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The hourglass law math problem is urgent! ! !
The hourglass theorem is similar to a triangle to be learned in junior high school.

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.