5. The diagram shows two circles and a square with sides of length $10 \mathrm{~cm}$. One vertex of the square is at the centre of the large circle and two sides of the square are tangents to both circles. The small circle touches the large circle. The radius of the small circle is $(a-b \sqrt{2}) \mathrm{cm}$.

What is the value of $a+b$ ?
5. 50 Let $O$ and $P$ be the centres of the large and small circles respectively and label points $Q$ and $S$ as shown in the diagram. Let the radius of the small circle be $r \mathrm{~cm}$. Draw line $P R$ so that $R$ is on $Q S$ and $P R$ is parallel to $O S$. Draw in line $O Q$. Since triangle $O Q S$ is right-angled and isosceles, $O Q^{2}=10^{2}+10^{2}$ by Pythagoras. Hence $O Q=10 \sqrt{2} \mathrm{~cm}$. Similarly, since triangle $P Q R$ is rightangled and isosceles, $P Q=r \sqrt{2} \mathrm{~cm}$. Note that angle $O Q S=$ angle $P Q S=45^{\circ}$ so $O P Q$ is a straight line. Therefore $10 \sqrt{2}=10+r+r \sqrt{2}$. This has solution
$$
r=\frac{10(\sqrt{2}-1)}{\sqrt{2}+1}=\frac{10(\sqrt{2}-1)(\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2}-1)}=\frac{10(2+1-2 \sqrt{2})}{2-1}=30-20 \sqrt{2} .
$$

Hence the radius of the small circle is $(30-20 \sqrt{2}) \mathrm{cm}$ and the value of $a+b$ is $30+20=50$.