Rectangles that measure $4 \times 2$ are positioned in a pattern in which the top left vertex of each rectangle (after the top one) is placed at the midpoint of the bottom edge of the rectangle above it, as shown. When a total of ten rectangles are arranged in this pattern, what is the perimeter of the figure?
(A) 48
(B) 64
(C) 90
(D) 84
(E) 100

![](https://cdn.mathpix.com/cropped/2024_04_20_46ead6524a8d61e21c51g-030.jpg?height=409&width=339&top_left_y=1124&top_left_x=1294)
Solution 1

The first and tenth rectangles each contribute an equal amount to the perimeter.

They each contribute two vertical sides (each of length 2), one full side of length 4 (the top side for the first rectangle and the bottom side for the tenth rectangle), and one half of the length of the opposite side.

That is, the first and tenth rectangles each contribute $2+2+4+2=10$ to the perimeter. Rectangles two through nine each contribute an equal amount to the perimeter.

They each contribute two vertical sides (each of length 2), one half of a side of length 4, and one half of the length of the opposite side (which also has length 4).

That is, rectangles two through nine each contribute $2+2+2+2=8$ to the perimeter.

Therefore, the total perimeter of the given figure is $(2 \times 10)+(8 \times 8)=20+64=84$.

## Solution 2

One method for determining the perimeter of the given figure is to consider vertical lengths and horizontal lengths.

Each of the ten rectangles has two vertical sides (a left side and a right side) which contribute to the perimeter.

These 20 sides each have length 2 , and thus contribute $20 \times 2=40$ to the perimeter of the figure.

Since these are the only vertical lengths contributing to the perimeter, we now determine the sum of the horizontal lengths.

There are two types of horizontal lengths which contribute to the perimeter: the bottom side of a rectangle, and the top side of a rectangle.

The bottom side of each of the first nine rectangles contributes one half of its length to the perimeter.

That is, the bottom sides of the first nine rectangles contribute $\frac{1}{2} \times 4 \times 9=18$ to the perimeter. The entire bottom side of the tenth rectangle is included in the perimeter and thus contributes a length of 4 .

Similarly, the top sides of the second rectangle through to the tenth rectangle contribute one half of their length to the perimeter.

That is, the top sides of rectangles two through ten contribute $\frac{1}{2} \times 4 \times 9=18$ to the perimeter. The entire top side of the first rectangle is included in the perimeter and thus contributes a length of 4.

In total, the horizontal lengths included in the perimeter sum to $18+4+18+4=44$.

Since there are no additional lengths which contribute to the perimeter of the given figure, the total perimeter is $40+44=84$.

## Solution 3

Before they were positioned to form the given figure, each of the ten rectangles had a perimeter of $2 \times(2+4)=12$.

When the figure was formed, some length of each of the ten rectangles' perimeter was "lost" (and thus is not included) in the perimeter of the given figure.

These lengths that were lost occur where the rectangles touch one another.

There are nine such locations where two rectangles touch one another (between the first and second rectangle, between the second and third rectangle, and so on).

In these locations, each of the two rectangles has one half of a side of length 4 which is not included in the perimeter of the given figure.

That is, the portion of the total perimeter of the ten rectangles that is not included in the perimeter of the figure is $9 \times(2+2)=36$.

Since the total perimeter of the ten rectangles before they were positioned into the given figure is $10 \times 12=120$, then the perimeter of the given figure is $120-36=84$.

ANSWER: (D)