We assume the figure DEFG has vertices labeled clockwise from the top, so that diagonal EG spans the distance between vertices where sides of different length meet. Then half that diagonal (12) will be one leg of two different right triangles.
One right triangle will have a hypotenuse of 20 and a leg of 12. Its other leg is found using the Pythagorean theorem as √(20²-12²) = 16 The other right triangle will have a hypotenuse of 13 and a leg of 12. Its other leg is √(13²-12²) = 5