
Susan C. answered 11/14/15
Tutor
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(31)
I love math, and I love to teach it.
Dear Sanjay,
The title of this problem says, "Circular Cylinder," but the story problem says, "right circular cone." These are two different objects. The right cylinder is like a soda can, but the right circular cone is like the snow cone that holds the ice. I am going to assume that you mean the snow cone shape with a perpendicular height from the peak or vertex to the circular base. The circular base is one piece, which is silver. The curved piece that makes the cone part is the second part.
Given: r=21 h=28
Area of circular base, A=πr2 =π(21)2 = 138.23 sq. cm. 138.23 * $1.00= $138.23 to paint the circular base
$1= 1sq. cm for the silver base
Area of curved surface all the way up the height=lateral area= π*r*s, where s =slant height, which is the length
of a side. Since that side could be the hypotenuse of a right triangle, we have s=√(r2+h2).
Substitute values into the equation: s=√(212+(28)2) =35 cm lateral length
lateral area= π*r*s π (21)(35)= 735(π)=2,309.07 Sq. cm 2309.07*$2=__________
$2= 1 sq. cm for the golden part
cost of circular base + cost of curved area all the way up to 14 cm= ______________Total cost
I will let you complete the problem. If I helped you, please give me a thumbs up.
Susan C.