8. The diagram shows a square of perimeter $20 \mathrm{~cm}$ inscribed inside a square of perimeter $28 \mathrm{~cm}$.
What is the area of the shaded triangle?
A $6 \mathrm{~cm}^{2}$
B $7 \mathrm{~cm}^{2}$
C $8 \mathrm{~cm}^{2}$
D $9 \mathrm{~cm}^{2}$
E $10 \mathrm{~cm}^{2}$
8. A First note that the length of each side of the larger square is $7 \mathrm{~cm}$ and that of the smaller square is $5 \mathrm{~cm}$. So their areas are $49 \mathrm{~cm}^{2}$ and 25 $\mathrm{cm}^{2}$ respectively. Let $\angle B A C=x^{\circ}$. As the interior angles of a triangle sum to $180^{\circ}$ and $\angle A B C=90^{\circ}, \angle A C B=(180-90-x)^{\circ}=(90-x)^{\circ}$. The angles on a straight line sum to $180^{\circ}$, so $\angle E C D=[180-90-(90-x)]^{\circ}=x^{\circ}$.

Consider triangles $A B C$ and $C D E$. Note that $\angle B A C=\angle D C E=x^{\circ}$; $\angle A B C=\angle C D E=90^{\circ} ; A C=C E=5 \mathrm{~cm}$. Therefore the two triangles are congruent (AAS). By a similar method, it may be shown that all four triangles which lie between the two squares are congruent to each other.
So the area of the shaded triangle is $\frac{1}{4}(49-25) \mathrm{cm}^{2}=6 \mathrm{~cm}^{2}$.