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thehay95

Member

Registered: 2018-10-18

Posts: 1

Website

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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Chuyên cung cấp người giúp việc nhà, gioi thieu dich vu giup viec nha gia re

Bob

Administrator

Registered: 2010-06-20

Posts: 10,623

Re: Volume of irregular polyhedrons

hi thehay95

Welcome to the forum.

I've had a go at sketching this solid.  Trouble is that just having the letters PQRSTUV doesn't exactly tell me which points to join to which.  It looks like there's a prism PQRSTV, which has a triangular cross section QRV, and sitting on this is a rectangular based pyramid TVRSU, where the base is TVRS and the vertex U.

Volume of the prism should be fairly straight forward.  Calculate the area of the triangle QRV and multiply by the length, PQ.

The pyramid is more tricky.  You'll need the area of the base, TVRS.  This is a rectangle.  You need Pythag. to calculate VR.  Then the perpendicular height from TVRS to U.  If you draw the triangle UZY, you'll see that UZ is not perpendicular to ZY so that length won't do.  But you can calculate UY and the angle UYZ and so treating UY as a hypotenuse you can get the required height by opposite (height) = hyp x sin(UYZ).  Once you have this the volume of the pyramid is  one third of the area of base times the perpendicular height.

Then just add together the two volumes. 

Hope that helps,

image upload problem.  I have a diagram but I cannot log in to imgur to upload it.  Some problem with a Vodafone security certificate ??? I'm on BT broadband.  Anyone got any suggestions.  It happens with both IE11 and Chrome. My account still exists as old images are still showing on the forum

Bob

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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

Mathegocart

Member

Registered: 2012-04-29

Posts: 2,226

Re: Volume of irregular polyhedrons

hi thehay95

Welcome to the forum.

I've had a go at sketching this solid.  Trouble is that just having the letters PQRSTUV doesn't exactly tell me which points to join to which.  It looks like there's a prism PQRSTV, which has a triangular cross section QRV, and sitting on this is a rectangular based pyramid TVRSU, where the base is TVRS and the vertex U.

Volume of the prism should be fairly straight forward.  Calculate the area of the triangle QRV and multiply by the length, PQ.

The pyramid is more tricky.  You'll need the area of the base, TVRS.  This is a rectangle.  You need Pythag. to calculate VR.  Then the perpendicular height from TVRS to U.  If you draw the triangle UZY, you'll see that UZ is not perpendicular to ZY so that length won't do.  But you can calculate UY and the angle UYZ and so treating UY as a hypotenuse you can get the required height by opposite (height) = hyp x sin(UYZ).  Once you have this the volume of the pyramid is  one third of the area of base times the perpendicular height.

Then just add together the two volumes.  :)

Hope that helps,

image upload problem.  I have a diagram but I cannot log in to imgur to upload it.  Some problem with a Vodafone security certificate ??? I'm on BT broadband.  Anyone got any suggestions.  It happens with both IE11 and Chrome. My account still exists as old images are still showing on the forum

Bob

I'm not acquaintanced with Vodafone and how it all works in the UK,  but when I had these irritating issues, I logged into my router(type it's ip in your search bar), chose the Internet/DNS page, and set DNS to manual. Then enter the Google DNSs, i.e 8.8.8.8 and 8.8.4.4. That seemed to work for me but you should contact Vodafone to see what's up..

EDIT 1: These are the IPv6 addresses..  2001:4860:4860::8888 and 2001:4860:4860::8844.

Last edited by Mathegocart (2018-10-20 00:27:41)

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The integral of hope is reality.
May bobbym have a wonderful time in the pearly gates of heaven.
He will be sorely missed.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,623

Re: Volume of irregular polyhedrons

Hi Mathegocart,

Thanks for the help.

I'd forgotten I'm on someone else's wi fi at the moment. It's their content blocker.  Seems to be a common problem with Vodafone. So I'll wait and see if the OP wants my diagram before tinkering with the settings.

Bob

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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

mr.wong

Member

Registered: 2015-12-01

Posts: 252

Re: Volume of irregular polyhedrons

Hi  thehay95  ,

I  think  your  thread  was  taken  from  my  thread  with  the  same 
name  posted  2  years  ago .

This  problem  had  already  been  solved  with  the  help  of 
bob  bundy , you  can  search  the  thread  and  find  the  solution .

Still  thankful  to  bob  bundy  !