Question

After some effort, Virat drew a perfect circle and a square on a paper. Then he cut those
pieces out and weighed them in a sophisticated machine. He surprisingly noticed that
the weights were identical. If so, what would be the ratio of radius (r) of the circle and
edge (a) of the square (r:a)​

Answer 1

$Let \: radius \:of \:a\:circle = r \:units$

$edge\:of \:the \:square = a \:units$

$Thickness \:of \:the \:paper = h$

/* According to the problem given */

/* Weights of the papers are equal */

$Volume \:of \:a Circle = Volume \:of \:a\:square$

$\implies \pi r^{2} \times h = a^{2} \times h$

$\implies \frac{r^{2}}{a^{2}} = \frac{1}{\pi}$

$\implies \Big(\frac{r}{a}\Big)^{2} = \frac{1}{\frac{22}{7}}$

$\implies \Big(\frac{r}{a}\Big)^{2} = \frac{7}{22}$

$\implies \frac{r}{a} = \sqrt{\frac{7}{22}}$

$\implies r : a = \sqrt{7} : \sqrt{22}$

Therefore.,

$\red { r : a } \green { = \sqrt{7} : \sqrt{22} }$

•••♪