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Weekly Problem 20 - 2011

Stage: 2 and 3 Short Challenge Level: Challenge Level:1

 
 
The area of square BDFG is $6\times 6 = 36$ square units.
So the total area of the three triangles ABG, BCD and DEF is also $36$ square units.
These three triangles are congruent and so each has an area of $12$ square units.
 
The area of each triangle is $\frac{1}{2}\times base \times \ height$ and the base is $6$ units and hence we have $\frac{1}{2}\times 6 \times \ height = 12$, 
so the height is $4$ units.
 
Let $X$ be the midpoint of BD. Then CX is perpendicular to the base BD (since BCD is an isosceles triangle).
 
By Pythagoras' Theorem, $BC = \sqrt{3^2+4^2} = 5$ units.
 
Therefore the perimeter of ABCDEFG is $6\times 5+ 6 = 36$ units.
 
 
 
 

This problem is taken from the UKMT Mathematical Challenges.

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