In a circular table cover of radius 32cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in the figure. ____ is the area of the design (Shaded region).

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Detailed Solution

\Rightarrow cos θ = \frac{AD}{OA}

\Rightarrow \cos 30 ° = \frac{AD}{32}

Now, as we know that

\cos 30 ° = \frac{3}{\sqrt{2}}

hence,

\Rightarrow \frac{3}{\sqrt{2}} = \frac{AD}{32}

\Rightarrow AD = 16 \sqrt{3} cm

⇒AB = 2AD

\Rightarrow AB = 2 \times 16 \sqrt{3}

⇒AB = 32

\sqrt{3}

cm
= 12×32

\sqrt{3}

×3×16
= 768

\sqrt{3}

cm2
Shaded region = Area of circle - Area of triangle
=π×(32)2cm2

= \frac{22}{7} \times {(32)}^{2}

= (\frac{22578}{7}) c m^{2}

Now we must calculate the area of the shaded region, which can be calculated by subtracting the area of the equilateral triangle from the area of the circle.
Hence,

(\frac{22578}{7} - 768 \sqrt{3}

)cm2

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