A geometry problem by Francois Tardieu

We define: $JC=x$ and $JB=12-x$ so, $IJ=x$ and we have, $AI=IB=6$

with Pythagoras in triangle IBJ

$IJ^2=IB^2+JB^2$

$\Longleftrightarrow x^2=6^2+(12-x)^2 \Longleftrightarrow x^2= 36+144-24x+x^2 \Longleftrightarrow 24x= 180 \Longleftrightarrow x=\dfrac{15}{2}=7,5$

Then $IJ=x= 7,5$ and $JB=12-x=4,5$

we note, with the help of complimentary angles, that:

$\widehat{IJB}= \widehat{AIM} =\widehat{LKM}$ (1)

and

$\widehat{BIJ}= \widehat{AMI} =\widehat{LMK}$ (2)

Let's say that: $LM=y$ and $LK=z$ so we get $MI=12-y$

and with help of (1)

$\cos \widehat{AIM}=\cos \widehat{IJB}\Longleftrightarrow \dfrac{IA}{IM}= \dfrac{6}{12-y} \Longleftrightarrow \dfrac{4,5}{7,5}=\dfrac{6}{12-y} \Longleftrightarrow 3(12-y)=30 \Longleftrightarrow y=2$

and with help of (2)

$\tan \widehat{LMK} = \tan \widehat{BIJ} \Longleftrightarrow \dfrac{z}{2}=\dfrac{4,5}{6} \Longleftrightarrow z= 1,5$

we can compute areas now:

area(IJKL) $= \dfrac{1}{2} \times (LK+IJ) \times LI =\dfrac{1}{2} \times (1,5+7,5) \times 12 =54$

area(LKM) $= \dfrac{1}{2} \times LK \times LM = \dfrac{1}{2} \times 1,5 \times 2 = 1,5$

area(IJKM)=area(IJKL) -area(LKM) $=54-1,5=52,5$