# Solution to the Divided Square Puzzle +-- {.image} [[DividedSquare.png:pic]] > The diagonal of the square has been divided into three sections of length $4$. What's the total shaded area? =-- ## Solution by [[Area of a Triangle]] and [[Vertically Opposite Angles]] +-- {.image} [[DividedSquareLabelled.png:pic]] =-- With the points labelled as above, consider first triangle $A B E$. The length of $E B$ is given as $4$. The height of $A E$ above $E B$ is the length of $A O$, which is half a diameter and so is $6$. Therefore the area of triangle $A B E$ is $12$. Since $E$ is $\frac{2}{3}$rds of the way along $D B$, $E$ is also $\frac{2}{3}$rds of the way along $A I$, so $E I$ is half the length of $A E$. As angles $O \hat{E} I$ and $A \hat{E} B$ are equal, since they are [[vertically opposite]], the height of $I$ above $E F$ is then half the height of $A$ above $B E$, so the area of $E I F$ is half that of $A B E$, so is $6$. The line $H G$ is vertical through $E$, and so as $E$ is the [[midpoint]] of $F B$, $G$ is the [[midpoint]] of $F I$. This means that triangle $E H F$ has half the area of triangle $E I F$, so has area $3$. Then angles $D \hat{F} J$ and $E \hat{F} H$ are equal as they are [[vertically opposite]] and $D J$ is [[parallel]] to $E H$ and $D F$ and $F E$ are the same length, so triangles $D F J$ and $E F H$ are [[congruent]]. Hence triangle $D F J$ has area $3$. The shaded region therefore has area $12 + 6 + 3 = 21$.