The perimeter of the original photo is \(2\cdot\left(8+10\right)=36\left(inches\right)\)

The area of the original photo is \(8\cdot10=80\left(inches^2\right)\)

After reduced in size the new perimeter will equal \(\dfrac{27}{36}=\dfrac{3}{4}\) the original perimeter. So each measurements of the new photo will equal \(\dfrac{3}{4}\) the original measurements of the original photo.

The area of the new photo will equal \(\left(\dfrac{3}{4}\right)^2=\dfrac{9}{16}\) the area of the original photo.

So the area of the new photo will be \(80\cdot\dfrac{9}{16}=45\left(inches^2\right)\)