Perimeter

Geometry Level 3

Given a square A B C D ABCD of each side equal to 1 1 . F F and E E are points on A B AB and A D AD respectively such that ∠ F C E = 4 5 ∘ \angle FCE = 45^{\circ} . What is the perimeter of the △ A F E \triangle AFE ?

Note :Give your answer to 2 decimal places.


The answer is 2.00.

Vilakshan Gupta by Vilakshan Gupta , 19, India

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1 solution

Maria Kozlowska
Nov 5, 2017

Let's choose point F F on A B AB . Let's reflect B B about C F CF . C B F B ′ CBFB' and C B ′ E D CB'ED are two kites. C F CF is bisector of ∠ B C B ′ \angle BCB' and C E CE is bisector of ∠ D C B ′ \angle DCB' . This means that ∠ F C E = 4 5 o \angle FCE=45^o .

A F + F B ′ = A F + F B = A B = 1 , A E + E B ′ = A E + E D = A D = 1 AF+FB'=AF+FB=AB=1, AE+EB'=AE+ED=AD=1

This concludes that the perimeter of △ A F E = 1 + 1 = 2 \triangle AFE = 1+1=\boxed{2}

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