Question

Q.) In a circular table cover of radius 32 cm, a
design is formed leaving an equilateral
triangle ABC in the middle as shown in figure. Find the area of the design.​

Answer 1

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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

Answer 2

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${\huge{\boxed{\tt{\color {red}{❥ Explanation}}}}}$

768√3 cm^2

Now, /_ BOM= 1/2×120°=60°

So,From /\BOM,we have

Math

5 points

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

Area of design= area of O - area of /\ ABC

= (22528/7 - 768√3)cm^2

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