Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

(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 .

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

Offline

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,408

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

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

Offline

**Mathegocart****Member**- Registered: 2012-04-29
- Posts: 1,954

bob bundy wrote:

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 forumBob

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)*

The integral of hope is reality.

May bobbym have a wonderful time in the pearly gates of heaven.

He will be sorely missed.

Offline

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,408

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

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

Offline

**mr.wong****Member**- Registered: 2015-12-01
- Posts: 250

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 !

Offline

Pages: **1**