I notice that $\triangle ABC$ is a right triangle, since $6^2 + 8^2 = 10^2$ by the Pythagorean theorem. This means that the area of $\triangle ABC$ is half the product of the lengths of the legs, which are 6 and 8. So the area of $\triangle ABC$ is $\frac{1}{2} \times 6 \times 8 = 24$ square units. To increase each side length by $50\%$, I need to multiply it by $1.5$. So the new side lengths are $6 \times 1.5 = 9$, $8 \times 1.5 = 12$, and $10 \times 1.5 = 15$. This means that the area of $\triangle DEF$ is also half the product of the lengths of the legs, which are 9 and 12. So the area of $\triangle DEF$ is $\frac{1}{2} \times 9 \times 12 = 54$ square units. 
$\triangle ABC$ has side lengths 6, 8 and 10, as shown.  Each of the side lengths of $\triangle ABC$ is increased by $50\%$, forming a new triangle, $\triangle DEF$.  What is the area of $\triangle DEF?$