}} = {{{5sqrt(2)}}} cm.

</TD>
 <TD>
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            line (  0.0, 0.0, -2.0,  0.8),
      green(line ( -2.0, 0.8,  1.0,  0.8)),
      green(line (  1.0, 0.8,  3.0,  0.0)),

line (  0.0, 0.0,  0.5,  2.9),
            line (  3.0, 0.0,  0.5,  2.9),
      green(line (  1.0, 0.8,  0.5,  2.9)),
            line ( -2.0, 0.8,  0.5,  2.9),

locate ( 1.0,  0.0, 10),
          locate (-1.6,  0.5, 10),

locate (-1.4,  2.0, 10),
          locate ( 1.6,  1.9, 10),
          locate (-0.24, 1.7, 10)
)}}}

Figure 3a. To the Problem 3&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

</TD>
 <TD>
{{{drawing( 220, 200, -2.5, 3.5, -0.5, 3.8,

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

line (  0.0, 0.0,  0.5,  2.9),
            line (  3.0, 0.0,  0.5,  2.9),
      green(line (  1.0, 0.8,  0.5,  2.9)),
            line ( -2.0, 0.8,  0.5,  2.9),

red(line ( 0.5,  2.9,  0.5,  0.4)),

locate ( 1.0,  0.0, 10),
          locate (-1.6,  0.5, 10),

red(line ( 0.5,  0.4,  0.0,  0.0)),

locate ( 0.6,  0.5, O),
          locate ( 0.4,  3.2, P),
          locate (-0.1,  0.0, A),

locate (-0.24, 1.7, 10),

arc (0, 0, 0.8, 0.8, 280, 323)
)}}}

&nbsp;&nbsp;Figure 3b. To the solution 
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;of the Problem 3

</TD>
 </TR>
</TABLE>Now, &nbsp;the volume of the given pyramid is &nbsp;{{{V}}} = {{{1/3}}}{{{100}}}.{{{5sqrt(2)}}} = {{{500sqrt(2)/3}}} = 235.70 {{{cm^3}}} (approximately).

On the way, &nbsp;we proved that the right-angled triangle &nbsp;{{{DELTA}}}AOP&nbsp; is isosceles: |OP| = |AO|.

It means that the angle &nbsp;LOAP&nbsp; is of 45�.

Answer. &nbsp;The volume of the given pyramid is &nbsp;{{{500sqrt(2)/3}}} = {{{235.70}}} {{{cm^3}}} (approximately).

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;The angle between the lateral edge and the base of the pyramid is of &nbsp;45�.

<H3>Problem 4</H3>Find the volume of a regular hexagonal pyramid if the base edge has the measure of &nbsp;4 cm&nbsp; and the lateral edge of the pyramid is of &nbsp;8 cm&nbsp; long &nbsp;(Figure 4a).

Also find the angle between the lateral edge and the base plane of the pyramid.

<TABLE>
 <TR>
 <TD> 
Solution

Let the point &nbsp;O&nbsp; be the center of the regular hexagon at the base of the pyramid,

the point &nbsp;A&nbsp; be one of the hexagon's vertices, &nbsp;and the point &nbsp;P&nbsp; be the vertex

of the pyramid &nbsp;(Figure 4b).

The area of the regular hexagon at the base of the pyramid is

{{{S}}} = {{{6}}}*{{{1/2}}}.{{{4}}}*{{{4sqrt(3)/2}}} = {{{24}}}{{{sqrt(3)}}} {{{cm^2}}}.

Next, &nbsp;since the hexagon at the base is regular, &nbsp;the segment &nbsp;OA&nbsp; connecting the&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

hexagon's center with its vertex has the same length as the hexagon side, &nbsp;i.e. &nbsp;4 cm.

So, &nbsp;in the right-angled triangle &nbsp;{{{DELTA}}}AOP&nbsp; the leg &nbsp;OA&nbsp; is of &nbsp;4 cm&nbsp; long and the

hypotenuse &nbsp;AP&nbsp; is of &nbsp;8 cm&nbsp; long.

It implies, &nbsp;in particular, &nbsp;that the angle &nbsp;LOAP&nbsp; is of &nbsp;60�.

It implies also that the leg &nbsp;OP, &nbsp;which is the height of the pyramid, &nbsp;has the measure

|OP| = {{{sqrt(abs(AP)^2 - abs(OA)^2)}}} = {{{sqrt(8^2 - 4^2)}}} = {{{sqrt(64 - 16)}}} = {{{sqrt(48)}}} = {{{4}}}{{{sqrt(3)}}} cm.

</TD>
 <TD> 
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      green(line (  1.0,  2.3,  4.5,  2.3)),
      green(line (  4.5,  2.3,  6.5,  1.0)),
            line (  6.5,  1.0,  4.0,  0.0),

line (  0.0,  0.0,  2.25,  7.5),
            line ( -1.7,  1.5,  2.25,  7.5),
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            line (  4.0,  0.0,  2.25,  7.5),

red(line(   2.25, 1.2,  2.25,  7.5)),

locate ( 1.9, -0.05, 4),
            locate ( 5.3,  0.60, 4),

locate ( 2.1,  1.15, O),
            locate ( 2.1,  8.1, P),
            locate (-1.35, 0.95, 4),

locate ( 0.7,   4.0, 8),
            locate (-0.1,   5.0, 8),
            locate ( 3.24,  4.0, 8),
            locate ( 4.25,  5.0, 8)
)}}}

&nbsp;&nbsp;&nbsp;&nbsp;Figure 4a. To the &nbsp;Problem 4&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

</TD>
 <TD> 
{{{drawing( 238, 232, -2.5, 7.0, -0.8, 8.3,

red(line(   2.25, 1.2,  2.25,  7.5)),
        red(line(   2.25, 1.2,  0.00,  0.0)),

locate ( 1.9, -0.05, 4),
            locate ( 5.3,  0.60, 4),

locate ( 2.1,  1.15, O),
            locate ( 2.1,  8.1, P),
            locate (-1.35, 0.95, 4),

arc( 0.00, 0.00, 1.6, 1.6, 283, 330),

locate ( 0.7,   4.0, 8),
            locate (-0.1,   5.0, 8),
            locate ( 3.24,  4.0, 8),
            locate ( 4.25,  5.0, 8),

locate (-0.2,  0.0, A),
            locate ( 1.0,  1.2, 4)
)}}}

&nbsp;&nbsp;&nbsp;&nbsp;Figure 4b. To the solution 
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;of the &nbsp;Problem 4

</TD>
 </TR>
</TABLE>Therefore, &nbsp;the volume of the prism is &nbsp;{{{V}}} = {{{1/3}}}{{{Sh}}} = {{{1/3}}}.{{{24}}}{{{sqrt(3)}}}.{{{4}}}{{{sqrt(3)}}} = 96 {{{cm^3}}}.

Answer. &nbsp;The volume of the prism is &nbsp;96 {{{cm^3}}}.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;The angle between the lateral edge and the base plane of the pyramid is of &nbsp;60�.

<H3>Problem 5</H3>Find the volume of a composite solid body of a "diamond" shape which comprises of two regular rectangular pyramids with square bases joined base to base &nbsp;(Figure 5),

if all their edges are of &nbsp;4 cm.

<TABLE>
 <TR>
 <TD> 
Solution

We are given a 3D body of a "diamond" shape comprised of two regular rectangular&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;

pyramids whose two bases are joined and overposed each to the other &nbsp;(Figure 5).

In total, &nbsp;the volume of our solid body is two times the volume of the regular pyramid

with all edge measures of &nbsp;4 cm. &nbsp;The later is equal to one third the product of the

base square area and the measure of the pyramid's height.

The base square area is &nbsp;{{{S}}} = {{{4^2}}} = 16 {{{cm^2}}}.

To get the pyramid's height, &nbsp;note that the diagonal of the square at the base of the

pyramid is &nbsp;{{{4sqrt(2)}}} {{{cm}}} long, &nbsp;and half of the diagonal has half of this measure,

i.e. &nbsp;{{{2sqrt(2)}}} {{{cm}}}.

Therefore, &nbsp;the pyramid's height is &nbsp;{{{h}}} = {{{sqrt(4^2 - (2sqrt(2))^2)}}} = {{{sqrt(16-8)}}} = {{{sqrt(8)}}} = {{{2sqrt(2)}}} cm.

</TD>
 <TD>
{{{drawing( 180, 220, -2.5, 3.5, -2.5, 3.3,

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

line (  0.0, 0.0,  0.5,  2.9),
            line (  3.0, 0.0,  0.5,  2.9),
      green(line (  1.0, 0.8,  0.5,  2.9)),
            line ( -2.0, 0.8,  0.5,  2.9),

locate ( 1.2,  0.0, 4),
          locate (-1.3,  0.5, 4),

locate (-1.2,  2.0, 4),
          locate ( 0.93, 1.9, 4),
          locate (-0.07, 1.7, 4),

line (  0.0, 0.0,  0.5, -2.1),
            line (  3.0, 0.0,  0.5, -2.1),
      green(line (  1.0, 0.8,  0.5, -2.1)),
            line ( -2.0, 0.8,  0.5, -2.11),

locate (-1.1,  -0.5, 4),
          locate (-0.1,  -0.8, 4),
          locate ( 0.83, -0.6, 4),
          locate ( 1.66, -1.0, 4)
)}}}

Figure 5. To the Problem 5

</TD>
 </TR>
</TABLE>For the detailed explanation on this calculation see the solution of the &nbsp;Problem 3 above.

Now, &nbsp;the volume of the single pyramid is &nbsp;{{{V}}} = {{{1/3}}}{{{Sh}}} = {{{1/3}}}.{{{16}}}.{{{2sqrt(2)}}} = {{{(32sqrt(2))/3}}}, &nbsp;and the volume of the composite solid body under consideration is &nbsp;{{{2}}}.{{{(32sqrt(2))/3}}} = {{{(64sqrt(2))/3}}} = 30.17 {{{cm^3}}} (approximately).

Answer. &nbsp;The volume of the composite body under consideration is &nbsp;{{{(64sqrt(2))/3)}}} = 30.17 {{{cm^3}}} (approximately).

<H3>Problem 6</H3>Find the volume of a body obtained from a regular rectangular pyramid with the edge measures of &nbsp;10 cm &nbsp; for all edges after cutting off the part of the pyramid by the plane parallel to the base in a way that the cutting plane bisects the four lateral edges of the original pyramid &nbsp;(truncated pyramid, &nbsp;Figure 6a).

<TABLE>
 <TR>
 <TD> 
Solution

The strategy solving this problem is to find first the volume of the regular rectangular&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

pyramid with the edge measures of &nbsp;10 cm&nbsp; and then to distract the volume of the

regular rectangular pyramid with the edge measures of &nbsp;5 cm.

The base area of the original regular rectangular pyramid is &nbsp;{{{10^2}}} = {{{100}}} {{{cm^2}}}.

The height of the original regular rectangular pyramid is

{{{sqrt(10^2 - ((10sqrt(2))/2)^2)}}} = {{{sqrt(100 - (5sqrt(2))^2)}}} = {{{sqrt(100 - 50)}}} = {{{sqrt(50)}}} = {{{5sqrt(2)}}} {{{cm}}}.

This calculation is based on consideration of the right-angled triangle &nbsp;{{{DELTA}}}OAP

(Figure 6b). &nbspHence, &nbsp;the volume of the original pyramid is &nbsp;{{{1/3}}}{{{100}}}.{{{5sqrt(2)}}} = {{{(500sqrt(2))/3}}} {{{cm^3}}}.

</TD>
 <TD> 
{{{drawing( 220, 200, -2.5, 3.5, -0.5, 3.8,

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      green(line (  1.0, 0.8,  3.0,  0.0)),

line (  0.0,  0.0,  0.25, 1.45),
            line (  3.0,  0.0,  1.75, 1.45),
      green(line (  1.0,  0.8,  0.5,  2.9)),
            line ( -2.0,  0.8, -0.75, 1.85),

locate ( 1.0,  0.0, 10),
          locate (-1.6,  0.5, 10),

line (  0.25, 1.45,  1.75, 1.45),
            line (  0.25, 1.45, -0.75, 1.85),
            line ( -0.75, 1.85,  0.75, 1.85),
            line (  0.75, 1.85,  1.75, 1.45),

green(line (  0.25, 1.45,  0.5,  2.9)),
      green(line (  1.75, 1.45,  0.5,  2.9)),
      green(line (  0.75, 1.85,  0.5,  2.9)),
      green(line ( -0.75, 1.85,  0.5,  2.9)),

locate (-0.1,  1.2, 5),
          locate (-1.5,  1.7, 5),
          locate ( 2.2,  1.2, 5),

locate ( 0.1,  2.4, 5),
          locate (-0.4,  2.6, 5),
          locate ( 1.2,  2.4, 5)
)}}}

Figure 6a. To the &nbsp;Problem 6&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

</TD>
 <TD> 
{{{drawing( 220, 200, -2.5, 3.5, -0.5, 3.8,

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

line (  0.0,  0.0,  0.25, 1.45),
            line (  3.0,  0.0,  1.75, 1.45),
      green(line (  1.0,  0.8,  0.5,  2.9)),
            line ( -2.0,  0.8, -0.75, 1.85),

locate ( 1.0,  0.0, 10),
          locate (-1.6,  0.5, 10),

line (  0.25, 1.45,  1.75, 1.45),
            line (  0.25, 1.45, -0.75, 1.85),
            line ( -0.75, 1.85,  0.75, 1.85),
            line (  0.75, 1.85,  1.75, 1.45),

green(line (  0.25, 1.45,  0.5,  2.9)),
      green(line (  1.75, 1.45,  0.5,  2.9)),
      green(line (  0.75, 1.85,  0.5,  2.9)),
      green(line ( -0.75, 1.85,  0.5,  2.9)),

locate (-0.1,  1.2, 5),
          locate (-1.5,  1.7, 5),
          locate ( 2.2,  1.2, 5),

locate ( 0.1,  2.4, 5),
          locate (-0.4,  2.6, 5),
          locate ( 1.2,  2.4, 5),

red(line ( 0.5,  2.9,  0.5,  0.4)),
        red(line ( 0.5,  0.4,  0.0,  0.0)),

locate ( 0.6,  0.5, O),
          locate ( 0.4,  3.2, P),
          locate (-0.1,  0.0, A)
)}}}

&nbsp;&nbsp;&nbsp;&nbsp;Figure 6b. To the solution 
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;of the &nbsp;Problem 6

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

Calculation of the volume of the small pyramid is similar to this. &nbsp;The only difference is in reducing all linear dimensions in two times.

Thus, &nbsp;the base area of the small regular rectangular pyramid is &nbsp;{{{5^2}}} = {{{25}}} {{{cm^2}}}.

The height of the small regular rectangular pyramid is

{{{sqrt(5^2 - ((5sqrt(2))/2)^2)}}} = {{{sqrt(25 - ((5sqrt(2))/2)^2)}}} = {{{sqrt(25 - 12.5)}}} = {{{sqrt(12.5)}}} = {{{(5sqrt(2))/2}}} {{{cm}}}.

So, &nbsp;the volume of the smaller rectangular pyramid is &nbsp;{{{1/3}}}{{{25}}}{{{(5sqrt(2))/2}}} = {{{(125sqrt(2))/6}}} {{{cm^2}}}.

Now, &nbsp;the volume of the truncated pyramid is &nbsp;{{{(500sqrt(2))/3}}} - {{{(125sqrt(2))/6}}} = {{{(875sqrt(2))/6}}} = 206.24 {{{cm^3}}} (approximately).

Answer. &nbsp;The lateral surface area of the body under consideration is &nbsp;206.24 {{{cm^3}}} (approximately).

My lessons on volume of pyramids and other 3D solid bodies in this site are

<TABLE>
 <TR>
 <TD>
Lessons on volume of prisms

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

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

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

</TD>
 <TD>
Lessons on volume of pyramids

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

Solved problems on volume of pyramids

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

</TD>
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</Table><TABLE>
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Lessons on volume of cylinders

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

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

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

</TD>
 <TD>
Lessons on volume of cones

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

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

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

</TD>
 <TD>
Lessons on volume of spheres

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

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

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