A few steps required here:

1) Find the leg of the right isosceles triangle from the area. Since area = 1/2*base*height, and the two equal perpendicular legs can be considered the base and height, you get that 1/2*leg*leg = 25

leg^2 = 50
leg = sqrt(50) = 5sqrt(2)

2) Draw a height from the hypotenuse of the big triangle, dividing the triangle into two smaller 45:45:90 degree right isosceles triangles. In these smaller triangles, the legs are the hypotenuse. Using the 45:45:90 1:1:sqrt (2) ratio of a right isosceles triangle, you get that the height to the hypotenuse in the larger triangle is 5sqrt(2) / sqrt(2) = 5.

3) Now you know that the cut was made 2 inches from the hypotenuse. If the original height is 5, then the height of the cut triangle is 5-2=3. You can use this to figure out the legs of the small triangle and do the same process above in reverse to find the area, but you don't need to, if you remember this rule: In similar triangles, area ratio is equal to (linear ratio squared).
The cut triangle and the larger triangle are similar - they're both 45:45:90 triangles. If the linear ratio (the ratio between any two respective lines ) is 3:5 - the ratio of the heights - then the area ratio is the square of 3:5, or 9:25.

Since we know that the original triangle is of area 25, this means that the area of the cut triangle is 9, to maintain the 9:25 ratio.

SFtraveler wrote:A paper isoceles right triangle with an area of 25 square inches is cut along a line parallel to the hypotenuse of the triangle, leaving a smaller triangle. What is the area of the new triangle if the cut was made 2 inches from the hypotenuse?

In the picture below ABC is the triangle with area 25 Sq inch.

A'C' is the parallel line to AC, creating the new Triangle A'BC' which is also a Isoceles right triangle.

In triangle ABC,

Area = 1/2 (AB) (BC)
=> 25 = 1/2 (a)^2 (Say a=AB=BC)
=> a = sqrt(50)

Thus the Hyportaneous = Sqrt(2)*a = 10

Draw a height from B to AC named BD in the pic

Area = 1/2 (BD) (AC)
=> 25 = (1/2) (BD) 10
=> BD = 5

Thus, BD' = 5 - 2 = 3.

We know that Triangle ABC and Triangle A'BC' are similar. (Use the A-A-A property, all corresponding angles are same)

Using the similarity of triangle property,

Area of ABC/Area of A'BC' = (AB/A'B)^2 = (BC/BC')^2 = (AC/A'C')^2 = (BD/BD')^2

=> 25/Area of A'BC' = (5/3)^2

=> Area of A'BC' = 9