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Math help ** 10 points**?
A piece of wire x m long is cut into two pieces, the length of the first piece being 8 m. The first piece is bent into a circle, and the other is bent into a rectangle with length twice the width. Give an expression for the total area A enclosed in the two shapes in terms of x.
Please help - will give 10 points to the best answer!!
2 Answers
- DavidLv 79 years agoFavorite Answer
x - 8 = perimeter of the rectangle
2(L + W) = X - 8
L + W = (X - 8) / 2
L = 2W
2W + W = (X - 8) / 2
3W = (X - 8) / 2
W = (X - 8) / 6
LW = A
2W * W = A
2W^2 = A
2[(X - 8) / 6]^2 = A
2[(X^2 - 16X + 64) / 36] = A
(X^2 - 16X + 64) / 18 = A of rectangle
C = 2πr
C / 2π = r
8 / 2π = r
4 / π = r
A = πr^2
A = π(4 / π)^2
A = 16π / π^2
A = 16 / π
A of both
(16 / π) + [(x^2 - 16x + 64) / 18]
[288 + π(x^2 - 16x + 64)] / 18π
or
(288 + πx^2 - 16πx + 64π) / 18π
- VampireDogLv 69 years ago
the area of the circle will be given by pir^2, where r is the radius of the circle.
we know the cirumference of the circle is 8m and that C=2pir....this gives us 8=2pir or r=4/pi
thus the area of the circle is: pi(4/pi)^2 = 16/pi
the perimiter of the rectangle must be x-8 meters which is always 2(l+w) and here l=2w, thus the perimeter = 2(2w+w)=6w. so 6w=x-8 and w= (x-8)/6 and that means l= (x-8)/3
area of the rectangle =l*w or (x-8)/6 times (x-8)/3 or (x-8)^2/18
so the total area is 16/pi + (x-8)^2/18
