3D Trigonometry and Planes

A waste paper basket has a top ABCD that is a square of side 30cm and a base that is a square of side 20. The line joining the centres of the top and base PQRS is perpendicular to both and is 40cm long.
I've found the diagonals AC and PR of each base.
How would one find the length AP ??

Reply 1

TeeEm

Original post by Mr Pussyfoot

Iimagine the point E which lies on the diagonal AC. directly above P.
Draw the "vertical" triangle AEP
Pythagoras on it

Reply 2

Mr Pussyfoot

Thanks a lot, however I found that this was not possible as AEP would not be a right angles triangle as length AP is not vertical. Instead, I found the length below A by doing 30-20 and since the perpendicular height Is 40 I could use Pythagoras.

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Reply 3

ghostwalker

Original post by Mr Pussyfoot

Thanks a lot, however I found that this was not possible as AEP would not be a right angles triangle as length AP is not vertical.

It's EP that is vertical, and AE horizontal, hence a rightangled triangle.

Instead, I found the length below A by doing 30-20 and since the perpendicular height Is 40 I could use Pythagoras.

Can't see how that would work.

Reply 4

Mr Pussyfoot

ImageUploadedByStudent Room1432482966.636902.jpg

really?

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Reply 5

ghostwalker

Original post by Mr Pussyfoot

really?

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Yes, really.

(edited 8 years ago)

ImageUploadedByStudent Room1432482966.636902.jpg63.9KB

Reply 6

Mr Pussyfoot

So what would AE be?

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Reply 7

ghostwalker

Original post by Mr Pussyfoot

So what would AE be?

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By Pythagoras

AC=30\sqrt{2}

And similar triangles (drop a perpendicular from E to AB if you need)

AE=5\sqrt{2}

Added a diagram to my previous post.

Reply 8

Mr Pussyfoot

That does seem to make more sense, but which two triangles are similar??? Can't seem to picture it in my mind

Reply 9

ghostwalker

Original post by Mr Pussyfoot

That does seem to make more sense, but which two triangles are similar??? Can't seem to picture it in my mind

Draw an overhead view and drop the perpendicular I suggest.

ABC and A(foot of perpendicular)E.

PS: Quote if you want a reply, as it then comes up under "Notifications"

(edited 8 years ago)

Reply 10

Mr Pussyfoot

Original post by ghostwalker

Draw an overhead view and drop the perpendicular I suggest.

ABC and A(foot of perpendicular)E.

PS: Quote if you want a reply, as it then comes up under "Notifications"

😂😂 Thanks for the tip, and you mean that AE forms a triangle when you imagine a perpendicular to AB? if so this is similar to ABC but I still don't see how that came to 5 root 2

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Reply 11

ghostwalker

Original post by Mr Pussyfoot

Thanks for the tip, and you mean that AE forms a triangle when you imagine a perpendicular to AB? if so this is similar to ABC but I still don't see how that came to 5 root 2

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A, E, together with the foot of the perpendicular from E to AB - I'll call it F - forms one triangle.

So, AFE and ABC are similiar.

Hopefully you have drawn a diagram with the overhead view, and the length of AF should be clear.

So, AF: AB :: AE : AC

Reply 12

Mr Pussyfoot

Original post by ghostwalker

ImageUploadedByStudent Room1432486585.479491.jpg

So this would be what it would look like?

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Reply 13

ghostwalker

Original post by Mr Pussyfoot

So this would be what it would look like?

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Not quite. You need the smaller lower square in as well to see what's happening clearly - note that it has the same centre as the upper square. AF will be 5.

Reply 14

Mr Pussyfoot

Original post by ghostwalker

Not quite. You need the smaller lower square in as well to see what's happening clearly - note that it has the same centre as the upper square. AF will be 5.

Oh I think understand what you meant: AF is 5cm because of (30-20)/2 and this gives AF:AB---> 1:6. Similarly AE will be 1/6 of 30√2??

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Reply 15

ghostwalker

Original post by Mr Pussyfoot

Oh I think understand what you meant: AF is 5cm because of (30-20)/2 and this gives AF:AB---> 1:6. Similarly AE will be 1/6 of 30√2??

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Yep.