Problem 538

The perimeter of $\triangle ABC$ is equal to $(3x+4)+(3x+4)+2x=8x+8$. The perimeter of rectangle $DEFG$ is equal to $2 \times (2x-2)+2 \times (3x-1)=4x-4+6x-2=10x-6$. Since these perimeters are equal, we have $10x-6=8x+8$ which gives $2x=14$ and so $x=7$. Thus, $\triangle ABC$ has $AC=2 \times 7=14$ and $AB=BC=3 \times 7+4=25$. We drop a perpendicular from $B$ to $T$ on $AC$. Since $\triangle ABC$ is isosceles, then $T$ is the midpoint of $AC$, which gives $AT=TC=7$. By the Pythagorean Theorem, $BT=\sqrt{BC^{2}-TC^{2}}=\sqrt{25^{2}-7^{2}}=\sqrt{625-49}=\sqrt{576}=24$. Therefore, the area of $\triangle ABC$ is equal to $\frac{1}{2} \cdot AC \cdot BT=\frac{1}{2} \times 14 \times 24=168$.