Find the perimeter of the given figure. AE = 50 m.

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Solution

$Δ A B E$ is a right angled triangle
So apply pythagoras theorem,
Hypotenuse $= \sqrt{b a s e^{2} + (p e r p e n d i c u l a r)^{2}} S o, B E = \sqrt{(50)^{2} + (120)^{2}} = 130 m$
$Δ B C D$ is a right angled triangle
So, BD $= \sqrt{(40)^{2} + (30)^{2}} = 50 m$
Now, perimeter $= A B + B C + C D + D B + B E + E A$
$= 120 m + 40 m + 30 m + 50 m + 130 m + 50 m = 420 m$

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