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#1 2016-03-01 23:54:28

mr.wong
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Registered: 2015-12-01
Posts: 252

Volume of irregular polyhedrons

(I) Let  X  denotes  a  polyhedron  with  vertices   PQRSTUV  where
the  base  PQRS  is  a  square  with  sides   1/2  unit  . TP  and  VQ  are 
straight  lines  each  with  length  1/8 unit   and  perpendicular   to  PQRS .
W  is  the  mid-point  of  PQ  while  Y  is  the  mid-point  of  RS . UW with
length  1/4  unit  is  perpendicular  to  PQRS  also . Z  is  the  mid-point  of
TV .( also  of  UW )
Find  the  volume  of  X .

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#2 2016-03-02 20:51:00

Bob
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Registered: 2010-06-20
Posts: 10,053

Re: Volume of irregular polyhedrons

hi mr.wong

I cannot get a sensible polyhedron here.  I've drawn the square PQRS and constructed the other points.  P, T, U, V, Z, W and Q are all in a plane perpendicular to PQRS. Y lies in PQRS.

I've joined P to Q to R to S to T to U to V to P.  That doesn't make a solid.  And why tell me about Y as it doesn't feature in the problem?  I think I'm missing something here. dizzy

As any polygon from some of these points will make a pyramid, I think I can get a volume when I know what the polyhedron is.

Please reply.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#3 2016-03-03 16:47:49

mr.wong
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Registered: 2015-12-01
Posts: 252

Re: Volume of irregular polyhedrons

Hi  bob  bundy , thanks  for  your  reply !

You  should  not  join  V  to  P  , instead  you  should  join  V  to  R , and 
also  S  to  T . If  you  like  you  may  also  join  U  to  S  and  U  to  R .
Really  the  point  Y  is  not  necessary  in  the  problem . I  am  sorry 
for  my  statement  has  not  been  clear  enough , is  it  better  to  denote
the  polyhedron  by  PQRSTUVQP  ?

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#4 2016-03-03 20:44:44

Bob
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Registered: 2010-06-20
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Re: Volume of irregular polyhedrons

hi mr.wong

Thanks.  That's helped a lot.  I think this is the correct diagram.  I haven't made it to scale.  As well as the 'perspective view' on the left, I've also shown a side view looking along the line PQ.

i370JgK.gif

I think you can treat this a two separate shapes:  PQVTRS is a triangular prism with cross section QRV and length PQ; and a pyramid TVRSU with base TVRS and vertex U.

For the prism, work out the area of the cross section and multiply by the length.

For the pyramid, work out the area of the base and multiply by 1/3 of the perpendicular height.  You need to do some calculations to get the measurements for this.  In the rectangle TVRS calculate VR.  The perpendicular height is the length of a line that is at right angles to ZY and measured from U.  Take care to use the right distances here.  On the side view the true length of U to 'RYS' will be UY so you will need the point Y after all smile .  You can use trig. to get angles ZYW and UYW and hence calculate UYZ.  From there you can calculate the distance UX where X is the 'foot' of the perpendicular from U to the line YZ extended, as you will know UY and angle UYZ.   

Hope that's enough to get you through this. 

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#5 2016-03-05 20:30:33

Bob
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Re: Volume of irregular polyhedrons

Were you hoping I'd check this?

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#6 2016-03-05 20:47:57

mr.wong
Member
Registered: 2015-12-01
Posts: 252

Re: Volume of irregular polyhedrons

Hi  bob  bundy ,

Thanks  much  to  your  analysis  to  the  problem  and  the  nice  diagrams !
Let  us  consider  the  prism  firstly . Since  the  area  of  Δ TPS  = 1/2 *
1/2 * 1/8  sq.unit  while  the  height  = 1/2  unit , thus  the  volume
of  the  prism = 1/64 ( about  0.0156 ) cu.unit 
Then  for  the  pyramid , ST  can  be  found  to  be  √(17/64 ) unit ,
Thus  the  area  of  the  base  STVR = 1/2 * √(17/64 ) =  (√17)/16  sq.unit .
But  for  the  height , it  is  too  bad  for  me  to  find  its  value  with  trigonometry  for  I  have  forgotten  most  of  the   trigonometry !  I  have  to  find  the  volume  of  the  pyramid  by  other  means .
Let  us  cut  the  pyramid  into  2  halves  through  Δ UYZ , we  choose  the
portion  UYSTZ , also  a  pyramid , and  denoted  by  H .
For  H  we  shall  cut  it  into  2  parts  again   through   Δ USZ ,
part  1  , the  pyramid  YSUZ  will  be  denoted  by  H1 , while
part  2 , the  pyramid  TSUZ , will  be  denoted  by  H2 .
For  H1 , the  point  S  will  be  considered  as  the  new  vertex  with  Δ UYZ  as  base . As  SY  being  perpendicular  to  Δ UYZ ,
thus  the  height  of  H1  will  be  1/4  unit  .
Since  the  area  of  Δ  YWU = 1/2 * 1/2 * 1/4  = 1/16  sq.unit 
while  the  area  of  Δ  YWZ = 1/32  sq.unit , thus  the  area  of
Δ UYZ = 1/32  sq.unit  also . (  In  fact  since  YW  ⊥ UW , the
area  of  Δ UYZ  may  be  found  directly  to  be  1/2 * 1/2 * 1/8 =
1/32  sq.unit .) Thus  the  volume  of  H1 = 1/3 * 1/4 * 1/32 =
1/384  cu. unit .
For  H2 , the  point  U  will  be  considered  as  the  new  vertex 
with  Δ STZ  as  base .
The  height  of  H2 , UZ = 1/8 unit , while  the  area  of  Δ STZ  =
1/2 * 1/4 * √(17/64 )  = (√17) / 64   sq. unit . Thus  the  volume  of  H2  = 1/3 * 1/8 * (√17) / 64   = (√17) / 1536  cu.unit  .
So  the  volume  of  H  =  H1 + H2  = 1/384 + (√17) / 1536 
= ( 4 + √17)/ 1536  cu.unit
Therefore  the  volume  of  the  original  pyramid  = ( 4 + √17)/ 768  cu .unit
( about   0.01  cu.unit )
( Since  the  area  of  the  original  pyramid = (√17)/16  sq.unit . Thus  the 
height  with  U  as  vertex  = 1/( 4√17)  + ( 1/16 ) unit  ( about  0.12 unit ) )
Lastly  the  volume  of  the  original  polyhedron  will  be  found  to  be
1/64 + 0.01 = 0.016 + 0.01 =  0.026  cu.unit  (approximately )

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#7 2016-03-05 23:45:40

Bob
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Registered: 2010-06-20
Posts: 10,053

Re: Volume of irregular polyhedrons

That's what I get too!  smile

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#8 2016-03-06 14:54:42

mr.wong
Member
Registered: 2015-12-01
Posts: 252

Re: Volume of irregular polyhedrons

Thanks  bob  bundy  !

Here  is  another  problem , wish  you  will  help .

Problem (II) 
Let  H  denotes  a  polyhedron  PQRSTUVWXYZ  where  the  base  is  a
rt. ∟.   Δ PRT   with  PR = PT  = 1/2  unit .  Q, S  and  U  are 
mid-points  of  PR , RT  and  TP  respectively .VP , XQ , WR ,YS  and
ZU  are  all  ⊥ to  Δ PRT , with  VP = WR = YS = ZU = 1/4 un. and
XQ = 1/2 un.
Find  the  volume  of  H .

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#9 2016-03-06 20:21:27

Bob
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Registered: 2010-06-20
Posts: 10,053

Re: Volume of irregular polyhedrons

hi

When the question just gives a list of the vertices like this, it is not obvious which points are meant to join to which.  Here is my guess at a diagram:

VhmbkGy.gif

It looks to me like a prism PUSR/VZYW, a pyramid VZYW/X and a pyramid USYZ/T

P doesn't look like a right angle here because of the perspective.

So get the areas of the shaded regions and it should be straight forward from there.

Note.  The line XT lies in the plane of the two triangles so I could have omitted it to simplify.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#10 2016-03-07 21:41:23

mr.wong
Member
Registered: 2015-12-01
Posts: 252

Re: Volume of irregular polyhedrons

Hi  bob  bundy ,

Your  diagram  is  indeed  what  I  want . Originally  I  expect  you  will 
arrange  the  vertices  in  an  anti-clockwise  direction ,  and  with  the 
pentagon PVXWR  facing  me . Now  with  your  diagram  I  can  look  from
the  back  and  obtain  more  integral  impression  of  the  polyhedron .
The  whole  polyhedron  , as  you  said , can  be  divided  into  3  portions :
(i)  pyramid USYZ/T ,  will  be  denoted  by  H1 .
(ii)  pyramid VZYW/X ,  will  be  denoted  by  H2 .
(iii)  prism  PUSR/VZYW ,  with  cross - section  being  a  trapezium , will  be
       denoted  by  H3 .
I  shall  calculate  their  volumes  one  by  one .
Thanks  again  and  I  promise  next  time  I  will  make  the  statement  more
clearly .

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#11 2016-03-08 15:20:50

mr.wong
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Registered: 2015-12-01
Posts: 252

Re: Volume of irregular polyhedrons

(i) For  H1 , area  of  the  base = 1/4* 1/4 = 1/16 sq. unit   while
  height = 1/4 un  , thus  volume  of  H1 = 1/3* 1/4 * 1/16 = 1/192 cu.un
(ii) For  H2 , area  of  the  base = ( 1/4 + 1/2 )/2  * 1/4 = 3/32 sq.un  while
   height = 1/4 un  , thus  volume  of  H2 = 1/3 * 1/4 * 3/32 = 1/128 cu.un
(iii) For  H3 , area  of  the  base = 3/32 sq.un  while  height =  1/4 un ,
    thus  volume  of  H3 = 1/4 * 3/32 =  3/128 cu. un
To  sum  up , volume  of  H = 1/192 + 1/128 + 3/128 = 7/192 cu.un
(about  0.036)

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#12 2016-03-08 20:46:12

Bob
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Registered: 2010-06-20
Posts: 10,053

Re: Volume of irregular polyhedrons

hi,

Yes I get that too.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#13 2016-03-09 14:42:46

mr.wong
Member
Registered: 2015-12-01
Posts: 252

Re: Volume of irregular polyhedrons

Hi  bob  bundy ,

Another  problem  is  waiting ! Thanks  in  advance .
Problem  (III)
Let  H  denotes  a  polyhedron  PQRSTUP/VWXYZ  where  the  base  is  a
rt. ∟.   Δ PRT   with  PR = PT  = 1/2  unit .  Q, S  and  U  are 
mid-points  of  PR , RT  and  TP  respectively .VQ,  WR ,XS  ,YT and
ZU  are  all  ⊥ to  Δ PRT , with  VQ = WR = YT = ZU = 1/4 un.
and  XS = 1/2 un. ; X and Z , Z and P , P and  V , V  and  X  are 
joined . Notice  that  XP  and  ZV  are  co-planed . XZPV  seems  to  be  a
kite  with  XZ = XV  and  ZP = PV . ( While  XZTY  in  Problem (II) should  be
a  parallelogram  with  unequal  adjacent  sides .)
Find  the  volume  of  H .

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#14 2016-03-09 21:04:50

Bob
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Registered: 2010-06-20
Posts: 10,053

Re: Volume of irregular polyhedrons

GHRq0zM.gif

US = ZM = PU = UZ = MX (by midpoints )

so PZ = ZX.

That kite is therefore a rhombus (special case of a kite).

The method is similar to the previous question.

How many more of these have you got hidden away?

Have you got geogebra yet?  You can download it free and then make your own diagrams.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#15 2016-03-10 17:50:29

mr.wong
Member
Registered: 2015-12-01
Posts: 252

Re: Volume of irregular polyhedrons

Hi  bob  bundy ,

You  are  right  , XVPZX  is  a  rhombus , in  addition ,
both  XVZ  and  VPZ  are  equilateral  Δs .
Let  (i)  H1  denotes  the  pyramid  P/QUZV  ;
      (ii)  H2  denotes  the  pyramid  X/VZYW    and
     (iii) H3  denotes  the  prism  QUTR/VZYW
For  (i)  , height = 1/2 *√1/8  (=  √1/32  un.) ,
               area  of  base  = √1/8 * 1/4  sq . un .
Thus  vol.  of  H1 =  1/3 * 1/2* √1/8  * √1/8 * 1/4
      =  1/192  cu.un.
For (ii)  , height  = 1/4 un. ,  area  of  base = (√1/8 +√1/2)/2  *  1/2 *√1/8 
         =  3/32  sq.un.
Thus  vol. of  H2 = 1/3 * 1/4* 3 * √1/8 * 1/2*1/2 *√1/8
= 1/128  cu un .   
For  (iii) , height = 1/4 un , area  of  base = 3/32  sq un .
Thus  vol. of  H3 = 1/4 * 3/32 = 3/128  cu un
  To  sum  up , volume  of  H = 1/192 + 1/128 + 3/128
= 7/192 cu.un (about  0.036)  [  exactly  the  same  as  Problem (II) ]

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#16 2016-03-12 15:45:27

mr.wong
Member
Registered: 2015-12-01
Posts: 252

Re: Volume of irregular polyhedrons

Hi  bob  bundy ,

Here  is  another  problem . Before  I  have  learnt  the  Geogebra , please  help  me  once  more !

Problem ( 3 T ) :
Let  PQRSTUP / VWXYZKV  denotes  a  polyhedron  with 
rt. ∟.   Δ PRT  as  base   where  PR = PT  = 1/2  unit Q, S  and  U  are  mid-points  of  PR , RT  and  TP  respectively .VP , WQ , XR ,YS  , ZT  and  KU are  all  ⊥ to  Δ PRT , with  VP = XR = ZT = 1/6 un. and  WQ = YS = KU
= 7/24  un.  WYK forms  a  triangle.
Find  the  volume  of  the  polyhedron .

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#17 2016-03-14 20:49:34

Bob
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Registered: 2010-06-20
Posts: 10,053

Re: Volume of irregular polyhedrons

VT5p6Zp.gif

I've added three more points, L, M and N on the same level as V, X and Z (see red dotted lines)

This looks to me like a triangular prism PRT/VXZ, another LMN/KWY, and three pyramids WKLM/V, KYNL/Z and WYNM/X.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#18 2016-03-15 14:49:50

mr.wong
Member
Registered: 2015-12-01
Posts: 252

Re: Volume of irregular polyhedrons

Hi  bob  bundy ,

Many  thanks  again !
Now  let  us  cut  the  polyhedron  into  2  layers  along  VMXNZLV ,
the  lower  part  is  a  triangular  prism  with  height  1/6  un  and  area  1/2* 1/2* 1/2
= 1/8  sq un . Thus  its  vol  =  1/6 * 1/8 = 1/48  cu un .
We  shall  denote  the  upper  part  by  H , which  can  further  be  dissected  into  4  portions .
The  1st  one  being  a  pyramid , V/WKLM , will  be  denoted  by  H1 .
The  2nd  one  also  a  pyramid , Z/KYNL , will  be  denoted  by  H2 .
The  3rd  one  also  a  pyramid , X/ WYNM , will  be  denoted  by  H3 .
The  4th  portion  remained  is  a  prism , LMN/KWY, will  be  denoted  by  H4 .
The  height  of  H1  is  found  to  be  √1/32 = 1/2 * √1/8  un , while  the  base
is  a  rectangle  with  area  1/8 * √1/8  sq un ,  thus  vol  of  H1 = 1/3 * 1/2 * √1/8  *1/8 * √1/8 
=  1/6 * 1/8 * 1/8 = 1/384  cu un
The  height  of  H2  is  1/4  un , while  the  base  is  also  a  rectangle  with  area  1/4 * 1/8  sq un ,
thus  vol  of  H2 = 1/3 * 1/4  * 1/4 * 1/8  = 1/384  cu un
The  height  and  area  of  H3  is  the  same  as  H2 , thus  the  vol  of  H3 = 1/384 cu un .
The  height  of  H4  is  1/8 un , while  the  base   being  a  triangle  with  area  1/2 * 1/4 * 1/4
=  1/ 32  sq un , thus  the  vol  of  H4 = 1/ 256  cu  un .
Thus  the  total  volume  of  the  polyhedron  = 1/48 + 1/384 + 1/384 + 1/384 + 1/256
= 1/48 + 1/128 + 1/256  = 16/768 + 6/768 + 3/768  = 25/ 768  cu un .
Divided  by  the  area  of  the  base  of  the  polyhedron , we  also  get  its  average  height
= (25/768)/ (1/8) =  25 / 96 un . ( about  0.26 )                                                                                          while  the  square  of  0.26  is  about  0.0677 .

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#19 2016-03-25 15:48:33

mr.wong
Member
Registered: 2015-12-01
Posts: 252

Re: Volume of irregular polyhedrons

Volume  of  polyhedron  consisting  of  non-linear  curves .

Related  problem  ( S1 ) :
Let   T/ PQRS  denotes  a  pyramid  where  PQRS  is  a  square  with 
each  side  1/2  unit .  TP  =  1/4  un.  being  perpendicular  to  the  base .
Both  TQ  and  TS  are  straight  lines  while  TR  is  a  quadratic  curve
with  equation   z =  1/4 ( 1-2x) ( 1-2y)  where  PQ  denotes  the  x-axis ,
PS  denotes  the  y-axis  and  PT  denotes  the  z-axis .
How  to  find  the  volume  of  the  pyramid ?
( Perhaps  we  can  cut  the  pyramid  into  2  halves  at  TPR  to  get
2  pyramids  S/ TPR  and  Q/TPR ,  but  how  to  find  the  area  of  the
base  TPR ? )

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#20 2016-03-29 21:32:56

mr.wong
Member
Registered: 2015-12-01
Posts: 252

Re: Volume of irregular polyhedrons

To  find  the  area  of  the  quasi-triangle  TPR , let  us  at  first  re-arrange
the  equation  z =  1/4 ( 1-2x) ( 1-2y)  to  z  =   1/4 (2x-1) (2y-1) .
Since  PR  is  symmetric  in  x  and  y , thus  z =  1/4 (2x-1) (2y-1)  may
be  simplified  to   z =  1/4 (2x-1) (2x-1)  =  1/4 ( 2x-1) ^ 2 .
Let  PR  denotes  the  w-axis  ,  we  can  see  that  w = √2 * x ,
i.e. x = w / √2 .
Thus  z = 1/4 ( 2w/ √2  -  1) ^ 2 .  i.e.  z = 1/4 [(√2) w -1] ^2 .
The  length  of   PR  may  be  found  to  be  √1/2  un . Thus
the  area  of  the  quasi-triangle  TPR  may  be  found  by  ∫z dw  from
0  to  √1/2  = 1/4 * 1/3 ( √2  w - 1)^ 3
                    = 1/12 * [ (√2 * √1/2  - 1) ^ 3  -( √2 * 0 - 1)  ^ 3  ]
                    = 1/12 * [( 1-1) ^ 3  -  (-1)^ 3  ]
                    = 1/12 * 1
                    =  1/12  sq. un.
Assume  that  the  volume  of pyramid  S/ TPR   may  be  calculated  as  1/3 * 1/4 * 1/12  = 1/144 cu. un . Thus  twice  the  volume = 1/72 cu.un .
being  the  volume  of  pyramid  T/ PQRS  .
Dividing  by  the  base  PQRS  , we  also  get  its  average  height 
to  be ( 1/72 ) / (1/4) =  1/18 un . ( about  0.056  )

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#21 2016-06-27 17:18:56

thickhead
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Registered: 2016-04-16
Posts: 1,086

Re: Volume of irregular polyhedrons

mr.wong wrote:

Hi  bob  bundy ,

Thanks  much  to  your  analysis  to  the  problem  and  the  nice  diagrams !
Let  us  consider  the  prism  firstly . Since  the  area  of  Δ TPS  = 1/2 *
1/2 * 1/8  sq.unit  while  the  height  = 1/2  unit , thus  the  volume
of  the  prism = 1/64 ( about  0.0156 ) cu.unit 
Then  for  the  pyramid , ST  can  be  found  to  be  √(17/64 ) unit ,
Thus  the  area  of  the  base  STVR = 1/2 * √(17/64 ) =  (√17)/16  sq.unit .
But  for  the  height , it  is  too  bad  for  me  to  find  its  value  with  trigonometry  for  I  have  forgotten  most  of  the   trigonometry !  I  have  to  find  the  volume  of  the  pyramid  by  other  means .
Let  us  cut  the  pyramid  into  2  halves  through  Δ UYZ , we  choose  the
portion  UYSTZ , also  a  pyramid , and  denoted  by  H .
For  H  we  shall  cut  it  into  2  parts  again   through   Δ USZ ,
part  1  , the  pyramid  YSUZ  will  be  denoted  by  H1 , while
part  2 , the  pyramid  TSUZ , will  be  denoted  by  H2 .
For  H1 , the  point  S  will  be  considered  as  the  new  vertex  with  Δ UYZ  as  base . As  SY  being  perpendicular  to  Δ UYZ ,
thus  the  height  of  H1  will  be  1/4  unit  .
Since  the  area  of  Δ  YWU = 1/2 * 1/2 * 1/4  = 1/16  sq.unit 
while  the  area  of  Δ  YWZ = 1/32  sq.unit , thus  the  area  of
Δ UYZ = 1/32  sq.unit  also . (  In  fact  since  YW  ⊥ UW , the
area  of  Δ UYZ  may  be  found  directly  to  be  1/2 * 1/2 * 1/8 =
1/32  sq.unit .) Thus  the  volume  of  H1 = 1/3 * 1/4 * 1/32 =
1/384  cu. unit .
For  H2 , the  point  U  will  be  considered  as  the  new  vertex 
with  Δ STZ  as  base .
The  height  of  H2 , UZ = 1/8 unit , while  the  area  of  Δ STZ  =
1/2 * 1/4 * √(17/64 )  = (√17) / 64   sq. unit . Thus  the  volume  of  H2  = 1/3 * 1/8 * (√17) / 64   = (√17) / 1536  cu.unit  .
So  the  volume  of  H  =  H1 + H2  = 1/384 + (√17) / 1536 
= ( 4 + √17)/ 1536  cu.unit
Therefore  the  volume  of  the  original  pyramid  = ( 4 + √17)/ 768  cu .unit
( about   0.01  cu.unit )
( Since  the  area  of  the  original  pyramid = (√17)/16  sq.unit . Thus  the 
height  with  U  as  vertex  = 1/( 4√17)  + ( 1/16 ) unit  ( about  0.12 unit ) )
Lastly  the  volume  of  the  original  polyhedron  will  be  found  to  be
1/64 + 0.01 = 0.016 + 0.01 =  0.026  cu.unit  (approximately )

The volume of pyramid is 1/96 and that of polyhedron =5/192 to be exact.


{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}

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#22 2016-06-27 23:02:53

mr.wong
Member
Registered: 2015-12-01
Posts: 252

Re: Volume of irregular polyhedrons

Hi  thickhead ,

Does  this  mean  ( 4 + √17)/ 768  can  be  simplified  to 
1/96  ?

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#23 2016-06-28 01:55:24

thickhead
Member
Registered: 2016-04-16
Posts: 1,086

Re: Volume of irregular polyhedrons

The volume of pyramid=


{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}

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#24 2016-06-28 04:32:35

thickhead
Member
Registered: 2016-04-16
Posts: 1,086

Re: Volume of irregular polyhedrons

Your solution seemed too long for a simple solution and I had not gone through it. But now I find you have taken the height of pyramid as UZ but you have to take perpendicular distance of U from the inclined plane i.e draw a perpendicular from U to ST and take its length.


{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}

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#25 2016-06-28 15:54:45

mr.wong
Member
Registered: 2015-12-01
Posts: 252

Re: Volume of irregular polyhedrons

Thanks  thickhead,

Due  to  my  weakness  in  trigonometry  , I  solve  the 
problem  in  that  clumsy  method  instead  of  your  concise 
way  with  trigonometry  thus  error  will  easily  occur !

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