19. The figure shows four smaller squares in the corners of a large square. The smaller squares have sides of length $1 \mathrm{~cm}, 2 \mathrm{~cm}, 3 \mathrm{~cm}$ and $4 \mathrm{~cm}$ (in anticlockwise order) and the sides of the large square have length $11 \mathrm{~cm}$.
What is the area of the shaded quadrilateral?

A $35 \mathrm{~cm}^{2}$
B $36 \mathrm{~cm}^{2}$
C $37 \mathrm{~cm}^{2}$
D $38 \mathrm{~cm}^{2}$
E $39 \mathrm{~cm}^{2}$
19. A The small numbers in the figure show the lengths in $\mathrm{cm}$ of each line segment. The larger numbers inside the figure show the areas in $\mathrm{cm}^{2}$ of each square or trapezium. (The area of a trapezium is $\frac{1}{2}(a+b) h$ where $a$ and $b$ are the lengths of the parallel sides and $h$ is the perpendicular distance between them.)
So the area of the shaded portion in $\mathrm{cm}^{2}$ is $11 \times 11-(1+12+4+15+9+14+16+15)=35$. (See the extended solutions for a beautifully elegant solution of this problem.)