Problem 713

We begin by constructing rectangle $ABCD$ around the given quadrilateral $PQRS$, as shown.

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The vertical sides $AB$ and $DC$ pass through points $Q$ and $S$, respectively.

The horizontal sides $AD$ and $BC$ pass through points $P$ and $R$, respectively.

We determine the area of $PQRS$ by subtracting the areas of the four right-angled triangles, $AQP$, $QBR$, $CSR$, and $SDP$, from the area of $ABCD$.

To determine the horizontal side lengths of the right-angled triangles we count units along the $x$-axis, or we subtract the $x$-coordinates of two vertices.

For example, since $AB$ is vertical and passes through $Q(-5,1)$, the $x$-coordinates of $A$ and $B$ are equal to that of $Q$, which is $-5$.

Thus, the length of $AP$ is determined by subtracting the $x$-coordinate of $A$ from the $x$-coordinate of $P$, which is 7.

Therefore the length of $AP$ is $7-(-5)=12$.

Similarly, the length of $BR$ is $-2-(-5)=3$.

Since $DC$ is vertical and passes through $S(10,2)$, the $x$-coordinates of $D$ and $C$ are equal to that of $S$, which is $10$.

Thus, the length of $PD$ is $10-7=3$, and the length of $RC$ is $10-(-2)=12$.

To determine the vertical side lengths of the right-angled triangles we may count units along the $y$-axis, or we may subtract the $y$-coordinates of two vertices.

For example, since $AD$ is horizontal and passes through $P(7,6)$, the $y$-coordinates of $A$ and $D$ are equal to that of $P$, which is $6$.

Thus, the length of $AQ$ is determined by subtracting the $y$-coordinate of $Q$ (which is 1) from the $y$-coordinate of $A$.

Therefore the length of $AQ$ is $6-1=5$.

Similarly, the length of $DS$ is $6-2=4$.

Since $BC$ is horizontal and passes through $R(-2,-3)$, the $y$-coordinates of $B$ and $C$ are equal to that of $R$, which is $-3$.

Thus, the length of $QB$ is $1-(-3)=4$, and the length of $SC$ is $2-(-3)=5$.

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The area of $\triangle AQP$ is $\frac12\times AQ\times AP=\frac12\times5\times12=30$.

The area of $\triangle CSR$ is also 30.

The area of $\triangle QBR$ is $\frac12\times QB\times BR=\frac12\times4\times3=6$.

The area of $\triangle SDP$ is also 6.

Since $AB=AQ+QB=5+4=9$ and $BC=BR+RC=3+12=15$, the area of $ABCD$ is $9\times 15=135$.

Finally, the area of $PQRS$ is $135-30\times 2-6\times 2=135-60-12=63$.