Solutions; As shown in figure ①
Let the length AD of rectangular ABCD be 3k and the width CD be 2k.
From 3k×2k=6
K= 1。
AE=3/2，CG=3/3= 1。
Press rectangle
AC=√[(3)？ +(2)? ]
=√ 13.
And AE = CG.
have
∠EAI=∠GCH，
and
∠CHG=∠AHE，
∴△CHG∽△AHE.
Similarity rate:
CG/AE= 1/(3/2)=2/3。
With CH/AH=2/3,
CH/(CH+AH)=2/(2+3)=2/5。
that is
CH/AC=2/5。
∴CH=2AC/5=2√ 13/5.
In the same way; In a similar way
CI = 3AC/5 = 3ì 13/5。
Passing points H and I lead the vertical line to BC respectively, and the vertical feet are J and K, as shown in the figure below.
Then sin ∠ BCA = AB/AC = 2/(√13) = 2 √13.
Author JH/BCA
JH = ch×sin∠BCA =(2√ 13/5)×(2√ 13/ 13)= 4/5。
S△CGH=CG×HJ/2= 1×4/5=4/5。
In the same way; In a similar way
Author KI/CI=sin∠BCA
ki = ci×sin∠BCA =(3√ 13/5)×(2√ 13/ 13)= 6/5。
S△CIF=CF×KI/2=(2×6/5)/2=6/5。
The area of the quadrilateral fghid = s △ cif-s △ CGH = (6/5)-(4/5) = 2/5.
s△EFG = fg×AB/2 = 1×2/2 = 1。
∴ area of shaded part =S△EFG- area of quadrilateral FGHID
= 1-2/5
=3/5.
Only one tenth of the rectangular area.