19. In quadrilateral $A B C D, \angle A B C=\angle A D C=90^{\circ}, A D=D C$ and $A B+B C=20 \mathrm{~cm}$.

What is the area in $\mathrm{cm}^{2}$ of quadrilateral $A B C D$ ?
19. 100 Let the lengths of $B C, A B$ and $A C$ be $x, y$ and $z$ centimetres respectively. Let the area of $\triangle A C D$ be $U \mathrm{~cm}^{2}$ and let the area of $\triangle A B C$ be $V \mathrm{~cm}^{2}$. Note that $\triangle A C D$ is one quarter of the square which has $A C$ as an edge. Hence $U=\frac{1}{4} z^{2}$. Next, using Pythagoras, $z^{2}=x^{2}+y^{2}=(x+y)^{2}-2 x y=20^{2}-4 V$. Hence $U=\frac{1}{4}(400-4 V)=100-V$. Therefore the area in $\mathrm{cm}^{2}$ of $A B C D$ is $U+V=100$.
(Note: Since the answer to the problem is independent of $x$ and $y$, one could observe that the given properties of quadrilateral $A B C D$ are satisfied by a square of side $10 \mathrm{~cm}$ which has area $100 \mathrm{~cm}^{2}$ and conclude that this is therefore the required answer.)