A7. The diagram on the right shows a 4 by 4 square placed on top of a 5 by 5 square, so that they have one vertex in common as shown. One diagonal of each square is also drawn. What is the area of the shaded region that is inside the 4 by 4 square and between the two diagonals?
SOLUTION
3.5

Referring to the diagram on the right, $J L$ is a diagonal of the square $J K L M$ therefore $\angle P J A=45^{\circ}$. The given configuration of the two squares means that $\angle A P J=90^{\circ}$ therefore $\angle J A P=180^{\circ}-45^{\circ}-90^{\circ}=$ $45^{\circ}$. Hence triangle $A P J$ is isosceles and $A P=J P$. As $J P$ is the difference between the side lengths of the two squares, then $A P=J P=5-4=1$. Similarly, we can deduce that $C Q=1$. The shaded area is equal to the area of triangle $P B Q$ minus the area of triangle $A B C$. As $A B=B C=4-1$, this is $\frac{1}{2} \times 4 \times 4-\frac{1}{2} \times 3 \times 3=3.5$ square units.