Question

plzz help me guys....

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Answer 1

Answer:

hope it helps you ..............

Answer 2

Answer:

(22528/7 - 768√3)$cm^{2}$

Step-by-step explanation:

Let ABC be the eq.Δand let O be the center of the circle of r=32cm

Area of circle =π$r^{2}$

=(22/7×32×32)$cm^{2}$

=22528/7$cm^{2}$

Draw OM⊥BC

Now, ∠BOM= 1/2×120°=60°

So, From ΔBOM,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 ΔBOC = 1/2 BC×OM

=1/2×32√3×16

area of ΔABC = 3× area of ΔBOC

= 3×1/2×32√3×16

= 768√3$cm^{2}$

Area of design= area of O - area of ΔABC

= (22528/7 - 768√3)$cm^{2}$