# Solution to the [[A Square and a Triangle]] Puzzle +-- {.image} [[ASquareandaTriangle.jpeg:pic]] > The square has area 36. What’s the area of the right-angled triangle? =-- ## Solution by [[Similar Triangles]] +-- {.image} [[ASquareandaTriangleLabelled.jpeg:pic]] =-- 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: $$ \tan(2\theta) = \frac{2 \tan(\theta)}{1 - \tan^2(\theta)} = \frac{4}{3} $$ Using [[Pythagoras' theorem]], the length of $D B$ is: $$ \sqrt{3^2 + 6^2} = \sqrt{45} = 3\sqrt{5} $$ 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: $$ \frac{1}{2} \times 3 \sqrt{5} \times 4 \sqrt{5} = 30 $$