Two circles of radius 4 cm and 9 cm respectively touch each other externally at a point and a common tangent touches them at the points P and Q respectively. Then the area of a square with one side PQ, is

$r_{1}+r_{2}=13 mathrm{~cm}$

$r_{2}-r_{1}=9-4=5 mathrm{~cm}$

$mathrm{PQ}=sqrt{left(egin{array}{c} ext { distance between } \ ext { centres }end{array} ight)^{2}-left(r_{2}-r_{1} ight)^{2}}$

$=sqrt{left(13^{2}-5^{2} ight)}=12 mathrm{~cm}$

$herefore$ Area of square $=12 imes 12=144$ sq. $mathrm{cm}$.