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Thread: Summing areas of squares

  1. #1
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    Summing areas of squares

    A square S1 has a perimeter of 40 inches. The vertices of a second square S2 are the midpoints of the sides of S1. The vertices of a third square S3 are the midpoints the sides of S2. Assume the process continues indefinitely, with the vertices of S K+1 being the midpoints of the sides of Sk for every positive integer k. What is the sum of the areas, in square inches, of S1, S2, S3,...?

    My instructor says the answer is 200 but I still don't understand how???
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  2. #2
    Super Member angel.white's Avatar
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    Quote Originally Posted by donnagirl View Post
    A square S1 has a perimeter of 40 inches. The vertices of a second square S2 are the midpoints of the sides of S1. The vertices of a third square S3 are the midpoints the sides of S2. Assume the process continues indefinitely, with the vertices of S K+1 being the midpoints of the sides of Sk for every positive integer k. What is the sum of the areas, in square inches, of S1, S2, S3,...?

    My instructor says the answer is 200 but I still don't understand how???
    Your perimeter is 40, so each side is 10

    Then the first square has an area of 10*10 = 100

    Now lets find the area of the second square. It will be equal to the area of the first square, minus the four triangles around the edges.

    So Area = 100-4*(area of triangle)

    We know the length and height, because they go to the midpoint, so the length will be 5, and the height will be 5.
    So the area of the triangle is (1/2)bh = (1/2)5*5
    Area = 100-4*(1/2)*5*5
    Area = 100-50 = 50

    Notice that this triangle's area is half the area of the first triangle. Continuing with this pattern, you will see that each triangle's area will be half of the area of the triangle it is bounded by.

    so our total area = 100 + 50 + 25 + 12.5 + ...

    = 100 + (1/2)100 + (1/2)(1/2)100 + (1/2)(1/2)(1/2)100 + ....

    = (1/2)^0 *100 + (1/2)^1 *100 + (1/2)^2 *100 + (1/2)^2 *100 + ...

    so we can write this as a geometric series (see Geometric series - Wikipedia, the free encyclopedia for more about the formulas)

    $\displaystyle = \sum_{n=0}^\infty 100(1/2)^n$

    Now in a geometric series, if |r| < 1, the series converges to $\displaystyle \frac a{1-r}$

    In our case, r = 1/2, and a =100.
    1/2 <1 so our series converges

    $\displaystyle = \frac {100}{1-1/2}$

    $\displaystyle = \frac {100}{1/2}$

    $\displaystyle = 200 ~in^2$
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  3. #3
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    Hello, donnagirl!

    Another approach . . . same answer.


    A square $\displaystyle S_1$ has a perimeter of 40 inches.
    The vertices of a second square $\displaystyle S_2$ are the midpoints of the sides of $\displaystyle S_1.$
    The vertices of a third square $\displaystyle S_3$ are the midpoints the sides of $\displaystyle S_2.$
    Assume the process continues indefinitely, with the vertices of $\displaystyle S_{k+1}$
    being the midpoints of the sides of $\displaystyle S_k$ for every positive integer $\displaystyle k.$
    What is the sum of the areas, in square inches, of $\displaystyle S_1,\,S_2,\,S_3,\,\hdots$ ?
    Code:
          *-------*-------*
          |     *:|:*     |
          |   *:::|:::*   |
          | *:::::|:::::* |
          *:-:-:-:+:-:-:-:*
          | *:::::|:::::* |
          |   *:::|:::*   |
          |     *:|:*     |
          *-------*-------*
    The sides and diagonals of $\displaystyle S_{k+1}$ divides $\displaystyle S_k$ into eight congruent triangles.

    Since $\displaystyle S_{k+1}$ is composed of four of these triangles,
    . . the area of $\displaystyle S_{k+1}$ is one-half of $\displaystyle S_k$

    That is, each square has half the area of the preceding square.


    $\displaystyle S_1$ has a perimeter of 40 inches; its side is 10 inches.
    . . Its area is: .$\displaystyle 10^2 \,=\,100$ inē

    Then the total area is: .$\displaystyle A \;=\;100 + \frac{100}{2} + \frac{100}{4} + \frac{100}{8} + \cdots$

    . . $\displaystyle \text{And we have: }\;A \;=\;100\underbrace{\left(1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots\right)}_{\text{geometric series}} $

    The geometric series has the sum: .$\displaystyle S \:=\:\frac{1}{1-\frac{1}{2}} \:=\:2$

    . . Therefore: .$\displaystyle T \;=\;100(2) \;=\;200\text{ in}^2$

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