# Four Squares Around a Quadrilateral +-- {.image} [[FourSquaresAroundaQuadrilateral.png:pic]] > The four squares enclose a quadrilateral of area $10$. What's the total shaded area? =-- ## Solution by [[Area of a Triangle]] +-- {.image} [[FourSquaresAroundaQuadrilateralLabelled.png:pic]] =-- In the above diagram, the point $N$ is obtained by shifting $L$ parallel to $I J$ so that $N I$ has the same length as $L I$. This means that triangles $I L J$ and $I N J$ have the same area. Angles $J \hat{I} N$ and $J \hat{I} L$ add up to $180^\circ$. This can be seen in various ways, one of which is that reflecting rectangle $L N P M$ through the line through $I$ takes triangle $N I P$ to $L I M$. Angles $A \hat{I} B$ and $J \hat{I} L$ also add up to $180^\circ$ since the [[angles at a point]] add up to $360^\circ$. Since $A I$ and $N I$ have the same length, and $B I$ and $J I$ also have the same length, this means that triangles $A I B$ and $N I J$ are [[congruent]] and so have the same area. Putting that together, triangles $A I B$ and $L I J$ have the same area. Following round the same argument for the other grey triangles, each grey triangle has the same area as a triangle formed by cutting the central quadrilateral along a diagonal. So the four grey regions have total area $20$. ## Solution by the [[Sine Rule]] This argument follows a similar line to the above, except that showing that triangles $L I J$ and $A I B$ have the same area is deduced from the [[sine rule]]. Since angles $L \hat{I} J$ and $A \hat{I} B$ add up to $180^\circ$, they have the same $sine$ and so triangles $A I B$ and $L I J$ have the same area.