19. C Since the non-shaded squares are congruent and since $M N=6 \mathrm{~cm}$, both $S N$ and $M R$ have length $(10-6) \mathrm{cm} \div 2=2 \mathrm{~cm}$. Therefore the areas of the four non-shaded squares are each $(2 \times 2) \mathrm{cm}^{2}=4 \mathrm{~cm}^{2}$.

Label the point $X$ on $S P$ as shown. Since all of the non-shaded squares are congruent, the lengths of $S M$ and $S X$ are equal and hence triangle $M S X$ is an isosceles, right-angled triangle with angles of $90^{\circ}, 45^{\circ}$ and $45^{\circ}$. Therefore, since the non-shaded triangles are all isosceles and have an angle of $45^{\circ}$, they are also right-angled. Therefore these four triangles can be fitted together to form a square of side-length $6 \mathrm{~cm}$. Hence the total area of the non-shaded triangles is $(6 \times 6) \mathrm{cm}^{2}$. Therefore the area of the shaded region is $(10 \times 10-4 \times 4-6 \times 6) \mathrm{cm}^{2}=(100-16-36) \mathrm{cm}^{2}=48 \mathrm{~cm}^{2}$.
19. In the diagram, $P Q R S$ is a square of side $10 \mathrm{~cm}$. The distance $M N$ is $6 \mathrm{~cm}$. The square is divided into four congruent isosceles triangles, four congruent squares and the shaded region.
What is the area of the shaded region?
A $42 \mathrm{~cm}^{2}$
B $46 \mathrm{~cm}^{2}$
C $48 \mathrm{~cm}^{2}$
D $52 \mathrm{~cm}^{2}$
E $58 \mathrm{~cm}^{2}$