The correct answer is $StartFraction 135 Over 8 EndFraction$ . It's given that triangles $A B C$ and $D E F$ are similar and each side length of triangle $A B C$ is $4$ times the corresponding side length of triangle $D E F$ . For two similar triangles, if each side length of the first triangle is $k$ times the corresponding side length of the second triangle, then the area of the first triangle is $k^{2}$ times the area of the second triangle. Therefore, the area of triangle $A B C$ is $4 squared$ , or $16$ , times the area of triangle $D E F$ . It's given that the area of triangle $A B C$ is $270$ square inches. Let $a$ represent the area, in square inches, of triangle $D E F$ . It follows that $270$ is $16$ times $a$ , or $270 equals 16 a$ . Dividing both sides of this equation by $16$ yields $\frac{270}{16} = a$ , which is equivalent to $\frac{135}{8} = a$ . Thus, the area, in square inches, of triangle $D E F$ is $StartFraction 135 Over 8 EndFraction$ . Note that 135/8, 16.87, and 16.88 are examples of ways to enter a correct answer.