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. Find the area of the design.

Let us given that, circular table cover of radius 32 cm

Here, we mark O  as the center of the given circle and then we join BO and CO

so, our new figure becomes,

and here the triangle is equilateral triangle then each side of triangle ABC will subtend equal angles at the center.

Thus,  ∠BOC = 360⁰/3 = 120⁰

Consider ΔBOC. Drop a perpendicular from OM to BC

We know perpendicular from the center of circle to a chord bisects it.

⇒ BM = MC

OB = OC (radius)

OM = OM (common)

Hence,  ΔOBM ≅ ΔOCM

⇒ ∠BOM = ∠COM (by CPCT)

 2∠BOM = ∠BOC = 120⁰

∠BOM = 120°/2 = 60⁰

sin 60⁰ = BM/BO = $\sqrt{3}$/2

∴ BM = $\sqrt{3}$/2 $\times$ BO = $\sqrt{3}$/2 $\times$ 32 = 16$\sqrt{3}$

⇒ BC = 2BM = 32$\sqrt{3}$

Using the formula of area of equilateral triangle = $\sqrt{3}$/4 (side)²

Area of the design = Area of a circle - Area of ΔABC = ${\mathrm{\pi r}}^2$$-$ $\sqrt{3}$/4 (BC)²

= ${\mathrm{\pi r}}^2$ $-$ $\sqrt{3}$/4 (BC)²

= 22/7 $\times$ (32)² $-$$\sqrt{3}$/4 $\times$ (32$\sqrt{3}$)²

= 22/7 $\times$ 1024 $-$ $\sqrt{3}$/4 $\times$ 1024 $\times$ 3

= 22528/7$-$ 768$\sqrt{3}$

Area of design = (22528/7 $-$ 768$\sqrt{3}$) cm²