Question

In a circular table cover of radius $ 32 \mathrm{cm}$, a design is formed leaving an equilateral triangle $ \mathrm{ABC}$ in the middle as shown in Fig. Find the area of the design.

Answer 1

Step 1: Find the area of the circle.

Given that: Radius of the circle is $32\mathrm{cm}$.

Area of circle$={\mathrm{\pi r}}^2$

$$\begin{gathered} =\frac{22}{7}\times {\left(32\right)}^2 \\ =3218.28{\mathrm{cm}}^2 \end{gathered}$$

Step 2: Find the area of triangle $\mathrm{ABC}$.

Area of a triangle $\mathrm{ABC}$ $=3\times$Area of the small triangle

The formula of the triangle when one angle and two opposite sides are given $\mathbf{=}\mathbf{}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\times }\mathbf{side}\mathbf{1}\mathbf{\times }\mathbf{side}\mathbf{2}\mathbf{\times }\mathbf{sin}\mathbf{\left(}\mathbf{A}\mathbf{\right)}$

$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{triangle}=3\times \frac{1}{2}\times \mathrm{side}1\times \mathrm{side}2\times \mathrm{sinA} \\ =3\times \frac{1}{2}\times 32\times 32\times \sin (120°) \\ =3\times \frac{1}{2}\times 32\times 32\times \frac{\sqrt{3}}{2}\mathbf{\left[}\mathbf{sin}\mathbf{(}\mathbf{120}\mathbf{°}\mathbf{)}\mathbf{}\mathbf{=}\mathbf{}\frac{\sqrt{\mathbf{3}}}{\mathbf{2}}\right] \\ =1330.21{\mathrm{cm}}^2 \end{gathered}$$

Step 3: Find the area of the designed part.

Area of the designed part $=$Area of the circle $-$Area of the triangle

$$\begin{gathered} =3218.28-1330.21 \\ =1888.07{\mathrm{cm}}^2 \end{gathered}$$

Hence, the area of the design is $1888.07{\mathrm{cm}}^2$ .