maths question

[SPIRIT]

what is the formula to find out the height of a pyramid? (USING PYTHAGORAS)


the length of each side is 10
and the length of each side of the square is 8:?

[watson]

what is the formula to find out the height of a pyramid? (USING PYTHAGORAS)


the length of each side is 10
and the length of each side of the square is 8:?
Sorry arty,
That all doesn't make sense to me ....yet

[SPIRIT]

a drink container, in the shape of a box, measures 20cm by 10 cm by 4 cm. how long must a drinking straw be so as to reach 5 cm out of the box???:?:?

[Sleeping Dog]

I'm reading that the the square base measures 8 on a side, and the length from the base corner to the apex is 10...? is this correct Arty??

[Ivan in Oz]

How about 8.246

[Ivan in Oz]

Or is this for an assignment,
then you will require the workings:p

[watson]

G'day sleepingDog,
Arty, is this what you mean???

[Blocklayer]

Have a go at this for the trig calculations

http://www.visualtrig.com

To get the height of a 4 sided pyramid from side lengths and length up corner, work out the triangles.

Draw the 2 diagonals across the square base. The distance from a corner to the center point, is the Base of the triangle (use visualtrig). The Diagonal of the triangle is the length up the pyramid at the corner (10 ?)
Enter these 2 values into visualtrig and it will show the height and formulas to calculate it.

.

[ian]

what is the formula to find out the height of a pyramid? (USING PYTHAGORAS)


the length of each side is 10
and the length of each side of the square is 8:?you have a square pyramid.
you measure the height using triangles
let the distance to the centre of the square from one corner be A, you have a right triangle where
A(squared) + A(squared) = 8(squared) = 64, so A(squared) = 32

then determine the height B, again using a triangle where the base is A, and the hypotenuse is 10,
again
A(squared) + B(squared) = 10(squared),
32 + B(squared) = 100
B(squared) = 100-32 = 68, therefore B is approximately 8.1


ian

[Ivan in Oz]

Where was this going?
Other than to give a few headaches to some:doh:

Oh!!:~


B(squared) = 100-32 = 68, therefore B is approximately 8.1
ian

Square Root of 68 is Closer to 8.24621125123532:rolleyes:

Rounds to 8.246:D

C'Mon someone else do the Straw, P L E A S E !

[ian]

a drink container, in the shape of a box, measures 20cm by 10 cm by 4 cm. how long must a drinking straw be so as to reach 5 cm out of the box???:?:?if the straw is placed in one corner, 25cm

ian

[Wongo]

What they are trying to say is....:D

[Ashore]

Where was this going?
Other than to give a few headaches to some:doh:

Oh!!:~


Square Root of 68 is Closer to 8.24621125123532:rolleyes:

Rounds to 8.246:D

C'Mon someone else do the Straw, P L E A S E !
Spot on :2tsup:

[Ivan in Oz]

Bit more than 25cm me thinks

27.7156cm:)

[Ashore]

but if the box is lying flat ie 4cm high 20cm long and 10cm wide and the hole was in the centre of the top face then the straw could be 9cm long to reach the bottom with 5cm out or 17.247 cm long to reach the furtherest corner with 5cm sticking out :D

[Wongo]

9cm:p

[autogenous]

the length of each side is 10
and the length of each side of the square is 8If half base is at the centre of the pyramid is 4

The hypotenuse is 10

The centre of the pyramid is 90 degrees

c² = 10*10;
b² = 4*4;
a² = rise*rise;

c²=a²+b²;

100=c²;
16=b²;

84=100-16;

9.165151389911680013176094387456 = √ 84;

height = 9.165151389911680013176094387456


its too late to write this properly and do it properly:C

[autogenous]

the length of each side is 10if your talking about the hip and not the jack Ill kill you!:o:U

[Ashore]

Mate he asked for the height of the pryamid not the length from the centre of one side to the apex , check out wongo's drawing
Rgds

[Gingermick]

Now wongo, if the pyramid was attached to a lathe 20mm offset at the apex and also spinning on its own, what is the rate of change of the angle between the lathe centre and one corner of the square base. (lathe going at 100rpm and box spinning at 1 rpm)

[Ivan in Oz]

> Now wongo, if the pyramid was attached to a lathe 20m offset at the apex
< SNIP >
and box spinning at 1 rpm)

OK!!!

NOW I DO have that Headache I was thinking of earlier.

[Gingermick]

20mm offset twas meant to be:)

[Wongo]

Dude I don't have a lathe so how would I know? :shrug:

[Gingermick]

come on now, that's an easy question. :D
The harder question would be the rate of change of the volume of the pyramid defined by base of the spinning pyramid with the spindle of the lathe as its apex. (in mm³/s)

that almost made my head explode just thinking about it:oo:

[Wongo]

Mick, stop talking with your backside. :smack:

[Gingermick]

:Cwhats wrong with applied calculus. If ever a student asks 'What will we ever need this for?' you can tell em to work out the rate of change of volume on a spinning offset pyramid. It very useful information if you want to know that sort of thing. Even if you're not a teacher

[Wongo]

I'll let you do it Mick.:cool:

:D

[Gingermick]

wus

[SPIRIT]

How about 8.246the winner for the maths quizz ,sorry all l had to go after posting question ,no tell the truth l wasn't home and my little pincess rang me on the mobile so l told her to start a math thread to help her with her maths homework :)
not to talk to wongo is the only advice l was able give her on the matter of maths, turn her into some pen in top shirt pocket pen pusher, button pushing clerk:U

[autogenous]

Care to post the math or formula? :)

[ian]

Bit more than 25cm me thinks

27.7156cm:)Ivan
read the question again, carefully,

there's no mention of where the straw leaves the top of the box,
if the box is open at the top, my answer is correct
and if the box is sealed like a "popa" with the hole essentially in one corner 25cm is still correct

also it's only an assumption (on the part of the reader) that the straw must reach to the bottom of the box
so the straw could be 11cm long, (6 in the box, 5 out of it)


ian

[Incoming!]

Just did a quick google for the formula, and noticed that someone caused similar headaches and confusion on the abc.net forum last night as well, with this same question. Troublemakers!