Notes
a square and a triangle solution

Solution to the A Square and a Triangle Puzzle

Solution by Similar Triangles

In the diagram above, $E$ is the midpoint of $F D$ and $G$ is such that angle $G \hat{F} D$ is a right-angle. Triangle $D E B$ has area one quarter that of the square, which is $9$.

Since angle $H \hat{D} B$ is a right-angle, angles $H \hat{D} F$ and $E \hat{D} B$ add up to $90^\circ$. Since triangle $D E B$ is also right-angled, so also angles $E \hat{D} B$ and $D \hat{B} E$ add up to $90^\circ$. Therefore, angles $G \hat{D} F$ and $D \hat{B} E$ are equal. Then $F D$ and $E B$ are the same length, so triangles $G F D$ and $D E B$ are congruent.

Angles $G \hat{F} H$ and $B \hat{F} E$ add up to $90^\circ$, since together with $D \hat{F} G$, which is a right-angle, they are angles at a point on a straight line. So angle $G \hat{F} H$ is the same as $G \hat{D} F$. This means that triangles $H G F$ and $H F D$ have the same angles (they also both share $G \hat{H} F$) and so are similar. The edges $G F$ and $F D$ correspond, and since $G F$ has the same length as $E D$, this means that the length scale factor from $H G F$ to $H F D$ is $2$. The area scale factor is therefore $4$.

Hence triangle $H F D$ has four times the area of triangle $H G F$. This means that triangle $G F D$ has three times the area of triangle $H G F$. Since $G F D$ has area $9$, this means that triangle $H G F$ has area $3$.

The total shaded area is then $9 + 9 + 9 + 3 = 30$.

Solution by the Double Angle Formula of Trigonometry and Pythagoras' Theorem

Let $\theta$ be angle $B \hat{D} C$. Since the area of the square is $36$, its side length is $6$ and so then $B C$ has length $3$. Therefore, $\tan(\theta) = \frac{3}{6} = \frac{1}{2}$.

Angle $F \hat{B} D$ is $2 \theta$ and so by the double angle formula its tangent is given by:

Using Pythagoras' theorem, the length of $D B$ is:

So then the length of $H D$ is $3 \sqrt{5} \times \tan(2\theta) = 4 \sqrt{5}$. The area of the shaded triangle is then: