Notes
parallelogram area solution

Parallelogram Area

Solution by Similar Triangles and Pythagoras' Theorem

In the diagram above point $C$ is on the continuation of $A B$ so that angle $D \hat{C} B$ is a right-angle. Since angles in a triangle add up to $180^\circ$, angle $B \hat{D} C$ is therefore $90^\circ - b$ which is $a$. Triangle $A C D$ is also a right-angled triangle with angle $a$ and so triangles $A C D$ and $D C B$ are similar. Since $A D$ has length $10$ and $B D$ has length $5$, triangle $D C A$ is double triangle $B C D$.

Let $x$ be the length of $B C$. Then $D C$ has length $2 x$ and $A C$ has length $4 x$, so $A B$ has length $3 x$. The area of the parallelogram is then $3 x \times 2 x = 6 x^2$. Applying Pythagoras' theorem to triangle $D B C$ yields the equation:

and so $x^2 = 5$. Therefore, the area is $30$.