In the diagram, each of the three squares has a side length of 3. Two of the squares have a common vertex $O$, and $O$ is the centre of the square labelled $A B C D$. The perimeter of the entire figure is closest to
(A) 21.5
(B) 22.0
(C) 22.5
(D) 24.0
(E) 30.0

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-099.jpg?height=344&width=333&top_left_y=2246&top_left_x=1297)
We label the remaining vertices in the figure.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-197.jpg?height=415&width=401&top_left_y=1538&top_left_x=924)

Since each of the squares has a side length of 3, then $P Q=Q R=B C=X Y=Y Z=D A=3$, so the perimeter of the figure equals $18+A P+R B+C X+Z D$.

Since $O$ is the centre of square $A B C D$, then $O A=O B=O C=O D$.

Since $O P=O R=O X=O Z$, then $A P=R B=C X=Z D$.

Therefore, the perimeter equals $18+4 A P$.

Since $O$ is centre of square $A B C D$, then $O A=O B$ and $\angle A O B=90^{\circ}$, so $\triangle A O B$ is an isosceles right-angled triangle, so has angles $45^{\circ}, 45^{\circ}$ and $90^{\circ}$.

Therefore, $A O=\frac{A B}{\sqrt{2}}=\frac{3}{\sqrt{2}}=\frac{3 \sqrt{2}}{2}$ and so $A P=O P-A O=3-\frac{3 \sqrt{2}}{2}$.

Thus, the perimeter is $18+4\left(3-\frac{3 \sqrt{2}}{2}\right)=18+12-6 \sqrt{2}=30-6 \sqrt{2} \approx 21.515$.

ANSWER: (A)

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