, 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 $x=16$.

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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$$$|PQRU|=|\triangle PQT|-|\triangle$ URT|=81 $\text{}$
cm}^2 - 9 $\text{}$ cm}^2=72 $\text{}$ cm}^2.$$$$

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