Find the shaded area

Geometry Level 4

A = A C B = A B C \angle A = \angle ACB = \angle ABC ,
G J = J F GJ = JF ,
D I = I F DI =IF ,
H C = C F = F B = B E HC=CF=FB=BE ,
G H = D E = 8 cm GH = DE = 8\text{ cm} ,
A C = 12 cm AC = 12\text{ cm} ,
area ( J C F ) = 1 4 area ( G H F ) \text{area}(JCF) = \dfrac14 \text{area}(GHF) ,
area ( I B F ) = 1 4 area ( D E F ) \text{area}(IBF) = \dfrac14\text{area}(DEF) .

Find the shaded area up to 3 decimal places.


The answer is 41.569.

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2 solutions

Yahia El Haw
Apr 23, 2016

A F i s t h e l i n e o f s y m m e t r y . . . . . . . . . . ( 1 ) A F i s a l t i t u d e o f e q u i l a t e r a l Δ A C B . B F = 1 2 B C = 6.......... ( 2 ) A l s o Δ A C B = 3 4 1 2 2 = 36 3 . . . . . . . . . . . . . ( 3 ) I n Δ F G H , J a n d C a r e m i d p o i n t s o f t h e s i d e s . J C = 1 2 G H = 4. S o b y ( 1 ) a r e a o f Δ s I F B = J C F = 1 2 H G B F S i n 60 = 6 3 . A r e a s A J F I = Δ A B C Δ J C F Δ I F B = 24 3 = 41.569. AF~ is~ the~ line~ of~ symmetry..........(1)\\ ~~~~~\\ AF~is~altitude~of~equilateral~\Delta ~ACB.\\ ~~~~~\\ \implies~BF=\frac1 2 BC=6..........(2)\\ ~~~~~\\ Also ~\Delta~ACB=\dfrac {\sqrt3} 4 *12^2=36*\sqrt3.............(3)\\ ~~~~~\\ In~\Delta~FGH,~J~ and~ C~are~midpoints~of~the~sides.\\ ~~~~~\\ \therefore~JC=\frac 1 2 GH=4.\\ ~~~~~\\ So~by~(1)~area~of~\Delta s~IFB=JCF=\frac 1 2 *HG*BF*Sin60=6*\sqrt3.\\ ~~~~~\\ \therefore~Area s~AJFI=\Delta~ABC - \Delta~JCF - \Delta~ IFB=24*\sqrt3=\color{#D61F06}{41.569}.

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