Question

13. If the diameter of a circle and the side of
a square are equal, then show that the ratio
of the areas of the circle and the square is
equal to the ratio of the perimeters of the
circle and the square. ​

Answer 1

$\underline{\sf\ \ Given:-}$

• Side of square is equal to the Diameter of circle

$\underline{\sf\ \ To \ Prove }$

• We have to prove that the ratio of the area of circle and square is equal to ratio of their perimeters .

$\underline{\sf\ \dag \ Solution :- }$

• Let the side of square be a
• So diameter of circle also be a

$\underline {\sf{ \: \: Square \: \: \: \: }}\begin{cases}\sf{Area= (side)^2}\\ \sf{Perimeter= 4\times side}\end{cases}\\ \\ \underline{\sf{ \: \: \: circle \: \: \: }}\begin{cases}\sf{Area= \pi r^2}\\ \sf{Perimeter=2 \pi r}\end{cases}$

• Now find the ratio of their Areas-

$:\implies\sf \dfrac{Ar(Square)}{Ar(Circle)}= \dfrac{(a)^2}{\pi (^a\!/_2)^2}\ \ \ \ \big\lgroup \ a= Diameter,\ r= a\!/_2\big\rgroup\\ \\ \\ :\implies\sf \dfrac{Ar(square)}{Ar(Circle)}= \dfrac{\cancel{a^2}}{\pi\ \cancel{a^2}\!/_4}}\\ \\ \\ :\implies\sf \dfrac{Ar(Square)}{Ar(Circle)}= \dfrac{1}{\pi\!/_4}\\ \\ \\ :\implies\sf \dfrac{Ar(Square)}{Ar(Circle)}= \dfrac{4}{\pi}\\ \\ \\ :\implies\sf \ Ratio_{areas} = \underline{\boxed{\purple{\sf\ 4 : \pi}}}\ \ -----\ eq.(i)$

• Now find the ratio of their perimeters

$:\implies\sf \dfrac{Perimeter (Square)}{Perimeter (Circle)}= \dfrac{4a}{2 \pi r}\ \ \ \ \big\lgroup r= a\!/_2\big\rgroup\\ \\ \\ :\implies\sf \dfrac{P(Square)}{P(Circle)}= \dfrac{4a}{\not2\times \pi \times \frac{a}{\not2}}\\ \\ \\ :\implies\sf \dfrac{P(Square)}{P(Circle)}= \dfrac{4\not a}{\pi \not a}\\ \\ \\ :\implies\sf \dfrac{P(Square)}{P(Circle)}= \dfrac{4}{\pi}\\ \\ \\ :\implies\sf \ Ratio_{Perimeters} = \underline{\boxed{\red{\sf\ 4 : \pi}}}\ \ -----\ eq.(ii)$

• From Eq.(i) & (ii)
• Ratios of their areas = Ratios of their perimeters

$\underline{\boxed{\sf\ \ \big[4 : \pi\big]\ =\big[4: \pi \big]}}$