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A wire is cut into two pieces, one of length and the other of length b. The piece of length a is bent to form an equilateral triangle, and the piece of length b is bent to form a regular hexagon. The triangle and the hexagon have equal area. What is a/b?

 Dec 14, 2019
 #1
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The  side of the equilateral triangle  =   a/3

So....the area  of the equilateral triangle  =  (1/2)(a/3)^2 sqrt (3) / 2 = a^2  [ sqrt (3)/ 36 ]

 

The side of the  hexagon = b / 6

So...it's area  =   6 (1/2) (b/6)^2 sqrt (3) / 2  = b^2  [ 3 sqrt (3) / 72]  =   b^2 [sqrt (3) / 24]

 

So

 

a^2 [ sqrt (3) / 36  ]  =  b^2 [ sqrt (3) / 24]

 

So

 

a^2 / b^2   =  [ sqrt (3) /24 ]  / [ sqrt (3) / 36 ]

 

a^2 / b^2  =  [ 36 / 24 ]

 

a^2/ b^2  =  3/2

 

a/ b  =   sqrt (3)  / sqrt (2)   =    sqrt (6) / 2

 

 

cool cool cool

 Dec 14, 2019
 #2
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+1

thank you so much CPhil i was on that question for soooo long. 

 Dec 15, 2019

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