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The original rectangle had a perimeter of $2 \times 30\text{cm} + 2\times 40\text{cm} = 140\text{cm}$.
Each of the new squares that is cut out adds three sides of $5\text{cm}$ each (shown in black on the diagram), and removes one portion of length $5\text{cm}$ (shown in green).
This means the overall effect of cutting out each square is to increase the perimeter by $10\text{cm}$. Since are ten squares, the overall increase is $10 \times 10 \text{cm} = 100\text{cm}$.
Therefore the final perimeter is $140\text{cm} + 100\text{cm} = 240\text{cm}$.
Christmas Cut-out
Age 11 to 14
ShortChallenge Level 
The original rectangle had a perimeter of $2 \times 30\text{cm} + 2\times 40\text{cm} = 140\text{cm}$.
Each of the new squares that is cut out adds three sides of $5\text{cm}$ each (shown in black on the diagram), and removes one portion of length $5\text{cm}$ (shown in green).This means the overall effect of cutting out each square is to increase the perimeter by $10\text{cm}$. Since are ten squares, the overall increase is $10 \times 10 \text{cm} = 100\text{cm}$.
Therefore the final perimeter is $140\text{cm} + 100\text{cm} = 240\text{cm}$.
This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.
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