6. A cone is formed by a circle and a three-quarter circle and the surface area is \( 4224 \mathrm{~cm}^{2} \). Find the radius of the three-quarter circle. (Take \( \pi=\frac{22}{7} \) )

To solve for the radius of the three-quarter circle that forms the cone, we need to analyze the surfaces involved. The surface area of a cone is given by the formula: \[ \text{Surface Area} = \pi r l + \pi r^2 \] where \( r \) is the radius of the base of the cone, and \( l \) is the slant height. In this case, the slant height \( l \) is equal to the radius of the three-quarter circle. Given that the total surface area \( S = 4224 \, \text{cm}^2 \) and substituting \( \pi = \frac{22}{7} \): \[ 4224 = \frac{22}{7} r l + \frac{22}{7} r^2 \] Since \( l = r \) (the slant height equals the radius of the three-quarter circle), we can replace \( l \) with \( r \): \[ 4224 = \frac{22}{7} r^2 + \frac{22}{7} r^2 \] \[ 4224 = \frac{44}{7} r^2 \] Now, to isolate \( r^2 \), we multiply both sides by \( 7 \): \[ 4224 \cdot 7 = 44 r^2 \] \[ 29568 = 44 r^2 \] Next, divide both sides by 44: \[ r^2 = \frac{29568}{44} \] \[ r^2 = 672 \] Now, taking the square root of both sides gives: \[ r = \sqrt{672} \] \[ r = 8\sqrt{105} \, \text{cm} \quad (\text{approx} \, 20.8 \text{ cm, but it's expressed in simplest radical form as } 8\sqrt{105}) \] So, the radius of the three-quarter circle is \( r = 8\sqrt{105} \, \text{cm} \).