Rectangle $A B C D$ is made up of six squares. The areas of two of the squares are shown. The perimeter of rectangle $A B C D$, in centimetres, is
(A) 50
(B) 44
(C) 46
(D) 52
(E) 48

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-111.jpg?height=442&width=486&top_left_y=1140&top_left_x=1188)
Since the area of the top left square is $36 \mathrm{~cm}^{2}$, then its side length is $6 \mathrm{~cm}$.

Since the area of the bottom left square is $25 \mathrm{~cm}^{2}$, then its side length is $5 \mathrm{~cm}$.

Therefore, the height $A B$ of rectangle $A B C D$ is $5+6=11 \mathrm{~cm}$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-217.jpg?height=439&width=480&top_left_y=2163&top_left_x=866)

Since the side length of the bottom left square is $5 \mathrm{~cm}$ and the side length of the upper left square is $6 \mathrm{~cm}$, then the side length of the small square in the middle must be $1 \mathrm{~cm}$.

This tells us that the side length of the top right square is $6+1=7 \mathrm{~cm}$.

Therefore, the width $A D$ of rectangle $A B C D$ is $6+7=13 \mathrm{~cm}$.

Thus, the perimeter of rectangle $A B C D$ is $2(13)+2(11)=48 \mathrm{~cm}$.

ANSwER: (E)