Okay, this is a very visual problem, and I will try to solve it the best I can with words.

As I understand, we have a rectangle, and a triangle inside of it. The dimensions of the rectangle are 8cm X 7cm. The triangle shares the side AD with the rectangle, so it has a base size of 7cm, and it stretches to the other side of the rectangle, so it
has a height of 8cm. Now we need to find three things:

1. The area of the triangle

2. The ratio of the area of triangle ABE to the triangle ADE

3. The ratio of the area of triangle CED to the rectangle(ABCD).

To answer the first one, we just need to know the equation for the area of a rectangle, which is 1/2*base*height. We know the base of the triangle is 7cm, and we know the height of the triangle is 8cm, so after plugging in those values, it looks like this:
1/2*7cm*8cm. This simplifies out to 28cm^2. The work is as follows:

1/2*base*height

1/2*7cm*8cm

1/2*56cm^2

28cm^2

For the next part, we look at the area of triangle ABE and the area of triangle ADE. Triangle ABE's base is exactly 1/2 of triangle ADE's base, so by knowing that alone, we also know the area is exactly 1/2 that of triangle ADE, but we can prove it with
work. We already found ADE's area. Now for ABE:

1/2*base*height

1/2*3.5cm*8cm (3.5cm is half of 7cm)

1/2*28cm^2

14cm^2

So the ratio of area of triangle ABE to ADE is 1 to 2, or 1:2.

The last part is simply CDE, which is the remaining triangle, to ABCD. By knowing that CDE has the same area as ABE which has half the area of ADE, which has half of the area of ABCD, we can already determine that the rectangle has four times the area of
the triangle, but we can prove it with work. The rectangle's area:

base*height

8cm*7cm

56cm^2

The area of CDE:

1/2*base*height

1/2*3.5cm*8cm

1/2*28cm^2

14cm^2

14 goes into 56 four times, so we can say that the ratio of triangle CDE to rectangle ABCD is exactly 1 to 4 or 1:4

Hope I answered your question!