Who is able to solve this difficult question!!! Open cones are made from nets cut from a large sheet of paper 1.2m x 1.0m. If a cone has a radius of 6cm and a slant height of 10cm, how many cones can be made from the sheet? (Assume there is negligible wastage of paper.)
Hint: the lateral surface area of a cone is LSA = pi*r*s where pi = 3.14 (roughly) r = radius of cone s = slant height of cone
R is given as 6cm S is given as 10cm
So in this question we aren't meant to use the formula SA=Pi(r)(s) + Pi(r)^2?
Only the lateral SA?
yes but this is an OPEN cone, so there is no base
so this is like an ice cream cone
Okay now I get it. So after we get the answer of the LSA we divide the area of the sheet by LSA to see how many cones are made from the sheet?
exactly
this is if you somehow manage to squeeze the nets of the cones in an optimal fashion wasting as little paper as possible
I've got another question that has been bugging me for quite a long time
ok
what is it
It's somehow related
If the surface area of a sphere to a cylinder is in the ratio 4:3 and the sphere has a radius of 3a, calculate the radius of the cylinder if the radius of the cylinder is equal to its height
The sphere has a radius of 3a, so r = 3a and the surface area is SA = 4*pi*r^2 SA = 4*pi*(3a)^2 SA = 4*pi*9a^2 SA = 36a^2*pi
the surface area of a sphere to a cylinder is in the ratio 4:3 so (SA of Sphere)/(SA of Cylinder) = 4/3 (36a^2*pi)/(SA of Cylinder) = 4/3 3*(36a^2*pi) = 4(SA of Cylinder) 108a^2*pi = 4(SA of Cylinder) SA of Cylinder = 108a^2*pi/4 SA of Cylinder = 27a^2*pi
SA of Cylinder SA = 2*pi*r^2 + 2*pi*r*h SA = 2*pi*r^2 + 2*pi*r*r .... height = radius SA = 2*pi*r^2 + 2*pi*r^2 SA = 4*pi*r^2 27a^2*pi = 4*pi*r^2 solve for r
Mate you don't know how much I appreciate your help
I posted these questions a few times but nobody knew how to answer them
I'm going to reward you a medal
you're welcome
To be sure is the answer r=3a square root of 3 /2 ???
27a^2*pi = 4*pi*r^2 27a^2 = 4*r^2 (27a^2)/4 = r^2 r^2 = (27a^2)/4 r = sqrt( (27a^2)/4 ) r = (3a/2)*sqrt(3) so you are correct
The answer in the textbook must be incorrect
what does it say
It has as an answer r=3*square root of 3a /2
\[\Large r = \frac{3\sqrt{3a}}{2}\] or \[\Large r = \frac{3\sqrt{3}a}{2}\]
basically I'm asking if the 'a' is in the square root or not
The second one
then you can rearrange the terms to get \[\Large r = \frac{3\sqrt{3}a}{2}\] \[\Large r = \frac{3a\sqrt{3}}{2}\]
which is the same as what we got
bad use of notation though
A is in it. I guess they made a slight mistake
oh 'a' is in the square root?
Yeh
maybe it's a typo or they didn't mean to write it in the root
it should be outside though
I agree
I'm going through a lot of these questions. They are sometimes tricky and hard to figure out
yeah i would just bring it up with the teacher and see what s/he says about it
Yeah I should. Thanks a lot any ways
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