Pyramid question

Victor

Can you solve the attached question? State what level you are at (higher, lower, 5th, 6th year).

Don't feel bad if you can't. Apparently its a difficult question.

mateo

I take one look at that and want to cry.

Nehpets

what are the area formulas? they're in the log tables right?

Creamy Goodness

yes, 2nd year comp science (bearly scraped a c3 (min req. for my course) though in 2005)

Mushy

Whats the answer? for the four sloping faces, its just area of triangle formula then that X4. so i got 24 metres squared. 6th year hons for me. In maths that is

PurpleFistMixer

Grr. I had written an entire explanation and my post disappeared...
Ahem.

What you want to find is the area of a face, which is 1/2 base by perpendicular height.
However, the H given in the quetsion isn't the perp. height of one of the faces (as they're slanted, so it'd actually be more)

You want to find what the length of the sloped edge of the pyramid is in order to find the perp. height of your triangles.

To do this (there may be a simpler way but this is what I did and it worked out okay...) look at the base of the triangle. 4 x 4 square. The midpoint of the square is where the H given is coming from.
Find out what the distance that point is from one of the corners (by constructing a triangle with the sides and pythagoras etc.)
This length will become the base of a new triangle which is perpendicular to the base of the pyramid, with height 3 as given in the question.

The hypotenuse of this triangle is the length of the sloped side, so find it.

Now with this sloped side, put it back onto the original triangle face. You have the sloped side and the base, so use pythagoras again to find the perpendicular height. Now you have the means to find the area. Multiply by 4 and you're done!


Hopefully that makes.. some sense. The answer I got for it was 8 root 13 m^2.

5th year honours maths for me. : )

Mushy

PurpleFistMixer, that would exactly be the way. And im in in 6th year:O. i just did it the first way i thought cos i really dont think of things

PurpleFistMixer

Yeah, I did it like that first but was then like.. no.. this is too easy.

PurpleFistMixer

PurpleFistMixer wrote:

Grr. I had written an entire explanation and my post disappeared...
Ahem.

What you want to find is the area of a face, which is 1/2 base by perpendicular height.
However, the H given in the quetsion isn't the perp. height of one of the faces (as they're slanted, so it'd actually be more)

You want to find what the length of the sloped edge of the pyramid is in order to find the perp. height of your triangles.

To do this (there may be a simpler way but this is what I did and it worked out okay...) look at the base of the triangle. 4 x 4 square. The midpoint of the square is where the H given is coming from.
Find out what the distance that point is from one of the corners (by constructing a triangle with the sides and pythagoras etc.)
This length will become the base of a new triangle which is perpendicular to the base of the pyramid, with height 3 as given in the question.

The hypotenuse of this triangle is the length of the sloped side, so find it.

Now with this sloped side, put it back onto the original triangle face. You have the sloped side and the base, so use pythagoras again to find the perpendicular height. Now you have the means to find the area. Multiply by 4 and you're done!


Hopefully that makes.. some sense. The answer I got for it was 8 root 13 m^2.

5th year honours maths for me. : )

Got the same answer, but was too chicken **** to put it up in case I made some stupid mistake!

Way I did it was that because the pyramid is symmetrical, you could take a cross-section connecting the midpoint of x on the left, to the bottom of 'h', to the midpoint of the side opposit it on the right (or the top/bottom, it doesn't matter but it has to be directly opposite it). This would give you a simple 2-D triangle like this:

/\

by which you could construct a smaller triangle inside with 2 as the base (half of x), 3 as the height, and the hypotenuse ending up as root13 (which is the sloping part), like this...

/|

Then because each of the faces don't bend, you could just use the simple 1/2 x base x height triangle area formula for one of the faces, so that would be

root13 x 4 x 0.5 = 2root13

Since there are 4 sides, just multiply by four.

Oh, and eh final year engineering.

Kwekubo

Repeat leaving, used the same method as you but I think I messed up the numbers. Got 6 root 17. Is there anything wrong with using the sine rule to find the distance from the centre of the pyramid's base to one of the corners?

Capt'n Midnight

Pity the base wasn't 8 instead of 4

then a side view of half the pyramid would give you a 3:4:5 triangle
so if each side would have been 8 wide and 5 up the side giving an area of
( 8*5/2 ) * 4 = 16*5 = 80

so each side of this one is 4 wide and sqrt( 2*2 + 3*3 )
so overall area is 4 x Sqrt(13) /2 *4 = 2 x Sqrt(13)

= Sqrt(54)

Limerick Dude

Questions like that make baby jesus cry

BobbyOLeary

The answer is (16 + 8 X root13) m^2. Don't forget about the bottom.

2nd Science (Physiology) A1 in the leaving though.

This is presuming this is a regular pyramid with the apex in the centre of the base. My calculations fall apart otherwise. This didn't seem all that comlicated, I've seen harder in the really old past papers.

:edit: My apologies I only spotted the 4 sloping faces thing now. The answer is then 8 X root13.

BobbyOLeary

Just a quick question. Who here is doing Technical Drawing, or has done Technical Graphics?

Victor

This is considered a difficult (by the students at least) higher Leaving Cert question.

PurpleFistMixer wrote:

The answer I got for it was 8 root 13 m^2.

Correct. As always.

Myth wrote:

Got the same answer, but was too chicken **** to put it up in case I made some stupid mistake!

Essentially the same method expressed differently

Capt'n Midnight wrote:

Pity the base wasn't 8 instead of 4

Sorry, my error.

BobbyOLeary wrote:

:edit: My apologies I only spotted the 4 sloping faces thing now. The answer is then 8 X root13.

Very important, always read the question asked.

Now who's up for a more difficult question?

declan_lgs

Yey for doing it in the head in a few secs while listening to music without looking at any other posts \o

The PRISM people didn't send the papers out to my school today. Now I know why .. they don't want competition!!

PurpleFistMixer

You win the award for not showing off. Congratulations.

Now to post in another thread Victor posted in, and my stalking is done for the night.

Capt'n Midnight

Victor wrote:

Sorry, my error.

would have been nice though, giving a nice round number as the answer, when everyone else was using sines / square roots. (third level maths btw)

Kwekubo

I got a thought-provoking question yesterday in an admissions test for medicine in England, so I thought I'd put it to the boardsies at large. It's meant to be based on GCSE knowledge of Maths (ie slightly above Junior Cert).

nicked from another forum wrote:

Two people doing a hill walk for a holiday. The walk consists of a flat piece of land, then uphill, then back downhill and then back across the flat piece of land. Total distance is 20 km.

Speed on flat land was 4km/h (i think)
Speed uphill is 3 km/h
Speed downhill is 6km/h

How long does the walk take in hours?

Oh, and if this were the real exam, you would have approximately 1 minute starting from... now

Victor

Hmmm....

First answer. How much of the journey is hill and how much is flat? If the flat is 0, then the walk takes 5 hours.

Second answer. The amount of flat is irrelevant as, even if the entire journey was flat, the the walk takes 5 hours. However, this is dependent on the starting and finishing altitudes being the same.

declan_lgs

Hunty wrote:

Sry for the septuple post but the moderater is definitely gay!!!

And?

Get the **** outta here.