The volumes of the two solids are equal, and the cross sections shown are taken at the same height above the bases. Find the missing length of the cross section of the rectangular pyramid. Round your answer to the nearest hundredth.A. 1.59mB. 1.62mC. 1.67mD. 1.76m

[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]