Radii of two circles:
\r1 = x + 2 cm
\r2 = 2x cm
\Given ratio of areas is A1 : A2 = ( x + 7 ) : x²
\A1 = π(r1)² = π( x + 2 )²
\A1 = π( x² + 2x + 4 )
\A2 = π(r2)² = π( 2x )²
\A2 = 4πx²
\Actual ratio of areas is A1 : A2 = π( x² + 2x + 4 ) : 4πx²
\A1 : A2 = ( x² + 2x + 4 ) : 4x²
\Given ratio of areas is equal to Actual ratio of areas
\( x + 7 ) / x² = ( x² + 2x + 4 ) / 4x²
\4( x + 7 ) = ( x² + 2x + 4 )
\4x + 28 = x² + 2x + 4
\x² - 2x - 24 = 0
\x² - 6x + 4x - 24 = 0
\x(x - 6) + 4(x - 6) = 0
\(x - 6)(x+ 4) = 0
\By using zero product property : If AB = 0 then A = 0 , B = 0
\(x - 6) = 0 and (x+ 4) = 0
\x = 6 and x = -4
\x = -4 is invalid due to negative symbol.
\x = 6 is valid.
\Substitute x = 6 in radii.
\r1 = x + 2 = 6+2 = 8
\r2 = 2x = 2*6 = 12
\The value of the radius of the smaller circle is 8 cm
\