1957 AHSME Problems/Problem 25

Problem

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:

$\textbf{(A)}\ \frac{ab + ac + bc + cd}{2} \qquad \textbf{(B)}\ \frac{ac + bd - ab}{2}\qquad \textbf{(C)}\ \frac{ab-ac-bd}{2}\qquad \textbf{(D)}\ \frac{ac+bd+ab}{2}\qquad \textbf{(E)}\ \frac{ac+bd-ab-cd}{2}$

Solution

$\boxed{\textbf{(B) }\frac{ac + bd - ab}{2}}$ .

1957 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Problem 26
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1957 AHSME Problems/Problem 25

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