Question

here is my question plzzz answer fast its urgent

Answer 1

$\huge{hello-dear}$
→→→→→→→→→→→→→→→→→→→→→
answer=> Let ABC be the eq./\and let o be the centre of the circle of r=32cm
Area of circle =πr^2
=(22/7×32×32)cm2
=22528/7 cm2

Draw OM_|_BC
Now, /_ BOM= 1/2×120°=60°
So,From /\BOM,we have
OM/OB=cos 60°(1/2)
i.e., OM= 16 cm
Also, BM/OB= cos60°(1/2)
i.e., BM= 16√3 cm
BC = 2 BM =32√3 cm
Hence, area of /\BOC = 1/2 BC .OM
=1/2×32√3×16
area of /\ ABC = 3× area of /\ BOC
= 3×1/2×32√3×16
= 768√3 cm^2
Area of design= area of O - area of /\ ABC
= (22528/7 - 768√3)cm^2


$<marquee >hope its help you$


Is your answer in photo also