, 2015

We can determine which triangle has the greatest area by using a fixed side length of 4 for each of the identical squares and using this to calculate the unknown areas.
We begin by constructing $\triangle PVU$ and noticing that it is contained within square $QABP$, as shown.

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The area of $\triangle PVU$ is determined by subtracting the areas of triangles $PQV$, $VAU$ and $PBU$ from the area of square $QABP$.

Since $QA=8$ and $AB=8$, then the area of square $QABP$ is $8\times8=64$.

Since $PQ=8$ and $QV=2$, then the area of $\triangle PQV$ is $\frac12\times8\times2=8$.

Since $VA=6$ and $AU=6$, then the area of $\triangle VAU$ is $\frac12\times6\times6=18$.

Since $PB=8$ and $UB=2$, then the area of $\triangle PBU$ is $\frac12\times8\times2=8$.
Therefore, the area of $\triangle PVU$ is $64-8-18-8=30$.

Next, we construct $\triangle PXZ$ and then construct rectangle $CDSP$ by drawing $CD$ parallel to $PS$ through $X$. Further, $X$ is the midpoint of the side of a square and so $C$ and $D$ are also midpoints of the sides of their respective squares.

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The area of $\triangle PXZ$ is determined by subtracting the areas of triangles $PCX$, $XDZ$ and $PSZ$ from the area of rectangle $CDSP$.

Since $CD=12$ and $DS=6$, then the area of rectangle $CDSP$ is $12\times6=72$.

Since $PC=6$ and $CX=8$, then the area of $\triangle PCX$ is $\frac12\times6\times8=24$.

Since $XD=4$ and $DZ=4$, then the area of $\triangle XDZ$ is $\frac12\times4\times4=8$.

Since $PS=12$ and $ZS=2$, then the area of $\triangle PSZ$ is $\frac12\times12\times2=12$.
Therefore, the area of $\triangle PXZ$ is $72-24-8-12=28$.

Construct $\triangle PVX$ and notice that it is contained within square $QABP$, as shown.

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The area of $\triangle PVX$ is determined by subtracting the areas of triangles $PQV$, $VAX$ and $PBX$ from the area of square $QABP$.

As we previously determined, the area of square $QABP$ is $64$ and the area of $\triangle PQV$ is $8$.

Since $VA=6$ and $AX=2$, then the area of $\triangle VAX$ is $\frac12\times6\times2=6$.

Since $PB=8$ and $XB=6$, then the area of $\triangle PBX$ is $\frac12\times8\times6=24$.
Therefore, the area of $\triangle PVX$ is $64-8-6-24=26$.

Construct $\triangle PYS$ and the perpendicular from $Y$ to $E$ on $PS$, as shown.

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Since $PS=12$ and $YE=4$ ($YE$ is parallel to $RS$ and thus equal in length to the side of the square), then the area of $\triangle PYS$ is $\frac12\times12\times4=24$.

Construct $\triangle PQW$, as shown.

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Since $PQ=8$ and $QW=6$, then the area of $\triangle PQW$ is $\frac12\times8\times6=24$.

The areas of the 5 triangles are $30,28,26,24,$ and 24. The triangle with greatest area, 30, is $\triangle PVU$.