A right-angled isosceles triangle is squared on the hypotenuse and on each of the other two sides. How many times larger than the triangle is the total figure?
Without loss of generality, assume the two short sides each have length 1. The hypotenuse has length sqrt(2). The area of the triangle is 1/2. The areas of the three resulting squares is 1, 1 and 2. The area of the resulting figure is 0.5+1+1+2 = 4.5 . 4.5/0.5 = 9. Q.E.D.
Since an isosceles triangle can only have a right angle at the angle opposite the hypotenuse, as all triangles, therefore all isosceles with a right angle must be of the 90, 45, 45, form, which means that all those having a 90 degree are the same triangle, therefore the relationships between the areas will always be the same. And since the 90, 45, 45, triangle gives areas of 1 squared plus 1 squared yielding an area on the hypotenuse of two. ( Hypotenuse equals square root of two. ) And since the area of the triangle itself equals one half of a rectangle with sides equal to 1 and is therefore = one half, and the areas on the sides equal 1+1+2=4 that means that the ratio is four to half, or eight to one. My only reason for voting one is that you did not define "larger" clearly.
Sorry for writing it out in longhand without math notation, but my keyboard does not do it very well.