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

$[asy] import geometry; point P = (0,3); point Q = (4,0); point R = (5,4); draw(P--Q--R--P); dot(P); label("P $(0,a)$",P,NW); dot(Q); label("Q $(b,0)$",Q,SE); dot(R); label("R $(c,d)$",R,NE); [/asy]$

To solve this problem, we could use the distance formula to find the lengths of the sides and then Heron's Formula to find the area of the triangle, but that solution seems messy and prone to mistakes with lots of square roots and polynomial expansions. Therefore, we look for a simpler, easier solution. Suppose $c=b$ and $a=d$ . This makes $\triangle PQR$ half of a rectangle with side lengths $a$ and $b$ , so it has an area of $\tfrac{ab}2$ . Plugging in $c=b$ and $d=a$ into all of the answer choices, the only one which returns $\tfrac{ab}2$ is $\boxed{\textbf{(B) }\frac{ac + bd - ab}{2}}$ .

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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