I've been studying geometry from 11am-10pm(now) and the result I got isn't the one it should be....Please help me!!! PLEASE ^_^ In the triangle ABC AB=8 and BC=12,the angle between AB and BC is 60 degrees.The triangle rotates around BC.Find the volume of the 'thing' formed from this rotation :)
rotates around bc? really
O.o yes it does...I can draw a pic if u want....
k here i will tell u how to do it when u rotate a triangle about 1 of its sides u get a figure with two cones of same radius attached now try to find the hight of each cone and u have the radius which is hight of the triange from side BC so u will have the volume by the formula V of cone = (1/3)(pi)(r^2)h
like in a circle?
yea i know that thnx :) but the problem is that i am not sure...we can't count the area of the circle formed twice can we?
its not gonna count twice
so it must be V=V1+V2-piR^2?
or not?
volume top cone + volume bottom cone = volume of rotation
amistre is correct
when you press your hands together; do you subtract the volume created by them touching?
you add the volume of one to the other;
how abt the circle? O.o
a circle has a area not a volume
the circle is just a way to measure the volume for the top and the volume for the bottom
yea but it will be counted twice since the same cones have the same base right?
when....bases..touch.... they dont count as ....twice anything
when you press your hands together; no EXTRA volume is created to take away
when you stack a can of corn on top of another can of corn; you havent add any EXTRA volume by staking
the place where they touch, is just that .... the place where they touch
prolly what your thinking of is that you are used to seeing area as a surface that has some depth to it; when area is infinitely thin has zero depth
How the hell do you form 2 cones by rotating about one of the edges? The tip of the triangle (point A) will make a circle through space...
an edge make an axel in space
ever play with those claker balls on a stick?
Right, so all points on edges AB, an AC will form circles when rotated all the way about the BC axis
right
oooh, ok i was thinking about it wrong, i was thinking 2 cones with their tips being point A
why is there 2 cones?
ABC, AB = 8, BC = 12 < ABC = 60
becasue the triangle is spun on an edge makind a top and bottom cone
The circle that A makes through space is the base of both cones, One cone is formed by each half of the triangle rotating 360 degrees through space
i get a radius of 7; and a height of 4 for the top i get a radius of 7 and a height of 8 for the bottom part
you mean a right and left cone, not top and bottom
does it really matter how we hold the edge?
lol
(1/3) pi 49(4) + (1/3) pi 49(8) should be the volume round abt
615.75.... is the best I can do with truncating numbers
i dont follow the rotation, also your triangle doesnt make sense, can you label things
one sec
ok i drew my triangle a little differently , so first maybe its smarter to actually find the other angles and sides, so we dont just guess the shape
mines not to scale :)
Think about a flag, that is a right triangle. rotate that flag around the flag pole. You get one cone. The triangle we are dealing with is not a right triangle but can be divided in half and considered as 2 right triangles. So 2 cones.
so i get 8 sin 60 for the radius , 8 sin pi/3
8 sin pi/3 ~ 6.92
if we want accuracy; it should go \[\frac{pi (8sin(30))^2*4}{3}+\frac{pi (8sin(30))^2*8}{3}\]
how do you get 4
oh 8 sin pi/6
lol.... i did sin instead of cos :)
8sin(30) = height of top = 4
12 - 4 = 8 for height of bottom
ok , so change sin to cos in your formula
yes ...
ok looks good !
we can clean it up by factoring: \[\frac{4\pi (8cos(30))^2}{3}(1+2)\] right?
16 /3 pi ( 8 cos (pi/6))^2
woops
\[4\pi\ (8 cos(30))^2\] seems to be the simplest form
4/3 pi ( 8 cos ( pi/6))^2
grrr
:) 603.185.....
this interface is not forgiving
can i ask a quick question, given a focus and directrix, there is only one e such that e = 1 , so there is only one parabola
we dint use eccentricity much, so Id have to look it up; but libraries closing so I gotta move to the veranda :)
but there are many ellipses and parabolas
ok bbl ?
:(:(:(:(I'm sooooo sorry guys..:(:( chrome wasn't responding :(:(I'm sorry :(
Hmmmm......the V=192pi tht's wht the answer must be acc.to my math book
Thanks for trying anyway :) ^_^
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