, 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$x=16$. [[IMAGE0]] 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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