In the diagram shown, each figure after Figure 1 is formed by joining two rectangles to the bottom of the previous figure. Each individual rectangle has dimensions $10 \mathrm{~cm}$ by $5 \mathrm{~cm}$. If Figure $n$ has a perimeter of $710 \mathrm{~cm}$, the value of $n$ is
(A) 29
(B) 43
(C) 66
(D) 172
(E) 65
The perimeter of Figure 1 consists of 4 rectangle side lengths of $10 \mathrm{~cm}$ (each of which is horizontal) and 4 rectangle side lengths of $5 \mathrm{~cm}$ (each of which is vertical).

Thus, the perimeter of Figure 1 is $(4 \times 10 \mathrm{~cm})+(4 \times 5 \mathrm{~cm})=40 \mathrm{~cm}+20 \mathrm{~cm}=60 \mathrm{~cm}$.

The perimeter of Figure 2 consists of 4 rectangle side lengths of $10 \mathrm{~cm}$ (each of which is horizontal) and 6 rectangle side lengths of $5 \mathrm{~cm}$ (each of which is vertical).

Thus, the perimeter of Figure 2 is $(4 \times 10 \mathrm{~cm})+(6 \times 5 \mathrm{~cm})=40 \mathrm{~cm}+30 \mathrm{~cm}=70 \mathrm{~cm}$.

The perimeter of Figure 3 consists of 4 rectangle side lengths of $10 \mathrm{~cm}$ (each of which is horizontal) and 8 rectangle side lengths of $5 \mathrm{~cm}$ (each of which is vertical).

Thus, the perimeter of Figure 3 is $(4 \times 10 \mathrm{~cm})+(8 \times 5 \mathrm{~cm})=40 \mathrm{~cm}+40 \mathrm{~cm}=80 \mathrm{~cm}$.

Each figure after Figure 1 is formed by joining two rectangles to the bottom of the previous figure.

The bottom edge of a figure (consisting of two $10 \mathrm{~cm}$ side lengths) is replaced by two $10 \mathrm{~cm}$ lengths when the two new rectangles are adjoined.

That is, the addition of two rectangles does not change the number of $10 \mathrm{~cm}$ side lengths contributing to the perimeter of the new figure and so the number of $10 \mathrm{~cm}$ lengths remains constant at 4 for each figure.

The addition of two new rectangles does not replace any of the previous $5 \mathrm{~cm}$ (vertical) side lengths.

Thus, the addition of the two rectangles does add two $5 \mathrm{~cm}$ vertical segments to the previous perimeter, increasing the perimeter of the previous figure by $2 \times 5 \mathrm{~cm}=10 \mathrm{~cm}$.

That is, the perimeter of Figure 1 is $60 \mathrm{~cm}$, and the perimeter of each new figure is $10 \mathrm{~cm}$ greater than the previous figure.

We need to add $10 \mathrm{~cm} 65$ times to get a total of $710 \mathrm{~cm}$ (that is, $60 \mathrm{~cm}+10 \mathrm{~cm} \times 65=710 \mathrm{~cm}$ ). Thus, Figure 66 has a perimeter of $710 \mathrm{~cm}$, and so $n=66$.

ANSWER: (C)