[ Pythagorean Theorem (direct and inverse) $]$ [ Circle, sector, segment, etc. $\quad]$

In the circular sector $OAB$, the central angle of which is $45^{\circ}$, a rectangle $KMPT$ is inscribed. The side $KM$ of the rectangle lies on the radius $OA$, the vertex $P$ is on the arc $AB$, and the vertex $T$ is on the radius $OB$. The side $KT$ is 3 units longer than the side $KM$. The area of the rectangle $KMPT$ is 18. Find the radius.
Let $K M=P T=x$. Then

$$
K T=M P=x+3, x(x+3)=18
$$

From this, we find that $x=3$.

From the right triangle $O K T$, we find that

$$
O K=K T=x+3=6,
$$

and from the right triangle $O M P-$

$$
O P^2=O M^2+M P^2=(O K+K M)^2+M P^2=(6+3)^2+6^2=9^2+6^2=117 .
$$

Therefore,

$$
O P=\sqrt{117}=3 \sqrt{13} .
$$

## Answer

$3 \sqrt{13}$.