1957 AMC 12 Question 25 - thichhoctoan.com

Question 25

The vertices of triangle $PQR$ have coordinates as follows: $P(0,a),$ $Q(b,0),$ $R(c,d),$ where $a,$ $b,$ $c,$ and $d$ are positive. The origin and point $R$ lie on opposite sides of $PQ.$ The area of triangle $PQR$ may be found from the expression:

$\dfrac{ab+ac+bc+cd}{2}$
$\dfrac{ac+bd-ab}{2}$
$\dfrac{ab-ac-bd}{2}$
$\dfrac{ac+bd+ab}{2}$
$\dfrac{ac+bd-ab-cd}{2}$