Let the radius of the base of the cone be \(r\) cm.
\(\therefore \pi r(r + 7) = 147.84\)
Or, \(\cfrac{22}{7} r(r + 7) = 147.84\)
Or, \(r(r + 7) = \cfrac{14784}{100} \times \cfrac{7}{22} = \cfrac{168 \times 7}{25}\)
So, \(r^2 + 7r = \cfrac{168 \times 7}{25}\)
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 \(5r + 56 = 0\), i.e., \(r = -\cfrac{56}{5}\)
Or, \(5r - 21 = 0\), i.e., \(r = \cfrac{21}{5} = 4.2\)
Since the radius of a cone cannot be negative, the radius of the base of the cone is 4.2 cm. .