CEMC Fryer, 2022

Solution 1

The area of $ABGH$ is equal to the
sum of the areas of $ABCD$ and $EFGH$ minus the area of overlap $EFCD$ since it is counted twice in this
sum.

Thus, the area of $ABGH$ is $(13\textrm{ cm}^2) + (13 \textrm{ cm}^2) - (5 \textrm{ cm}^2)= 21\textrm{ cm}^2$.

Solution 2

The area of $ABCD$ is 13 cm$^2$ and is equal to the sum of the areas
of $ABFE$ and $EFCD$.

Since the area of $EFCD$ is 5 cm$^2$, then the area of $ABFE$ is $(13\textrm{ cm}^2) - (5 \textrm{ cm}^2) = 8\textrm{ cm}^2$.

The area of $ABGH$ is equal to the
sum of the areas of $ABFE$ and $EFGH$ or $(8\textrm{ cm}^2) + (13 \textrm{ cm}^2) = 21\textrm{ cm}^2$.
Let the area of the overlapped region, $\triangle KLN$, be $x$ cm$^2$.

The area of $\triangle KLN$ is equal
to half of the area of $\triangle JKL$, and so the area of $\triangle JKN$ is also $x$ cm$^2$.

Since $\triangle JKL$ and $\triangle MLK$ are identical, then the
area of $\triangle MLN$ is also $x$ cm$^2$.

Thus, the area of the figure $JKLMN$
is $3x$ cm$^2$, and so $3x=48$ or $x=16$.

Since $\triangle JKL$ is
right-angled at $K$, then its area is
given by $\frac12(JK)(KL)$.

Since the area of $\triangle JKL$ is
$2x \text{ cm}^2=32$ cm$^2$, then $\frac12(JK)(KL)=32$ cm$^2$ or $\frac12(6 \text{ cm})(KL)=32$ cm$^2$, and so $KL=\frac{32}{3}$ cm.
We use the notation $|\triangle URT|$ to denote the area of
$\triangle URT$, and similar notation
for other areas.

Since $|PQRS|+|\triangle URT|=|PQTUS|$, then $$|\triangle URT|=|PQTUS|-|PQRS|=117 \text{ cm}^2 - 108 \text{ cm}^2=9 \text{ cm}^2.$$ Since $|PQRU|+|\triangle URT|=|\triangle PQT|$, then $$|PQRU|=|\triangle PQT|-|\triangle URT|=81 \text{ cm}^2 - 9 \text{ cm}^2=72 \text{ cm}^2.$$