​​​​please tell me its solution

Here, biggest triangle is triangle CED which is an isosceles triangle with CE= ED having area 36 cm^2.CD is a diameter of circle and PQSR is a square.
Since, CD is a diameter
so, angle CED= 90° [ Angles in semi circle]
ar( CED)= 1/2 × ED×CE [ Angle E= 90°]
Now, ED=CE and ar( CED)= 36 cm^2
so, ar( CED)= 1/2×ED×ED= 36 cm^2
=》ED= /72=CE
CD^2 = ED^2+CE^2 = (/72)^2 + (/72)^2
CD^2 = 144
=》CD=12cm
Again in triangle CED , CE= ED and angle E= 90°
so, angle ECD=45°= angle EDC
In triangle CPR,
we've angle R= 90° [ PQSR IS a square, hence, angle PRS = 90° = 180- angle PRC so, angle PRC = 90°]
angle C= angle ECD=45°
SO, Angle CPR= 45°
so, in tri. CPR,
angle C= angle P = 45°
hence, CR= PR ....(1)
similarly, in tri. QSD,
QS= SD ...(2)
PQ= RS ...(3) [ sides of square]
but PR , PQ and QS are sides of square PQRS and sides of square are equal
so, PR=QS = PQ ....(4)
From (1),(2),(3) and (4);
we've
CR= PR = QS= SD = PQ= RS
Hence, CR= RS=SD = 1/3 CD
RS = 1/3 CD
Now, CD = 12 cm
so, RS= 1/3×12=4 cm
Now, RS is one of the side of square , so, ar(sq.PQRS) = ( Side)^2 = (4)^2= 16 cm^2
hence, ar(PQRS)= 16 cm^2