Lesson Volume of pyramids

Source code of 'Volume of pyramids'

This Lesson (Volume of pyramids) was created by by ikleyn(52787)

: View Source, Show
About ikleyn:

<H2>Volume of pyramids</H2>

<B>Pyramid</B>&nbsp; is a &nbsp;<B>3D</B>&nbsp; solid body with flat faces which has one distinguished face of a polygonal shape, &nbsp;while all other faces are of a triangular shape with a common &nbsp;<B>vertex</B>&nbsp; for all triangles. &nbsp;The distinguished face is called the &nbsp;<B>pyramid base</B>. &nbsp;The remaining faces are called the &nbsp;<B>lateral faces</B>. &nbsp;The lateral faces are of triangular form.

Structurally a pyramid can be thought as a polygon on a plane and a point in a space out of the plane, &nbsp;which is connected with the polygon vertices by straight line segments - the &nbsp;<B>edges of the pyramid</B>. &nbsp;Figures &nbsp;<B>1a</B>&nbsp; - &nbsp;<B>1e</B>&nbsp; present the examples of pyramids.

&nbsp;&nbsp;{{{drawing( 200, 225,  -3.0, 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,  3.4),
            line (  3.0, 0.0,  0.5,  3.4),
      green(line (  1.0, 0.8,  0.5,  3.4)),
            line ( -2.0, 0.8,  0.5,  3.4),

locate ( -2.2, -0.1, base),
            line ( -1.6, -0.1, -0.4, 0.4),

locate ( -2.9,  2.6, lateral),
          locate ( -2.7,  2.3, faces),
            line ( -2.1,  1.9, -0.8, 1.5),
            line ( -1.8,  1.9, -0.6, 1.8)
)}}}

<B>Figure 1a</B>. Rectangular pyramid&nbsp;&nbsp;&nbsp;

</TD>
  <TD>

{{{drawing( 250, 225,  -5.0, 5.0, -0.5, 5.0,

line (  0.5, 0.0,  4.5,  0.0),
            line (  0.5, 0.0, -4.0,  1.5),

line (  0.5, 0.0, -0.5,  4.5),
            line (  4.5, 0.0, -0.5,  4.5),
            line ( -4.0, 1.5, -0.5,  4.5),
      green(line ( -0.5, 4.5, -0.5,  1.5)),

green(line ( -4.0, 1.5, -0.5,  1.5)),
      green(line ( -0.5, 1.5,  4.5,  0.0)),

locate ( -4.8,  3.5, lateral),
          locate ( -4.5,  3.2, edge),
            line ( -3.7,  2.7, -3.0, 2.6),

locate ( -4.8,  0.3, base_edge),
            line ( -3.4,  0.3, -2.3, 0.8)
)}}}

&nbsp;&nbsp;<B>Figure 1b</B>. Rectangular pyramid&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

</TD>
  <TD>
{{{drawing( 225, 225,  -4.0, 5.0, -0.5, 7.5,

line ( -0.8, 0.0,  4.0,  0.0),
            line ( -0.8, 0.0, -3.0,  1.5),
      green(line ( -3.0, 1.5,  4.0,  0.0)),

line ( -0.8, 0.0,  0.0,  7.0),

line (  0.0, 7.0,  4.0,  0.0),
            line (  0.0, 7.0, -3.0,  1.5)
)}}}

&nbsp;&nbsp;<B>Figure 1c</B>. Triangular pyramid&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

</TD>
  <TD>
{{{drawing( 225, 225,  -4.0, 5.0, -0.5, 8.5,

line ( -3.0,  1.5,  4.0,  0.0),
      green(line ( -3.0,  1.5,  2.0,  2.0)),
      green(line (  2.0,  2.0,  4.0,  0.0)),

line ( -3.0,  1.5,  2.0,  8.0),
            line (  4.0,  0.0,  2.0,  8.0),

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

locate ( -2.2,  0.2, base),
            line ( -1.6,  0.2, 1.2, 1.1),

locate ( -3.3,  5.6, lateral),
          locate ( -3.1,  5.2, faces),
            line ( -2.1,  4.4, -0.6, 3.5),
            line ( -1.8,  4.4, -0.6, 4.1)
)}}}

<B>Figure 1d</B>. Triangular pyramid&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

</TD>
  <TD>
{{{drawing( 238, 232,  -2.5, 7.0, -0.5, 8.6,

line (  0.0,  0.0,  4.0,  0.0),
            line (  0.0,  0.0, -1.7,  1.5),
      green(line ( -1.7,  1.5,  1.0,  2.3)),
      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,  7.15),
            line ( -1.7,  1.5,  2.25,  7.15),
      green(line (  1.0,  2.3,  2.25,  7.15)),
      green(line (  4.5,  2.3,  2.25,  7.15)),
            line (  6.5,  1.0,  2.25,  7.15),
            line (  4.0,  0.0,  2.25,  7.15)
)}}}

&nbsp;&nbsp;&nbsp;&nbsp;<B>Figure 1e</B>. Hexagonal pyramid

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

The base of a pyramid can be any type of polygon. &nbsp;Depending on the shape of this polygon, &nbsp;the prism can be called a &nbsp;<B>triangular prism</B>, &nbsp;or &nbsp;<B>rectangular prism</B>, &nbsp;or &nbsp;<B>pentagonal</B>, &nbsp;<B>hexagonal</B>&nbsp; and so on.

The &nbsp;<B>height of a pyramid</B>&nbsp; (sometimes called the &nbsp;<B>altitude of a pyramid</B>)&nbsp; is&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;

the perpendicular segment from the vertex, &nbsp;located out of the base plane, &nbsp;to

the base &nbsp;(Figures &nbsp;<B>2a</B>&nbsp; and &nbsp;<B>2b</B>).

If the polygon at the base of a pyramid is regular &nbsp;<B><U>and</U></B>&nbsp; all the pyramid lateral

edges have the same length then the pyramid is called a &nbsp;<B>regular pyramid</B>.

In a regular pyramid the altitude drops to the center of the regular polygon at

the base. &nbsp;In other words, &nbsp;in a regular pyramid the foot of the altitude

coincides with the center of the regular polygon at the base.

This lesson is focused on calculating the volume of pyramids.

</TD>
  <TD>
{{{drawing( 220, 220,  -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,  3.4),
            line (  3.0, 0.0,  0.5,  3.4),
      green(line (  1.0, 0.8,  0.5,  3.4)),
            line ( -2.0, 0.8,  0.5,  3.4),

red(line ( 0.5,  3.4,  0.5,  0.4)),

red(locate ( 0.6, 1.7, h_height))
)}}}

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<B>Figure 2a</B>. The height&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;of a pyramid

</TD>
  <TD> 
{{{drawing( 238, 220,  -2.5, 7.0, -0.8, 8.3,

red(line(   2.25, 1.2,  2.25,  7.15)),

red(locate ( 2.35, 3.5,  h_height))
)}}}

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<B>Figure 2b</B>. The height 
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;of a pyramid

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

<H3>Formula for calculating the volume of pyramids</H3><TABLE>
  <TR>
  <TD>

<B>The volume of a pyramid</B>&nbsp; is &nbsp;{{{V}}} = {{{1/3}}}{{{S[base]}}}.{{{h}}},

where &nbsp;{{{S[base]}}}&nbsp; is the area of the base of the pyramid and &nbsp;{{{h}}}&nbsp; are the pyramid's height.

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

<H3>Example 1</H3>Find the volume of a regular pyramid with the square base &nbsp;(<B>Figure 3</B>)&nbsp; if the height of the pyramid is of &nbsp;12 cm&nbsp; and the measure of the base edge is of &nbsp;10 cm.

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

The area of the base of the given pyramid  is&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;&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;&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;

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

Hence, &nbsp;the volume of the pyramid is &nbsp;{{{1/3}}}{{{100}}}.{{{12}}} = 400 {{{cm^3}}}.

<B>Answer</B>. &nbsp;The volume of the pyramid is &nbsp;400 {{{cm^3}}}.

</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,  3.4),
            line (  3.0, 0.0,  0.5,  3.4),
      green(line (  1.0, 0.8,  0.5,  3.4)),
            line ( -2.0, 0.8,  0.5,  3.4),

red(line ( 0.5,  3.4,  0.5,  0.4)),

locate ( 1.0,  0.0, 10),
          locate (-1.6,  0.5, 10),
      red(locate ( 0.54, 1.7, 12))
)}}}

<B>Figure 3</B>. To the <B>Example 1</B>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

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

<H3>Example 2</H3>Find the volume of a regular pyramid with the square base &nbsp;(<B>Figure 4a</B>)&nbsp; if the lateral edge of the pyramid has the same measure of &nbsp;12 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>
<B>Solution</B>

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;12 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]}}} = {{{12^2}}} = 144 {{{cm^2}}}.

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

For it, &nbsp;let us consider the triangle &nbsp;{{{DELTA}}}<B>AOP</B>&nbsp; (<B>Figure 4b</B>), &nbsp;where the point &nbsp;<B>A</B>&nbsp; is one

of the base vertices of the pyramid, &nbsp;the point &nbsp;<B>O</B>&nbsp; is the center of the square base

of the pyramid, &nbsp;and the point &nbsp;<B>P</B>&nbsp; is the pyramid vertex.

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

leg &nbsp;<B>AO</B>&nbsp; is half of the diagonal of the square base, &nbsp;and its measure is {{{12sqrt(2)/2}}} = {{{6sqrt(2)}}}.

Therefore, &nbsp;the measure of the height &nbsp;<B>OP</B>&nbsp; is

|<B>OP</B>| = {{{sqrt(abs(AP)^2 - abs(AO)^2)}}} = {{{sqrt(12^2 - (6sqrt(2))^2)}}} = {{{sqrt(144 - 72)}}} = {{{sqrt(72)}}} = {{{6sqrt(2)}}} cm.

</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),

locate ( 1.0,  0.0, 12),
          locate (-1.6,  0.5, 12),

locate (-1.4,  2.0, 12),
          locate ( 1.6,  1.9, 12),
          locate (-0.24, 1.7, 12)
)}}}

<B>Figure 4a</B>. To the <B>Example 2</B>&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, 12),
          locate (-1.6,  0.5, 12),

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, 12),

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

&nbsp;&nbsp;<B>Figure 4b</B>. To the solution 
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;of the <B>Example 2</B>

</TD>
 </TR>
</TABLE>Now, &nbsp;the volume of the given pyramid is &nbsp;{{{V}}} = {{{1/3}}}{{{144}}}.{{{6sqrt(2)}}} = {{{288sqrt(2)}}} = 407.29 {{{cm^3}}} (approximately).

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

It means that the angle &nbsp;<I>L</I><B>OAP</B>&nbsp; is of &nbsp;45�.

<B>Answer</B>. &nbsp;The volume of the given pyramid is &nbsp;{{{288sqrt(2)}}} = {{{407.29}}} {{{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>Example 3</H3>Find the volume of a regular hexagonal pyramid if its base edge is of &nbsp;4 cm&nbsp; and the height of the pyramid is of &nbsp;6 cm &nbsp;(<B>Figure 5</B>).

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

First, &nbsp;let us find the base area 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;&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;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

Since the base is a regular hexagon, &nbsp;its area is

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

Now, &nbsp;the volume of the pyramid is

{{{V}}} = {{{1/3}}}{{{S[base]}}}.{{{h}}} = {{{1/3}}}.{{{24}}}{{{sqrt(3)}}}.{{{6}}} = {{{48}}}{{{sqrt(3))}}} = {{{48}}}*{{{1.732}}} = 83.138 {{{cm^3}}} (approximately).

<B>Answer</B>. &nbsp;The volume of the pyramid is &nbsp;83.138 {{{cm^2}}} (approximately).

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

red(line(   2.25, 1.2,  2.25,  7.15)),

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

red(locate ( 2.35, 4.0,  6)),

locate ( 2.1,  1.15, O),
            locate ( 2.1,  7.65, P),
            locate (-1.35, 0.95, 4)
)}}}

&nbsp;&nbsp;&nbsp;&nbsp;<B>Figure 5</B>. To the &nbsp;<B>Example 3</B>

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

<H3>Example 4</H3>Find the volume of a regular tetrahedron if all its edges are of &nbsp;10 cm&nbsp; long &nbsp;(<B>Figure 6a</B>).

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

The base area of the given tetrahedron is the area of the equilateral triangle with the&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

side measure of &nbsp;10cm. &nbsp;So, &nbsp;the base area is equal to &nbsp;{{{S[base]}}} = {{{1/2}}}.{{{10}}}.{{{10}}}{{{sqrt(3)/2}}} = {{{25}}}{{{sqrt(3)}}}.

Next, &nbsp;let us find the measure of the height of the given tetrahedron &nbsp;(<B>Figure 6b</B>).

The height &nbsp;<B>OP</B>&nbsp; of the given tetrahedron drops to the center &nbsp;<B>O</B>&nbsp; of its base, &nbsp;which

is the intersection point of the base altitudes, &nbsp;medians and angle bisectors.

It is well known fact of &nbsp;<B>Planimetry</B>&nbsp; that the intersection point of medians of a

triangle divides them in proportion &nbsp;2:1&nbsp; counting from the vertices &nbsp;(see the lesson

<A HREF=http://www.algebra.com/algebra/homework/Triangles/Medians-of-a-triangle-are-concurrent.lesson>Medians of a triangle are concurrent</A>&nbsp; in this site).

Thus the segment &nbsp;<B>OA</B>&nbsp; in &nbsp;<B>Figure 6b</B>&nbsp; has the length of two third of the altitude

of the base triangle, &nbsp;i.e. &nbsp;|<B>OA</B>| = {{{2/3}}}.{{{10}}}{{{sqrt(3)/2}}} = {{{10*sqrt(3)/3}}}.

</TD>
  <TD> 
{{{drawing( 250, 225,  -2.5, 4.5, -0.8, 8.2,

line (  0.0, 0.0,  4.0,  0.0),
            line (  0.0, 0.0, -2.0,  1.5),
      green(line ( -2.0,   1.5,  4.0,  0.0)),

line (  0.667, 7.0,  0.0,  0.0),
            line (  0.667, 7.0,  4.0,  0.0),

line (  0.667, 7.0, -2.0,  1.5),

locate( 1.4, -0.1, 10),
            locate(-1.7,  0.9, 10),
            locate( 1.2,  1.3, 10),

locate( 0.4,  3.7, 10),
            locate( 2.5,  3.9, 10),
            locate(-1.3,  4.4, 10)
)}}}

&nbsp;&nbsp;&nbsp;&nbsp;<B>Figure 6a</B>. To the &nbsp;<B>Example 4</B>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

</TD>
  <TD> 
{{{drawing( 250, 225,  -2.5, 4.5, -0.8, 8.2,

line (  0.0, 0.0,  4.0,  0.0),
            line (  0.0, 0.0, -2.0,  1.5),
      green(line ( -2.0,   1.5,  4.0,  0.0)),

line (  0.667, 7.0,  0.0,  0.0),
            line (  0.667, 7.0,  4.0,  0.0),

line (  0.667, 7.0, -2.0,  1.5),

locate( 1.4, -0.1, 10),
            locate(-1.7,  0.9, 10),
            locate( 1.2,  1.3, 10),

locate(-0.2,  3.7, 10),
            locate( 2.5,  3.9, 10),
            locate(-1.3,  4.4, 10),

locate (-0.15, 0.0, A),
            locate ( 0.60, 7.6, P),
            locate ( 0.73, 0.7, O),

red(line (0.667, 0.5, 0.667, 7.0)),
        red(line (0.667, 0.5, 0.0, 0.0))
)}}}

&nbsp;&nbsp;&nbsp;&nbsp;<B>Figure 6b</B>. To the solution 
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;of the &nbsp;<B>Example 4</B>

</TD>
 </TR>
</TABLE>Now, &nbsp;the height of the pyramid is &nbsp;{{{h}}} = {{{sqrt(abs(AP)^2 - abs(OP)^2)}}} = {{{sqrt(10^2 - (10*sqrt(3)/3)^2)}}} = {{{sqrt(100 - 100/3)}}} = {{{10*sqrt(2)/sqrt(3)}}},

and the volume of our tetrahedron is {{{V}}} = {{{1/3}}}.{{{S[base]}}}.{{{h}}} = {{{1/3}}}.{{{25}}}{{{sqrt(3)}}}.{{{10*sqrt(2)/sqrt(3)}}} = {{{250*sqrt(2)/3}}} = {{{10^3*sqrt(2)/12}}} = 117.85 {{{cm^3}}} (approximately).

<B>Answer</B>. &nbsp;The volume of the given tetrahedron is &nbsp;{{{10^3*sqrt(2)/12}}} = 117.85 {{{cm^3}}} (approximately).

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(I intentionally presented the answer via the dimension of the tetrahedron's edge).

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;The general formula for the volume of a regular tetrahedron with the edge size &nbsp;{{{l}}}&nbsp; is &nbsp;{{{l^3*sqrt(2)/12}}}. &nbsp;You can prove this formula yourself using the same arguments).

<H3>Example 5</H3>Find the volume of a composite solid body of a "diamond" shape which comprises of two regular tetrahedrons joined face to face &nbsp;(<B>Figure 7</B>), &nbsp;if all their edges are

of &nbsp;4 cm long.

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

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

whose two faces are joined and overposed each to the other &nbsp;(<B>Figure 7</B>).

The total volume of our solid body is twice the volume of the single regular

tetrahedron with the side measure of &nbsp;4 cm. &nbsp;The later is equal to

{{{4^3*sqrt(2)/12}}}, &nbsp;according to the solution of the &nbsp;<B>Example 4</B>&nbsp; above.

Thus the volume of the composite body under consideration is

{{{2}}}.{{{4^3*sqrt(2)/12}}} = {{{128*sqrt(2)/12}}} = {{{32*sqrt(2)/3}}} = 15.085 {{{cm^3}}} (approximately).

</TD>
  <TD> 
{{{drawing( 160, 220,  -2.5, 4.5, -6.5, 8.0,

line (  0.0, 0.0,  4.0,  0.0),
            line (  0.0, 0.0, -2.0,  1.5),
      green(line ( -2.0,   1.5,  4.0,  0.0)),

line (  0.667, 7.0,  0.0,  0.0),
            line (  0.667, 7.0,  4.0,  0.0),

line (  0.667, 7.0, -2.0,  1.5),

locate( 1.4, -0.1, 4),
            locate(-1.1,  0.7, 4),
            locate( 1.2,  1.7, 4),

locate( 0.4,  3.7, 4),
            locate( 2.6,  3.9, 4),
            locate(-1.3,  4.4, 4),

line (  0.667, -5.4,  0.0,  0.0),
            line (  0.667, -5.4,  4.0,  0.0),
            line (  0.667, -5.4, -2.0,  1.5),

locate( 0.4, -2.0, 4),
            locate( 2.4, -2.4, 4),
            locate(-1.0, -2.0, 4)
)}}}

<B>Figure 7</B>. To the &nbsp;<B>Example 5</B>

</TD>
 </TR>
</TABLE><B>Answer</B>. &nbsp;The volume of the composite body under consideration is &nbsp;{{{2}}}.{{{4^3*sqrt(2)/12}}} = 15.085 {{{cm^3}}} (approximately).

<H3>Example 6</H3>Find the volume of a body obtained from the regular tetrahedron with the edge measure of &nbsp;10 cm&nbsp; after cutting off the part of the tetrahedron by the plane parallel to one of its faces in a way that the cutting plane bisects the three edges of the original tetrahedron &nbsp;(truncated tetrahedron, &nbsp;<B>Figure 7</B>).

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

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

with the edge measure of &nbsp;10 cm&nbsp; and then to distract the volume of the regular

tetrahedron with the edge measure of &nbsp;5 cm.

The volume of the original regular tetrahedron was found in the solution of the &nbsp;<B>Example 4</B>

above. &nbsp;It is equal to &nbsp;{{{10^3*sqrt(2)/12}}} = 117.85 {{{cm^3}}} (approximately).

The volume of the smaller regular tetrahedron with the edge size of 5 cm can be found using

similar formula with replacing the value of the edge size of &nbsp;10 cm&nbsp; by &nbsp;5 cm.

So, &nbsp;the volume of the smaller tetrahedron is &nbsp;{{{5^3*sqrt(2)/12}}} = 14.73 {{{cm^3}}} (approximately).

</TD>
  <TD> 
{{{drawing( 250, 225,  -2.5, 4.5, -0.8, 8.2,

line (  0.0, 0.0,  4.0,  0.0),
            line (  0.0, 0.0, -2.0,  1.5),
      green(line ( -2.0,   1.5,  4.0,  0.0)),

line (  0.667, 7.0,  0.0,  0.0),
            line (  0.667, 7.0,  4.0,  0.0),

line (  0.667, 7.0, -2.0,  1.5),

locate( 1.4, -0.1, 10),
            locate(-1.6,  0.9, 10),
            locate( 1.2,  1.3, 10),

locate( 0.3,  2.5, 5),
            locate( 3.0,  2.8, 5),
            locate(-1.6,  3.4, 5),

line (  0.333, 3.5,   2.333, 3.5),
            line (  0.333, 3.5,  -0.666, 4.25),
            line ( -0.666, 4.25,  2.333, 3.5),

green(line (  0.333, 3.5,   0.667, 7.0)),
      green(line ( -0.666, 4.25,  0.667, 7.0)),
      green(line (  2.333, 3.5,   0.667, 7.0)),

locate( 0.6,  5.5, 5),
            locate( 1.5,  5.8, 5),
            locate(-0.3,  6.1, 5)
)}}}

&nbsp;&nbsp;&nbsp;&nbsp;<B>Figure 7</B>. To the &nbsp;<B>Example 5</B>

</TD>
 </TR>
</TABLE>Now, &nbsp;the volume of the truncated tetrahedron is &nbsp;117.85 - 14.73 = 103.12 {{{cm^3}}} (approximately).

<B>Answer</B>. &nbsp;The volume of the body under consideration &nbsp;(truncated regular tetrahedron)&nbsp; is &nbsp;{{{10^3*sqrt(2)/12}}} - {{{5^3*sqrt(2)/12}}} = 103.12 {{{cm^3}}} (approximately).

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

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

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

Volume of pyramids

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

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

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

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

<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>.