document.write( "Question 1139006: A hemisphere of diameter 10cm is attached to a cylinder of equal diameter. If the total length of the shape is 20cm calculate:
\n" ); document.write( "A) the surface area of the hemisphere
\n" ); document.write( "B)length of cylinder
\n" ); document.write( "C) surface area of the whole shape
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Algebra.Com's Answer #756826 by MathLover1(20850)\"\" \"About 
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A hemisphere of diameter \"10cm\" is attached to a cylinder of equal diameter. If the total length of the shape is \"20cm+\"calculate:\r
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\n" ); document.write( "\n" ); document.write( "A) the surface area of the hemisphere \r
\n" ); document.write( "\n" ); document.write( "given: diameter \"d=10cm\"=> radius \"r=5cm\"\r
\n" ); document.write( "\n" ); document.write( "\"SA=2pi%2Ar%5E2\"
\n" ); document.write( "\"SA=2pi%2A%285cm%29%5E2\"
\n" ); document.write( "\"SA=2pi%2A25cm%5E2\"
\n" ); document.write( "\"SA=50pi%2Acm%5E2\"=> exact solution
\n" ); document.write( "\"+SA=157.14cm%5E2\"=>approximate solution\r
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\n" ); document.write( "\n" ); document.write( "B) length of cylinder \r
\n" ); document.write( "\n" ); document.write( " If the total length of the shape is \"20cm\" calculate, the length of cylinder will be difference between the total length of the shape and radius of cylinder \r
\n" ); document.write( "\n" ); document.write( "\"20cm-r=20cm-5cm=15cm\"\r
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\n" ); document.write( "\n" ); document.write( "C) surface area of the whole shape \r
\n" ); document.write( "\n" ); document.write( "will be of the hemisphere plus surface area of cylinder (without one base)
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\n" ); document.write( " surface area of cylinder is:
\n" ); document.write( "lateral surface area of cylinder is \"2pi%2Arh\"
\n" ); document.write( "area of bottom of cylinder (circle) is \"r%5E2%2Api\"\r
\n" ); document.write( "\n" ); document.write( "\"SA%5Bc%5D=2rpi%2Ah%2Br%5E2%2Api\"....plug in \"r=5cm\", \"h=15cm\"
\n" ); document.write( "\"SA%5Bc%5D=2%2A5cmpi%2A15cm%2B%285cm%29%5E2%2Api\"
\n" ); document.write( "\"SA%5Bc%5D=150pi%2Acm%5E2%2B25cm%5E2%2Api\"
\n" ); document.write( "\"SA%5Bc%5D=175pi%2Acm%5E2\"=> exact solution
\n" ); document.write( "\"SA%5Bc%5D=549.5cm%5E2\"=>approximate solution\r
\n" ); document.write( "\n" ); document.write( "=>surface area of the whole shape is: \r
\n" ); document.write( "\n" ); document.write( "\"total=50pi%2Acm%5E2%2B175pi%2Acm%5E2=225pi%2Acm%5E2\"=> exact solution\r
\n" ); document.write( "\n" ); document.write( "\"total=157.14cm%5E2%2B549.5cm%5E2=706.64cm%5E2\"=>approximate solution\r
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