# Joining the Midpoints of the Sides of a Quadrilateral

$ABCD$ is a quadrilateral. $P$, $Q$ and $R$ are the midpoints of $AB$, $BC$ and $CD$ respectively. If $PQ = 3$, $QR = 4$ and $PR = 5$; find the area of $ABCD$.

Since, $5^2 = 3^2+4^2$, So, $\angle PQR = 90^o$
I can't Find a way to solve this.

Note: This is a problem from BDMO $2010$ National.

• @imranfat Yes, I know. – Rezwan Arefin Feb 3 '16 at 14:36
• @imranfat I dont know please explain. – Max Payne Feb 3 '16 at 14:37
• Hold on, There are no right angles in quad ABCD...Let me take a cup of coffee first... – imranfat Feb 3 '16 at 14:41

Link $AC$, $BD$ and denote $O$ as their intersection point. Since $PQ \bot QR$ ,$PQ//AC,AC = 2 PQ$ and $QR//BD,BD = 2QR\Rightarrow AC \bot BD,AC = 6,BD =8$ then the quadrilateral is divided into to triangle $ABC$, $ACD$ which share the same edge $AC$.Then the area of the quadrilateral equals to the sum area of these two triangle. Solve the problem using following steps:
1. denote M as the area of triangle $ABC =\frac{1}{2} * AC *BO$
2. denote N as the area of triangle $ACD =\frac{1}{2} * AC *DO$
3. the answer is $M+N = \frac{1}{2} * AC *BD = \frac{1}{2} * 6 * 8 =24$