3
$\begingroup$

$ABCD$ 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.

$\endgroup$
  • 1
    $\begingroup$ @imranfat Yes, I know. $\endgroup$ – Rezwan Arefin Feb 3 '16 at 14:36
  • 2
    $\begingroup$ @imranfat I dont know please explain. $\endgroup$ – Max Payne Feb 3 '16 at 14:37
  • 1
    $\begingroup$ Hold on, There are no right angles in quad ABCD...Let me take a cup of coffee first... $\endgroup$ – imranfat Feb 3 '16 at 14:41
4
$\begingroup$

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 $M+N = \frac{1}{2} * AC *BD = \frac{1}{2} * 6 * 8 =24$
$\endgroup$
  • 2
    $\begingroup$ +1 Very cool! I went into using trigonometry stuff, but this is FAR quicker... $\endgroup$ – imranfat Feb 3 '16 at 15:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.