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The original square has area 180 + 4 x 4 = 196, so it has side length 14cm.
Therefore, the dimensions of the box are 10 x 10 x 2.
Therefore, the volume is $200$cm$^3$.
Alternatively:
The base of the open box is a square. Let its side be of length $x$ cm.
Then the total surface area of the box in cm$^2$ is $x^2+4\times 2x=x^2+8x$.
Hence, $x^2+8x=180$, that is $x^2+8x-180=0$.
Therefore, $(x+18)(x-10)=0$ which gives $x=-18$ or $x=10$.
As $x$ must be positive, it must be the case that $x=10$.
Now the open box has dimensions $10$ cm $\times 10$ cm $\times 2$ cm. So its volume is $200$cm$^3$.
Open the Box
Age 11 to 14
ShortChallenge Level 
The original square has area 180 + 4 x 4 = 196, so it has side length 14cm.
Therefore, the dimensions of the box are 10 x 10 x 2.
Therefore, the volume is $200$cm$^3$.
Alternatively:
The base of the open box is a square. Let its side be of length $x$ cm.
Then the total surface area of the box in cm$^2$ is $x^2+4\times 2x=x^2+8x$.
Hence, $x^2+8x=180$, that is $x^2+8x-180=0$.
Therefore, $(x+18)(x-10)=0$ which gives $x=-18$ or $x=10$.
As $x$ must be positive, it must be the case that $x=10$.
Now the open box has dimensions $10$ cm $\times 10$ cm $\times 2$ cm. So its volume is $200$cm$^3$.
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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