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'Weekly Problem 16 - 2011' printed from http://nrich.maths.org/

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Since the original triangle is isosceles and right-angled, folding it produces a smaller triangle, also isosceles and right-angled.

By Pythagoras' Theorem, the hypotenuse of the original triangle is $\sqrt{200}=10\sqrt{2}$ cm.
 
Hence the difference between the perimeters of the two triangles is $(10+10+10\sqrt{2})-(5\sqrt{2}+5\sqrt{2} +10)=10$ cm.
 
Alternatively: let the length of the shorter sides of the new triangle be x cm, shown below. Then the perimeter of the original triangle is $(20+2x)$ cm and the perimeter of the new triangle is $(10+2x)$cm. Hence the difference between the perimeters of the two triangles is $10$cm.
 
 
 
 

This problem is taken from the UKMT Mathematical Challenges.

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