Lesson Solved problems on volume of pyramids

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Solved problems on volume of pyramids


In this lesson you will find typical solved problems on volume of pyramids.
The theoretical base for these problems is the lesson  Volume of pyramids  under the topic  Area and surface area  of the section  Geometry  in this site.

Problem 1

Find the volume of a triangular pyramid  ABCD  if its edges issued from the vertex  A  are of  8 cm,  6 cm  and  6 cm  long and each of these three edges is perpendicular
to the two others  (Figure 1).

Solution

First,  let us find the area of the triangle  DELTAABC  at the base of the pyramid                            
(Figure 1).

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

S%5Bbase%5D = S%5BABC%5D = 1%2F2abs%28AB%29.abs%28AC%29 = 1%2F26.6 = 18 cm%5E2.

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

V = 1%2F3S%5Bbase%5D.h = 1%2F3.18.8 = 48 cm%5E3.

Answer.  The volume of the given pyramid is  48 cm%5E3.



  Figure 1. To the  Problem 1      


Problem 2

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

Solution

First,  let us find the area of the square  ABCD  at the base of the pyramid                                
(Figure 2).

This area is squared length of the base edge:

S%5Bbase%5D = S%5BABCD%5D = abs%28AB%29.abs%28AC%29 = 6.6 = 36 cm%5E2.

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

V = 1%2F3S%5Bbase%5D.h = 1%2F3.36.8 = 96 cm%5E3.

Answer.  The volume of the given pyramid is  96 cm%5E3.



  Figure 2. To the  Problem 2


Problem 3

Find the volume of a regular pyramid with the square base  (Figure 3a)  if the lateral edge of the pyramid has the same measure of  10 cm  as the the base edge has.
Also find the angle between the lateral edge and the base of the pyramid.

Solution

First,  let us find the area of the base of the given pyramid.
Since the base is a square with the side measure of  10 cm,  the area of the base is                    
S%5Bbase%5D = 10%5E2 = 100 cm%5E2.

Next,  let us find the height of the pyramid.
For it,  let us consider the triangle  DELTAAOP  (Figure 3b),  where the point  A  is one
of the base vertices of the pyramid,  the point  O  is the center of the square base
of the pyramid,  and the point  P  is the pyramid vertex.

It is a right-angled triangle  (the segment  OP  is the height of the pyramid).  Its
leg  AO  is half of the diagonal of the square base,  and its measure is 10sqrt%282%29%2F2 = 5sqrt%282%29.
Therefore,  the measure of the height  OP  is

|OP| = sqrt%28abs%28AP%29%5E2+-+abs%28AO%29%5E2%29 = sqrt%2810%5E2+-+%285sqrt%282%29%29%5E2%29 = sqrt%28100+-+50%29 = sqrt%2850%29 = 5sqrt%282%29 cm.


Figure 3a. To the Problem 3        



  Figure 3b. To the solution
      of the Problem 3
Now,  the volume of the given pyramid is  V = 1%2F3100.5sqrt%282%29 = 500sqrt%282%29%2F3 = 235.70 cm%5E3 (approximately).

On the way,  we proved that the right-angled triangle  DELTAAOP  is isosceles: |OP| = |AO|.
It means that the angle  LOAP  is of 45°.

Answer.  The volume of the given pyramid is  500sqrt%282%29%2F3 = 235.70 cm%5E3 (approximately).
               The angle between the lateral edge and the base of the pyramid is of  45°.


Problem 4

Find the volume of a regular hexagonal pyramid if the base edge has the measure of  4 cm  and the lateral edge of the pyramid is of  8 cm  long  (Figure 4a).
Also find the angle between the lateral edge and the base plane of the pyramid.

Solution

Let the point  O  be the center of the regular hexagon at the base of the pyramid,
the point  A  be one of the hexagon's vertices,  and the point  P  be the vertex
of the pyramid  (Figure 4b).

The area of the regular hexagon at the base of the pyramid is
S = 6*1%2F2.4*4sqrt%283%29%2F2 = 24sqrt%283%29 cm%5E2.

Next,  since the hexagon at the base is regular,  the segment  OA  connecting the                    
hexagon's center with its vertex has the same length as the hexagon side,  i.e.  4 cm.

So,  in the right-angled triangle  DELTAAOP  the leg  OA  is of  4 cm  long and the
hypotenuse  AP  is of  8 cm  long.

It implies,  in particular,  that the angle  LOAP  is of  60°.

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

|OP| = sqrt%28abs%28AP%29%5E2+-+abs%28OA%29%5E2%29 = sqrt%288%5E2+-+4%5E2%29 = sqrt%2864+-+16%29 = sqrt%2848%29 = 4sqrt%283%29 cm.



    Figure 4a. To the  Problem 4      




    Figure 4b. To the solution
        of the  Problem 4
Therefore,  the volume of the prism is  V = 1%2F3Sh = 1%2F3.24sqrt%283%29.4sqrt%283%29 = 96 cm%5E3.

Answer.  The volume of the prism is  96 cm%5E3.
               The angle between the lateral edge and the base plane of the pyramid is of  60°.


Problem 5

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  (Figure 5),
if all their edges are of  4 cm.

Solution

We are given a 3D body of a "diamond" shape comprised of two regular rectangular                          
pyramids whose two bases are joined and overposed each to the other  (Figure 5).

In total,  the volume of our solid body is two times the volume of the regular pyramid
with all edge measures of  4 cm.  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  S = 4%5E2 = 16 cm%5E2.
To get the pyramid's height,  note that the diagonal of the square at the base of the
pyramid is  4sqrt%282%29 cm long,  and half of the diagonal has half of this measure,
i.e.  2sqrt%282%29 cm.
Therefore,  the pyramid's height is  h = sqrt%284%5E2+-+%282sqrt%282%29%29%5E2%29 = sqrt%2816-8%29 = sqrt%288%29 = 2sqrt%282%29 cm.


Figure 5. To the Problem 5
For the detailed explanation on this calculation see the solution of the  Problem 3 above.

Now,  the volume of the single pyramid is  V = 1%2F3Sh = 1%2F3.16.2sqrt%282%29 = %2832sqrt%282%29%29%2F3,  and the volume of the composite solid body under consideration is  2.%2832sqrt%282%29%29%2F3 = %2864sqrt%282%29%29%2F3 = 30.17 cm%5E3 (approximately).

Answer.  The volume of the composite body under consideration is  %2864sqrt%282%29%29%2F3%29 = 30.17 cm%5E3 (approximately).


Problem 6

Find the volume of a body obtained from a regular rectangular pyramid with the edge measures of  10 cm   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  (truncated pyramid,  Figure 6a).

Solution

The strategy solving this problem is to find first the volume of the regular rectangular                      
pyramid with the edge measures of  10 cm  and then to distract the volume of the
regular rectangular pyramid with the edge measures of  5 cm.

The base area of the original regular rectangular pyramid is  10%5E2 = 100 cm%5E2.
The height of the original regular rectangular pyramid is
sqrt%2810%5E2+-+%28%2810sqrt%282%29%29%2F2%29%5E2%29 = sqrt%28100+-+%285sqrt%282%29%29%5E2%29 = sqrt%28100+-+50%29 = sqrt%2850%29 = 5sqrt%282%29 cm.
This calculation is based on consideration of the right-angled triangle  DELTAOAP
(Figure 6b).  Hence,  the volume of the original pyramid is  1%2F3100.5sqrt%282%29 = %28500sqrt%282%29%29%2F3 cm%5E3.



Figure 6a. To the  Problem 6      




    Figure 6b. To the solution
        of the  Problem 6

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

Thus,  the base area of the small regular rectangular pyramid is  5%5E2 = 25 cm%5E2.
The height of the small regular rectangular pyramid is
sqrt%285%5E2+-+%28%285sqrt%282%29%29%2F2%29%5E2%29 = sqrt%2825+-+%28%285sqrt%282%29%29%2F2%29%5E2%29 = sqrt%2825+-+12.5%29 = sqrt%2812.5%29 = %285sqrt%282%29%29%2F2 cm.
So,  the volume of the smaller rectangular pyramid is  1%2F325%285sqrt%282%29%29%2F2 = %28125sqrt%282%29%29%2F6 cm%5E2.

Now,  the volume of the truncated pyramid is  %28500sqrt%282%29%29%2F3 - %28125sqrt%282%29%29%2F6 = %28875sqrt%282%29%29%2F6 = 206.24 cm%5E3 (approximately).

Answer.  The lateral surface area of the body under consideration is  206.24 cm%5E3 (approximately).


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

Lessons on volume of prisms

Volume of prisms
Solved problems on volume of prisms
Overview of lessons on volume of prisms                    

Lessons on volume of pyramids

Volume of pyramids
Solved problems on volume of pyramids
Overview of lessons on volume of pyramids

Lessons on volume of cylinders

Volume of cylinders
Solved problems on volume of cylinders
Overview of lessons on volume of cylinders                

Lessons on volume of cones

Volume of cones
Solved problems on volume of cones
Overview of lessons on volume of cones                    

Lessons on volume of spheres

Volume of spheres
Solved problems on volume of spheres
Overview of lessons on volume of spheres


To navigate over all topics/lessons of the Online Geometry Textbook use this file/link  GEOMETRY - YOUR ONLINE TEXTBOOK.

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