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:

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