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Weekly Problem 49 - 2010

Stage: 3 Short Challenge Level: Challenge Level:1

Suppose that the triangle and the square are placed so that the area of the overlap is as large as possible.

The area of the square is $36{cm}^2$.

The area of overlap is $2\over 3$ of this, namely $24{cm}^2$.

This is $60$% of the area of the triangle.

So the area of the triangle is $\frac{10}{6}\times 24 = 40{cm}^2$.
 
 
 

This problem is taken from the UKMT Mathematical Challenges.

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