I am willing to give a MEDAL on this one
A. Determine the exact total surface area of a sphere with radius square root of 2 metres
B. an inverted cone with side length 4 metres is placed on top of the sphere so that the centre of its base is 0.5 metres above the centre of the sphere. Find the radius of the cone exactly?
A 11.8432
Most importantly I'm looking for answer B.
In addition A. Is 8Pi metres
surface area of a sphere = 4 pi r^2
Your right on that one
But I'm finding it hard to draw the whole figure
the hint about center of its base allows you to find the vertical height of the cone, so (sqrt 2) - 0.5m
Would that mean that the height of the cone is 0.5m?
How could one draw it?
so you have a side length and you have the vertical length, so use pythagoras theorum to get the radius of the base
will draw now, standby
Thanks a lot jack. Nobody was able to answer this question on this forum. It's been circulating around for two days :)
no, dammit im wrong, sorry, thats not the vertcal height, sorry, is only vertical height to top of sphere, not top of cone, my bad
Is there some other way were we could use trigonometric functions?
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What is the chord for?
I've re-read the q, think i have better understanding now so back to pythag thrm: r = sqrt 2 h = 0.5 m (above center of sphere) c = r - h, but pretty much irrelevant for what we're doing (its the vertical length from the chord to the surface of the sphere the chord length is the key
Does the chord split into the middle of h?
using pythag: h^2 = a^2 + b^2 => r^2 = 0.5 ^2 + x ^2 2 = 0.25 + x^2 x= sqrt 1.75 ~~approx 1.323m
Your answer is correct but could you please explain it more to me
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yep, no worries
Particularly the chord
Thanks jack
so we know that the cone has an opening with a radius less than that of the radius of the sphere. If not, instead of sitting on top of the the sphere like a party hat, the sphere would be engulfed by the cone until it reaches a point where the 2 circumference are equal ie a party hat on a shot put vs a party hat on a marble, the marble would roll up the hat till it hit a tight fit
so all that means is that the the side length of the cone is irrelevant for our purposes
Could you please draw it
next, it sits on top of the sphere, so as a sphere is a uniform object
sure
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Oh now I see logically the radius of the sphere is larger than the radius of the cone
In the first graph I mean
for example 2, the side length matters as we would have to do further calculations to work out how high up the cone it sits and how much further down the base of the cone is so therefore how much larger the hole is
yep, u got it
Mmmmmm
I'm starting to get it now
so its a uniform object, so we can mentally bisect the sphere where it joins the cone and work with that we're given that the base sits 0.5 m the center (one side of a triangle) the hypotenuse we get from the radius (the distance from the edge of any line bisecting a sphere from the origin/certer is the radius)
So when the inverted cone is sitting on the sphere the base of it is above the centre of the sphere by 0.5 m. How could you put that on the graph
so we split the chord line (where we bisected) knowing that see the 2nd image i drew: its not to scale but h=0.5
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Would you know of a graph that I can find off the web
hang on, ill google mathhelper
Your the best mate
http://mathhelpforum.com/geometry/184571-finding-diameter-slice-sphere.html this is pretty much bang on, and explains better than I do check out the pic down the bottom of the page
I drew it on a piece of paper
So basically the pointy side of the cone is standing in the centre of the sphere?
no, not at all man, the base of the cone is on top of the sphere like a hat, and we know how high up on the sphere it is resting (0.5m), then we use geometry and triangulation to work out that to be sitting at THAT height exactly, the circumference of the base of a cone is: x
so all those lines are just triangulation, not the cone itself
Can I find a video tutorial on this
if the cone had a base of 2 x pi x sqrt 2, it would sit on the sphere covering exactly half of it (circumference = 2 pi r)
and from circumference we work our radius and vice versa in ours we just worked out the radius of the base of the cone using triangulation
and not sure about videos, will aska a mod, standby @amistre64 @Mertsj
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