Problem 6 · Two areas on the sides of a rectangle
Central symmetry: $Q$ is the mirror image of $P$ and both areas follow from base times height.
Integer answer, at most 4 digitsIn a rectangle $ABCD$, the two diagonals meet at point $E$. We draw a line through $E$ that meets side $AB$ at $P$ and side $CD$ at $Q$, so that $PB = 7 \cdot AP$. If the area of rectangle $ABCD$ is 256 cm², what is the difference between the areas of triangles $PBE$ and $CQE$?
Reasoned solution
Key idea: $E$ is the centre of the rectangle. Any line through $E$ meets opposite sides at points symmetric about $E$, and both triangles have height $h/2$ from $E$.
Let $AB = w$ and $BC = h$ (area $wh = 256$). Since $PB = 7\,AP$, we get $AP = \dfrac{w}{8}$ and $PB = \dfrac{7w}{8}$.
$Q$ is the mirror image of $P$ about the centre, so $CQ = AP = \dfrac{w}{8}$.
Both triangles have their third vertex at $E$, at distance $\dfrac{h}{2}$ from sides $AB$ and $CD$: