Question

Daniel took a wire of 66 cm. He bent it into the the shape of a circle. If the same wire is re-bent into the shape of a square then what will be the length of its sides, which shapes encloses more area?

Answer 1

Dear student,
$$\begin{gathered} Weknow, \\ Circumferenceofcircle=lengthofwiretobemoulded \\ \Rightarrow 2\mathrm{\pi r}=66--[\mathrm{where}\mathrm{r}\mathrm{is}\mathrm{radius}] \\ \Rightarrow \mathrm{r}=\frac{66\times 7}{2\times 22} \\ \Rightarrow \mathrm{r}=\frac{3\times 7}{2\times 1} \\ \Rightarrow \mathrm{r}=\frac{21}{2}\mathrm{cm} \\ \mathrm{Also},\mathrm{Perimeter}\mathrm{of}\mathrm{square}=\mathrm{length}\mathrm{of}\mathrm{wire}\mathrm{to}\mathrm{be}\mathrm{moulded} \\ \Rightarrow 4\times \mathrm{s}=66--[\mathrm{where}\mathrm{s}\mathrm{is}\mathrm{side}\mathrm{of}\mathrm{square}] \\ \Rightarrow \mathrm{s}=\frac{66}{4} \\ \Rightarrow \mathrm{s}=\frac{33}{2}\mathrm{cm} \\ \mathrm{Area}\mathrm{of}\mathrm{circle}={\mathrm{\pi r}}^2 \\ =\frac{22}{7}\times \frac{21}{2}\times \frac{21}{2} \\ =\frac{11}{1}\times \frac{3}{1}\times \frac{21}{2} \\ =\frac{33\times 21}{2} \\ =\frac{693}{2} \\ =346.5{\mathrm{cm}}^2--\left[1\right] \\ \mathrm{Area}\mathrm{of}\mathrm{square}={\mathrm{s}}^2 \\ =\frac{33}{2}\times \frac{33}{2} \\ =\frac{1089}{4} \\ =272.25{\mathrm{cm}}^2--\left[2\right] \\ \mathrm{Thus},\mathrm{from}\left[1\right]\mathrm{and}\left[2\right] \\ \mathrm{Circle}\mathrm{encloses}\mathrm{more}\mathrm{area}. \end{gathered}$$
Regards