# Trapezium A **trapezium** is a [[quadrilateral]] in which two sides are parallel. +-- {.image} [[trapezium.png:pic]] =-- Writing $a$ and $b$ for the lengths of the parallel sides and $h$ for the perpendicular distance between them then the area of a trapezium is given by the formula: $$ A = \frac{a + b}{2} h $$ There are various ways to see that this is true. One is to take two trapeziums and place them alongside each other to make a [[parallelogram]]. The area of the parallelogram is $(a + b)h$, and the trapezium is half of that. ## Crossed Trapezium A **crossed trapezium** is a trapezium together with its diagonals. The four [[triangles]] formed by drawing in the diagonals have special properties. The two triangles which include the parallel sides are [[similar]] while the other two triangles have the same area. +-- {.image} [[crossedtrapezium.png:pic]] =-- In this picture, $ C\hat{A}B = D \hat{C } A $ as they are [[alternate angles]], similarly $ D \hat{ B} A = B \hat{ D } C $. So triangles $ A O B $ and $ C O D $ are [[similar]]. To see that the side triangles are the same area, note that triangles $A D B$ and $A C B$ have the same base and same perpendicular height so have the same area. Both consist of a side triangle together with triangle $A O B$, and hence the two side triangles have the same area. The areas of the four triangles are related by the scale factor between the similar triangles. Let $s$ be the (linear) scale factor from $C O D$ to $A O B$. Then $A O = s O C$. Triangles $A O D$ and $D O C$ have the same height and their bases are, respectively, $A O$ and $O C$. We therefore have that the area of $A O D$ is $s$ times that of $D O C$. Similarly, the area of $A O B$ is $s$ times that of $A O D$. This means that the total area of the trapezium is $1 + s + s + s^2 = (1 + s)^2$ times the area of $DOC$. [[!redirects crossed trapezium]] [[!redirects area of a trapezium]]