Let the radius of the cone's base be \(r\) cm.
\(\therefore \pi r(r + 7) = 147.84\)
Or, \(\cfrac{22}{7} r(r + 7) = 147.84\)
Or, \( r(r + 7) = \cfrac{\cancel{14784}\cancel{672}168}{\cancel{100}25} \times \cfrac{7}{\cancel{22}}\)
Or, \( r^2 + 7r = \cfrac{168 \times 7}{25}\)
Or, \( 25r^2 + 175r = 1176\)
Or, \( 25r^2 + 175r - 1176 = 0\)
Or, \( 25r^2 + 280r - 105r - 1176 = 0\)
Or, \( 5r(5r + 56) - 21(5r + 56) = 0\)
Or, \( (5r + 56)(5r - 21) = 0\)
\(\therefore\) Either \( 5r + 56 = 0\) or \( r = -\cfrac{56}{5}\)
Or, \( 5r - 21 = 0\) or \( r = \cfrac{21}{5} = 4.2\)
Since the radius of the cone's base cannot be negative, the radius of the base is 4.2 cm. (Proved).