A square and an equilateral triangle have the same perimeter.
A square and an equilateral triangle have the same perimeter. What is the ratio of the area of the circle circumscribing the square to the area of the circle inscribed in the triangle?
Answer/Solution
27/8
Steps/Work
let x be side of square
perimeter of square=4x=perimeter of triangle=3*side of triangle
so side of eq. triangle=(4/3)*x
diameter of circle circumscribing the square=sqrt(2)*x
area of circle circumscribing the square= pi*(sqrt(2)*x)^2/4=(pi/2)*x^2 ----(1)
to find radius of the circle inscribed in the triangle
area of triangle=r*s=sqrt(3)/4 * (4x/3)^2
now s=(4/3)*x+(4/3)*x+(4/3)*x/2=2x
so sqrt(3)/4 * (4x/3)^2=r*2x gives
r={2/3*(3^1/2)}*x
area of the circle inscribed in the triangle=pi*[{2/3*(3^1/2)}*x]^2
=pi*(4/27)*x^2 -------(2)
so reqd ratio= eqn(1)/eqn(2)
=[(pi/2)*x^2]/[pi*(4/27)*x^2]=27/8
so reqd ratio=27:8
ANSWER:C
perimeter of square=4x=perimeter of triangle=3*side of triangle
so side of eq. triangle=(4/3)*x
diameter of circle circumscribing the square=sqrt(2)*x
area of circle circumscribing the square= pi*(sqrt(2)*x)^2/4=(pi/2)*x^2 ----(1)
to find radius of the circle inscribed in the triangle
area of triangle=r*s=sqrt(3)/4 * (4x/3)^2
now s=(4/3)*x+(4/3)*x+(4/3)*x/2=2x
so sqrt(3)/4 * (4x/3)^2=r*2x gives
r={2/3*(3^1/2)}*x
area of the circle inscribed in the triangle=pi*[{2/3*(3^1/2)}*x]^2
=pi*(4/27)*x^2 -------(2)
so reqd ratio= eqn(1)/eqn(2)
=[(pi/2)*x^2]/[pi*(4/27)*x^2]=27/8
so reqd ratio=27:8
ANSWER:C