Question

In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in fig. 12.24. find the area of the design.

Answer 1

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

Answer:

The area of the design is approximately 1885.15sq.cm.

Step-by-step explanation:

Given the radius of a circle, $r=32 cm$

          and an equilateral triangle ABC in the circle.

Area of circle,

                            $a=\pi r^{2} =3.14 (32)^{2}$

                               $\approx 3215.36sq.cm$

The side of an equilateral triangle is given by $s=\sqrt{3} r$

And the area of an equilateral triangle is given by

                             $s^{2} \times \frac{\sqrt{3}}{4}=(32\sqrt{3}) ^{2} \times \frac{\sqrt{3}}{4}$

                                           $=(32\times 32\times 3) \times \frac{\sqrt{3}}{4}$

                                           $\approx 1330.21sq.cm$

Area of the design = area of circle - area of triangle

                               $=3215.36-1330.21=1885.15sq.cm$

Therefore, the area of the design is approximately 1885.15sq.cm.