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Mathematics 15 Online
OpenStudy (anonymous):

I am willing to give a MEDAL on this one

OpenStudy (anonymous):

A. Determine the exact total surface area of a sphere with radius square root of 2 metres

OpenStudy (anonymous):

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?

OpenStudy (anonymous):

A 11.8432

OpenStudy (anonymous):

Most importantly I'm looking for answer B.

OpenStudy (anonymous):

In addition A. Is 8Pi metres

OpenStudy (jack1):

surface area of a sphere = 4 pi r^2

OpenStudy (anonymous):

Your right on that one

OpenStudy (anonymous):

But I'm finding it hard to draw the whole figure

OpenStudy (jack1):

the hint about center of its base allows you to find the vertical height of the cone, so (sqrt 2) - 0.5m

OpenStudy (anonymous):

Would that mean that the height of the cone is 0.5m?

OpenStudy (anonymous):

How could one draw it?

OpenStudy (jack1):

so you have a side length and you have the vertical length, so use pythagoras theorum to get the radius of the base

OpenStudy (jack1):

will draw now, standby

OpenStudy (anonymous):

Thanks a lot jack. Nobody was able to answer this question on this forum. It's been circulating around for two days :)

OpenStudy (jack1):

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

OpenStudy (anonymous):

Is there some other way were we could use trigonometric functions?

OpenStudy (jack1):

|dw:1365813230801:dw|

OpenStudy (anonymous):

What is the chord for?

OpenStudy (jack1):

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

OpenStudy (anonymous):

Does the chord split into the middle of h?

OpenStudy (jack1):

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

OpenStudy (anonymous):

Your answer is correct but could you please explain it more to me

OpenStudy (jack1):

|dw:1365814024479:dw|

OpenStudy (jack1):

yep, no worries

OpenStudy (anonymous):

Particularly the chord

OpenStudy (anonymous):

Thanks jack

OpenStudy (jack1):

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

OpenStudy (jack1):

so all that means is that the the side length of the cone is irrelevant for our purposes

OpenStudy (anonymous):

Could you please draw it

OpenStudy (jack1):

next, it sits on top of the sphere, so as a sphere is a uniform object

OpenStudy (jack1):

sure

OpenStudy (jack1):

|dw:1365814523126:dw|

OpenStudy (jack1):

|dw:1365814617842:dw|

OpenStudy (anonymous):

Oh now I see logically the radius of the sphere is larger than the radius of the cone

OpenStudy (anonymous):

In the first graph I mean

OpenStudy (jack1):

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

OpenStudy (jack1):

yep, u got it

OpenStudy (anonymous):

Mmmmmm

OpenStudy (anonymous):

I'm starting to get it now

OpenStudy (jack1):

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)

OpenStudy (anonymous):

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

OpenStudy (jack1):

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

OpenStudy (jack1):

|dw:1365815168121:dw|

OpenStudy (anonymous):

Would you know of a graph that I can find off the web

OpenStudy (jack1):

hang on, ill google mathhelper

OpenStudy (anonymous):

Your the best mate

OpenStudy (jack1):

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

OpenStudy (anonymous):

I drew it on a piece of paper

OpenStudy (anonymous):

So basically the pointy side of the cone is standing in the centre of the sphere?

OpenStudy (jack1):

OpenStudy (jack1):

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

OpenStudy (jack1):

so all those lines are just triangulation, not the cone itself

OpenStudy (anonymous):

Can I find a video tutorial on this

OpenStudy (jack1):

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)

OpenStudy (jack1):

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

OpenStudy (jack1):

and not sure about videos, will aska a mod, standby @amistre64 @Mertsj

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