SOLUTION: A piece of wire measuring 48 feet long is cut into two pieces of equal length. One piece is bent into an equilateral triangle and the other is bent into a circle. What is the com

Algebra ->  Circles -> SOLUTION: A piece of wire measuring 48 feet long is cut into two pieces of equal length. One piece is bent into an equilateral triangle and the other is bent into a circle. What is the com      Log On


   

Question 1042997: A piece of wire measuring 48 feet long is cut into two pieces of equal length.
One piece is bent into an equilateral triangle and the other is bent into a circle.
What is the combined area of the two shapes?
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Each piece is 48/2 = 24 ft long

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The equilateral triangle has side length of 24/3 = 8 feet. The area of this triangle is

A = (sqrt(3))/4*s^2
A = (sqrt(3))/4*8^2
A = (sqrt(3))/4*64
A = 16*sqrt(3)

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The circle has circumference of 24 ft, so C = 24

C = 2*pi*r
24 = 2*pi*r
12 = pi*r
r = 12/pi

The area of the circle is then
A = pi*r^2
A = pi*(12/pi)^2
A = 144/pi

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The combined area is the result of adding the two areas

(16*sqrt(3))+(144/pi)

The exact combined area is %2816%2Asqrt%283%29%29%2B%28144%2Fpi%29

Using a calculator, (16*sqrt(3))+(144/pi) = 73.5494365315679

So the approximate combined area is 73.5494365315679