document.write( "Question 499276: A cylindrical tin of height h cm and radius r cm, has a surface area, including its top and bottom, A cm^2.
\n" ); document.write( "i) Write down an expression a A in terms of r, h and pi.
\n" ); document.write( "I got this to be A = 2pi r h + 2pi r^2\r
\n" ); document.write( "\n" ); document.write( "ii) A tin of height 6cm has surface area 54picm^2. What is the radius of the tin?
\n" ); document.write( "I'm not sure how to work this part out. \r
\n" ); document.write( "\n" ); document.write( "iii) Another tin has the same diameter as height. Its surface area is 150picm^2. What is its radius? \r
\n" ); document.write( "\n" ); document.write( "I'd really appreciate your help.
\n" ); document.write( "h = 6
\n" ); document.write( " \n" ); document.write( "
Algebra.Com's Answer #337561 by Theo(13342)\"\" \"About 
You can put this solution on YOUR website!
question 1:
\n" ); document.write( "height = h
\n" ); document.write( "radius = r
\n" ); document.write( "surface area = the area of the sides of the can plus the area of top and bottom of the can.
\n" ); document.write( "area of the top and bottom of the can is equal to 2*(pi*r^2)
\n" ); document.write( "area of the side of the can is equal to h*(2*pi*r)
\n" ); document.write( "S = surface area of the can.
\n" ); document.write( "S = 2*pi*r^2 + 2*pi*r*h\r
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\n" ); document.write( "\n" ); document.write( "question 2:
\n" ); document.write( "you are given that S = 54*pi square centimeters.
\n" ); document.write( "you are given that h = 6 centimeters.
\n" ); document.write( "you want to find the radius.
\n" ); document.write( "the formula used is the same formula you just derived in question number 1.
\n" ); document.write( "that formula is:
\n" ); document.write( "S = 2*pi*r^2 + 2*pi*r*h
\n" ); document.write( "you know h, so substitute for h in the equation to get:
\n" ); document.write( "S = 2*pi*r^2 + 2*pi*r*6
\n" ); document.write( "simplify to get:
\n" ); document.write( "S = 2*pi*r^2 + 12*pi*r
\n" ); document.write( "you know the value of S, so substitute for S in the equation to get:
\n" ); document.write( "54*pi = 2*pi*r^2 + 12*pi*r
\n" ); document.write( "divide both sides of the equation by pi to get:
\n" ); document.write( "54 = 2*r^2 + 12*r
\n" ); document.write( "subtract 54 from both sides of the equation to get:
\n" ); document.write( "2r^2 + 12r - 54 = 0
\n" ); document.write( "divide both sides of the equation by 2 to get:
\n" ); document.write( "r^2 + 6r - 27 = 0
\n" ); document.write( "this is a quadratic equation in standard form.
\n" ); document.write( "factor this equation to get:
\n" ); document.write( "(r + 9) * (r - 3) = 0
\n" ); document.write( "this equation is true if (r+9) = 0 or if (r-3) = 0 or if both are 0.
\n" ); document.write( "solve for r+9 = 0 to get r = -9
\n" ); document.write( "solve for r-3 = 0 to get r = 3
\n" ); document.write( "r can't be negative so your answer has to be r = 3.
\n" ); document.write( "let's see if that's true.
\n" ); document.write( "your original equation is:
\n" ); document.write( "S = 54*pi
\n" ); document.write( "the formula is:
\n" ); document.write( "S = 2*pi*r^2 + 2*pi*r*h
\n" ); document.write( "you now know that:
\n" ); document.write( "h = 6
\n" ); document.write( "r = 3
\n" ); document.write( "the formula becomes:
\n" ); document.write( "S = 2*pi*3^2 + 2*pi*3*6
\n" ); document.write( "simplify to get:
\n" ); document.write( "S = 2*pi*9 + 2*pi*18
\n" ); document.write( "simplify further to get:
\n" ); document.write( "S = 18*pi + 36*pi
\n" ); document.write( "simplify further to get:
\n" ); document.write( "S = 54*pi
\n" ); document.write( "the value of 3 for r is good.
\n" ); document.write( "your answer is:
\n" ); document.write( "r = 3 cm\r
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\n" ); document.write( "\n" ); document.write( "question 3:
\n" ); document.write( "you are given that S = 150*pi square centimeters
\n" ); document.write( "same formula is used again.
\n" ); document.write( "formula is:
\n" ); document.write( "S = 2*pi*r^2 + 2*pi*r*h
\n" ); document.write( "h = height
\n" ); document.write( "r = radius
\n" ); document.write( "S = surface area
\n" ); document.write( "d = diameter
\n" ); document.write( "you are given that the diameter is equal to the height.
\n" ); document.write( "you get:
\n" ); document.write( "d = h
\n" ); document.write( "diameter is equal to twice the radius.
\n" ); document.write( "this leads to:
\n" ); document.write( "d = 2r
\n" ); document.write( "since h = d, this leads to:
\n" ); document.write( "h = 2r
\n" ); document.write( "we can substitute for h in the equation by replacing h with 2r to get:
\n" ); document.write( "S = 2*pi*r^2 + 2*pi*r*h becomes:
\n" ); document.write( "S = 2*pi*r^2 + 2*pi*r*2r
\n" ); document.write( "simplify this to get:
\n" ); document.write( "S = 2*pi*r^2 + 4*pi*r^2
\n" ); document.write( "these are now like terms so we can combine them to get:
\n" ); document.write( "S = 6*pi*r^2
\n" ); document.write( "we are given that S = 150 square cm.
\n" ); document.write( "we replace S with 150 to get:
\n" ); document.write( "150 = 6*pi*r^2
\n" ); document.write( "divide both sides of the equation by 6 to get:
\n" ); document.write( "25 = pi*r^2
\n" ); document.write( "divide both sides of the equation by pi to get:
\n" ); document.write( "25/pi = r^2
\n" ); document.write( "take the square root of both sides of the equation to get:
\n" ); document.write( "r = +/- sqrt(25/pi)
\n" ); document.write( "we can simplify this a little further to get:
\n" ); document.write( "r = +/- 5/sqrt(pi)
\n" ); document.write( "since r can't be negative, this then becomes:
\n" ); document.write( "r = 5/sqrt(pi)
\n" ); document.write( "to confirm this is a good number we start over with the additional information that r = 5/sqrt(pi)
\n" ); document.write( "our formula is, once again:
\n" ); document.write( "S = 2*pi*r^2 + 2*pi*r*h
\n" ); document.write( "we replace h with 2r to get:
\n" ); document.write( "S = 2*pi*r^2 + 2*pi*r*2r
\n" ); document.write( "we combine like terms to get:
\n" ); document.write( "S = 2*pi*r^2 + 4*pi*r^2
\n" ); document.write( "we know that r^2 = 25/pi, so we can replace r^2 with that to get:
\n" ); document.write( "S = 2*pi*25/pi + 4*pi*25/pi
\n" ); document.write( "the pies in the numerator and denominator cancel out and we have:
\n" ); document.write( "S = 2*25 + 4*25 which becomes S = 50 + 100 which becomes S = 150
\n" ); document.write( "That's the number we are looking for, so the value of r = 5/sqrt(pi) is good.\r
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