Two circles touch externally. The sum of their areas is 130 sq.cm and the distance between their centres is 14 cm. The radius of the smaller circle is:

By given data,

\( \pi r_{1}^{2}+ \pi r_{2}^{2} = 130 \pi sq.cm. \)

\( r_{1}^{2}+r_{2}^{2} = 130 \)

\( r_{1}+r_{2}^{2} = 14 cm \)

\( \left(r_{1}+r_{2}\right)^{2} = \left(14\right)^{2} \)

\( r_{1}^{2}+r_{2}^{2}=2r_{1}.r_{2} = 196 \)

\( 2r_{1}r_{2} = 196 - 130 \)

\( r_{1}r_{2} = 33 \)

\( \Large r_{2} = \frac{33}{r_{1}} \)

\( \Large r_{1}+\frac{33}{r_{1}} = 14 \)

\( r_{1}^{2}+33-14r_{1} = 0 \)

=> \( r_{1}^{2} - 11r_{1} - 3r_{1} + 33 = 0 \)

=> \( r_{1} \left(r_{1}-11\right)-3 \left(r_{1-11}\right) = 0 \)

=> \( r_{1} = 3 \left(or\right) 11 \} \)

when \( r_{1} = 3 r_{2} = 11 \)

when \( r_{1} = 11 r_{2} = 3 \)

Therefore, radius of smaller circle is 3 cm