[tex]area(tri) = (1 \div 2)(base)(height) \\ a(t) = bh \div 2 \\ area(rect) = (length)(width) \\ a(r) = lw[/tex]
Since the triangle is right, and has sides of 4 & 5, then it is a 3-4-5 special right triangle. Now the height can be the side that's 3, and the base 4, since by definition the height us 90° from the base.
[tex]a(t) = bh \div 2 = 3m \times 4m \div 2 \\ = 12 {m}^{2} \div 2 = 6 {m}^{2} [/tex]
Now we know that the a(r) also = 6, because of the congruency of the two shapes, with equal volumes, at the same height. So now let's find our length (l):
[tex]a(r) = lw \: > > 6 {m}^{2} = l(3.6m) \\ l = 6 {m}^{2} \div 3.6m = 1.666m \\ = 1.67m[/tex]
