14. Two segments, each $1 \mathrm{~cm}$ long, are marked on opposite sides of a square of side $8 \mathrm{~cm}$. The ends of the segments are joined as shown in the diagram. What is the total shaded area?
A $2 \mathrm{~cm}^{2}$
B $4 \mathrm{~cm}^{2}$
C $6.4 \mathrm{~cm}^{2}$
D $8 \mathrm{~cm}^{2}$
E $10 \mathrm{~cm}^{2}$
14. B Let the height of the lower triangle be $h \mathrm{~cm}$. Therefore the height of the upper triangle is $(8-h) \mathrm{cm}$. Hence the shaded area in $\mathrm{cm}^{2}$ is $\frac{1}{2} \times 1 \times h+\frac{1}{2} \times 1 \times(8-h)=$ $\frac{1}{2} \times(h+8-h)=4$.