$25 \cdot 27 \triangle P Q R$ has vertex coordinates $P(0, a), Q(b, 0), R(c, d)$, where $a$, $b, c, d$ are all positive numbers, the origin and point $R$ are on opposite sides of $P Q$, the area of $\triangle P Q R$ can be expressed as
(A) $\frac{a b+a c+b c+c d}{2}$.
(B) $\frac{a c+b d-a b}{2}$.
(C) $\frac{a b-a c-b d}{2}$.
(D) $\frac{a c+b d+a b}{2}$.
(E) $\frac{a c+b d-a b-c d}{2}$.
(8th American High School Mathematics Examination, 1957)
［Solution］As shown in the figure, we consider two cases, first draw $R A \perp O x$.
(1) If $b<c$, then
$$
\begin{aligned}
S_{\triangle P Q R} & =S_{\text {quadrilateral } O P R A}-S_{\triangle Q A R}-S_{\triangle O P Q} \\
& =\frac{1}{2} c(a+d)-\frac{1}{2} d(c-b)-\frac{1}{2} a b \\
& =\frac{1}{2}(a c+b d-a b) ;
\end{aligned}
$$
(2) If $c<b$, then
$$
\begin{aligned}
S_{\triangle P Q R} & =S_{\text {quadrilateral } O P R A}+S_{\triangle Q A R}-S_{\triangle O P Q} \\
& =\frac{1}{2} c(a+d)+\frac{1}{2} d(b-c)-\frac{1}{2} a b \\
& =\frac{1}{2}(a c+b d-a b) .
\end{aligned}
$$

Therefore, the answer is $(B)$.