A circle with diameter before 6116cm is placed inside a rectangle so it touch the sides of the rectangle in three places. More circles are placed inside the rectangle. Each circle touches the circle to the left in one place and has centre on the line AB, with A and B midpoints of the sides of the rectangle on which they lie. The area of each new circle is one quarter of the area of the circle to its left. This process is continued indefinitely.
1.1 Determine the dimensions of the smallest rectangle that will allow all the circles to fit perfectly with no gaps between them on the left and right sides of the rectangle.
Please help, this is making my head hurt?
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Responses (2)
First you need to learn grammar so you can present the problem in an understandable fashion. Then you need to start proofreading what you type to be sure it is correct. ("Before" does not follow "diameter".) Then you need to express the numbers in a normal fashion, which means 61.16 meters instead of 6116cm. Then be sure the entire problem makes sense. "A and B midpoints of the sides of the rectangle on which they lie" does not make sense.
We know that the area of a circle c is proportional to the square of its radius:
c = πr^2.
Our first question is then: How to modify the radius in order to obtain a circle with the quarter of that area. With indeces
c.0 = area of original circle
c.1 = area of the circle with modified radius
c.1 = π (modified r)^2 = c.0 / 4
It would seem the thing to do is halving r.0 to establish the ratio of 1/4.
c.1 = π (r / 2)^2 = π * r^2 / 2^2 = π * r^2 / 4 = c.0 / 4 --> c.1 / c.0 = 1/4
If we go on halving the radius, we get the sequence (I think, that's the point where the headache started ;))
a.k = πr^2 * (1/ 2^k)^2 ;;; applying (p^m)^n = p^(m*n) yields:
a.k = πr^2 * 1/ 2^(2k)
a.k = πr^2 * 1/ 4^k
We got the area ratio R of 1/4 by halving the radius of the original circle. This ratio has to hold for any two consecutive terms of the sequence:
R = ((1 / 2^(k+1) / (1 / 2^k))^2 ;;; divide by multiplying the inverse:
R = ( 2^k / 2^(k+1))^2 ;;; expand the outer square into the terms of the fraction:
R = (2^k)^2 / (2^(k+1))^2 ;;; applying (p^m)^n = p^(m*n) yields:
R = 2^(2k) / 2^(2(k+1)) ;;; expand the exponent of the denominator:
R = 2^(2k) / 2^(2k+2) ;;; applying p^(m+n) = p^m * p^n yields:
R = 2^(2k) / (2^(2k) * 2^2) ;;; 2^(2k) cancels out, and 2^2 = 4 :
R = 1 / 4
We can extract the constant parts and write the geometric series s_a for the areas of circles:
s_a = πr^2 * sum of 1 / 4^k , for k=0 to ∞ ;;; IF n is NOT equal to 0 THEN n^0 = 1
If you work that out, you get the sum of the areas of the circles. (πr^2 * 1 / (1 - 1/4))
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We know the diameter of the original circle (which is also the length of the left and the right side of the rectangle)
d = 6116cm ;;; btw a decent 4-digit, palindromic number with a metric length unit attached to it; there's nothing to complain about that!
And, as we have worked out, the ratio of the diameters of two consecutive circles is R=1/2. All the circles are neatly lined up like pearls on a string with their centres on the connecting line AB, with A and B midpoints of the left and right side of the rectangle according to your description. All we've left to do, is figuring out the sum of the series of diameters s_d:
s_d = d * sum of (1/2)^k , for k=0 to ∞,
to get the length of the top and bottom side of the rectangle.
As 1/2 < 1, the sum of (1/2)^k , for k=0 to ∞ converges to 1/(1 - 1/2) = 2
(If you need help in order to derive this formula, feel free to ask.)
So, the area of the rectangle a_r you are looking for is:
a_r = 2d * d = 2 * d^2 ;;; The area of the rectangle all the circles live in
2 * 6116^2 = 74 810 912 cm^2, or if you prefer metres squared:
7481.0912 m^2