In the diagram, $\triangle A B C$ is right-angled. Side $A B$ is extended in each direction to points $D$ and $G$ such that $D A=A B=B G$. Similarly, $B C$ is extended to points $F$ and $K$ so that $F B=B C=C K$, and $A C$ is extended to points $E$ and $H$ so that $E A=A C=C H$. The ratio of the area of the hexagon $D E F G H K$ to the area of $\triangle A B C$ is
(A) $4: 1$
(B) $7: 1$
(C) $9: 1$
(D) $16: 1$
(E) $13: 1$

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-106.jpg?height=401&width=561&top_left_y=672&top_left_x=1194)
Solution 1

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-208.jpg?height=401&width=553&top_left_y=957&top_left_x=843)

Let the area of $\triangle A B C$ be $x$.

We break up hexagon $D E F G H K$ into a number of pieces and calculate the area of each piece in terms of $x$.

Consider $\triangle A D E$. Since $A D=A B, A E=A C$ and $\angle D A E=\angle C A B$, then $\triangle A D E$ is congruent to $\triangle A B C$, so the area of $\triangle A D E$ is $x$.

Similarly, the area of each of $\triangle B G F$ and $\triangle C K H$ is equal to $x$.

Consider quadrilateral $A E F B$.

If we join this quadrilateral to $\triangle A B C$, we form $\triangle C F E$.

Since $A E=A C$, then $C E=2 C A$; similarly, $C F=2 C B$.

Since $\triangle C F E$ and $\triangle C B A$ share an angle at $C$ and have two pairs of corresponding sides enclosing this angle in the same ratio, then $\triangle C F E$ is similar to $\triangle C B A$.

Now, the side lengths of $\triangle C F E$ are twice those of $\triangle C B A$, so the area of $\triangle C F E$ is $2^{2}=4$ times that of $\triangle C B A$, so is $4 x$.

Thus, the area of quadrilateral $A E F B$ is $3 x$.

Similarly, the areas of quadrilaterals $A D K C$ and $B C H G$ are $3 x$.

Therefore, the area of hexagon $D E F G H K$ equals the sum of the areas of triangles $A B C, A D E$, $B G F$, and $C K H$, and of quadrilaterals $A E F B, A D K C$ and $B C H G$, so equals

$4 x+3(3 x)=13 x$.

Hence, the ratio of the ratio of hexagon $D E F G H K$ to the area of $\triangle A B C$ is $13: 1$.

## Solution 2

We can triangulate hexagon $D E F G H K$ by drawing vertical line segments of length equal to that of $A B$, horizontal line segments of length equal to that of $B C$, and slanting line segments of length equal to that of $A C$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-209.jpg?height=393&width=569&top_left_y=150&top_left_x=840)

Thus, we have triangulated $D E F G H K$ into 13 congruent triangles. (We can argue that each of these triangles is congruent to $\triangle A B C$ by observing that each has two perpendicular sides and noting that each has at least two sides easily seen to be equal in length to the corresponding sides in $\triangle A B C$.)

Therefore, the area of $D E F G H I$ is 13 times that of the area of $\triangle A B C$, so the ratio of the areas is $13: 1$.

ANswer: (E)

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