A circle of radius 13 cm touches the adjacent sides AB and BC of a square ABCD atM and N, respectively. If AB = 18 cm and the circle intersects the other two sides CD and DA at P and Q, respectively, then the area, in sq. cm, of triangle PMD is

The radius of circle is 13 cm. Thus, BM = BN = 13 cm and MA = DT = CN = OT = 5 cm.

Let the length of CP be x cm. Since DT = 5 cm, thus PT = (13 - x) cm

In triangle OTP, $$OT^2+TP^2=OP^2$$

=> $$5^2+\left(13-x\right)^2=13^2$$

=> $$25+169-26x+x^2=169$$

=> $$x^2-26x+25=0$$

=> $$x=1$$ or $$x=25$$. But x cannot be 25 cm, because then the length of the side will be negative. Thus, $$x=1$$

Thus, PD = 13 - x + 5 = 17 cm, and MT = 18 cm

Thus, area of triangle PMD = $$\frac{1}{2}\times17\times18=153\ cm^2$$