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Weekly Problem 44 - 2009

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

In the diagram below, triangle $ABC$ represents the garden, $CD$ represents the fence and $E$ is the foot of the perpendicular from $D$ to $AC$.

The two sections of the garden have the same perimeter so $AD$ is $10$m longer than $DB$. Hence $AD=30$m and $DB=20$m.

Trangles $AED$ and $ACB$ are similar so $\frac{AE}{AC}=\frac{AD}{AB}=\frac{30}{50}$. Hence $AE=\frac{3}{5}\times 30$m $=18$m. So $EC=(30-18)$m $=12$m.

Also $\frac{ED}{CB}=\frac{AD}{AB}=\frac{30}{50}$. Hence $ED=\frac{3}{5}\times 40$m $=24$m.

Finally, by Pythagoras: $CD^2=EC^2+ED^2=(12^2+24^2)$m$^2=5\times12^2$m$^2$. So the length of the fence is $12\sqrt{5}$m.

This problem is taken from the UKMT Mathematical Challenges.

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