Question

please answer this question

Answer 1

diagonal of square ABCD = DIAMETER of outer circle ..

now by Pythagoras theorem diagonal of square ABCD :
${diagonal} = \sqrt{ ({2a })^{2} + ({2a})^{2} }$
so diagonal = 2√2 × a
so radius of outer circle = √2×a
the circumference of outer circle will be = 2×π×(√2×a)

now coming on to inner circle,
radius of inner circle = a
so, circumference of inner circle = 2×π×a


hence, the ratio of circumference of outer circle and inner circle will be = √2:1