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Question
Answer
sushi88
(a) Using Pythagoras theorem, height of cone = sqrt [(25)2 – (5)2 ] = sqrt(600)
Volume of cone = (1/3)πr2h
Volume of conical candle = (1/3)π (5)2 sqrt(600) = 641.27 cm3 (to 2 dec. places)
(b) 300 mins —> 641.27
In 30 mins, melted volume of candle = (641.27/300) x 30 = 64.127
Melted volume of candle / Original Candle = 64.127/641.27 = 0.1 or (1/10)
(c) Ratio of volumes of similar cones = (a/b) 3 where a and b are either both radius or both height of similar cones.
Ratio of surface areas of similar cones = (a/b) 2 where a and b are either both radius or both height of similar cones.
So, SA of shaded region of figure 2 = [cube-root (1/10)]2 x SA of original conical candle
= (1/10)2/3 x π x (5)2
= 16.92 cm2
(d) From (c), π( r1)2 = (1/10)2/3 x π x (5)2
( r1)2 = (1/10)2/3 x (5)2
We know that for similar cones, the volume ratio is:
(1/3)π( r1)2h1 / (1/3)π( r2)2h2 = 1/10
So, h1 = [ (1/10) x ( r2)2h2 ] / ( r1)2
= [ (1/10) x (5)2 sqrt(600) ] / (1/10)2/3 x (5)2
= 11. 3695
Hence height of cylinder = height of remaining conical candle = sqrt(600) – 11.3695
= 13.13 (to 2 decimal places)
(e) Volume of remaining conical candle = 641.27-64.127 = 577.143
Volume of cylinder = πr2h = π x (5)2 x (13.13)
Hence Volume of empty space in fig 3 = Vol of cylinder – Vol of remaining conical candle
= π x (5)2 x (13.13) – 577.143
= 454.08 cm2 (to 2 decimal places)
ChiefKiasu
OMG. This is beyond me.
sushi88
Chief, this is the type of tedious question that you spoke about. Doing math is like running a marathon, need to build stamina for such types of questions. The non-tedious types with typically less than 5 lines of solution need to use much more brain cells which you prefer but they are usually very difficult for most people.