I am unsure how to really go about the first problem as my e

I am unsure how to really go about the first problem... as my end result for part be is something like h= 3r/(sqr.(r^2-9)).. Can someone please help. And if possible, check problem #2? I am pretty sure I am doing it right. Thank you so much!

Solution

a)

lateral surface area of a right cone is L. S. A. = rl, where l is the slant height of the cone.

Volume of cone = lateral surface area:

(1/3)pi*r^2h = pi*r*l

(1/3)r*h = l

r = 3l/h

b) lateral surface area of a right cone is L. S. A. = rl, where l is the slant height of the cone.

l = sqrt(r^2 +h^2)

So Lateral surface area = rsqrt(r^2 + h^2)

Volume of cone = (1/3)r^2h

Volume of cone = lateral surface area:

(1/3)r^2h = rsqrt(r^2 + h^2)

(1/3)rh = sqrt(r^2 +h^2)

square both sides: (1/9)r^2h^2 = (r^2 +h^2)

r^2h^2 = 9r^2 + 9h^2

h^2 ( r^2 - 9) = 9r^2

h^2 = 9r^2/ ( r^2 - 9)

h = 3r/sqrt( r^2 -9)

2.. The solution written on your sheet is correct.