The smallest solution for the first problem has sides of lengths 7, 7, 12 and 4, 11, 11 with a common perimeter of 26 and an area of 6*Sqrt[13]. For the second problem the smallest solution has sides of lengths 17, 25, 28 and 20, 21, 29 with a common perimeter of 70 and an area of 210.
Denis Borris provided many solutions with more than two triangles. Here
are two with six triangles having the same perimeter and area:
Perimeter = 2340 Area = 196560 442 890 1008 450 876 1014 522 778 1040 533 765 1042 546 750 1044 624 666 1050 Perimeter = 2464 Area = 221760 462 962 1040 464 957 1043 476 932 1056 528 854 1082 572 800 1092 672 692 1100
Matt Hudelson gave the following parametric solution to the first part:
A more general answer for (a) can be obtained by fixing parameters s and t
and constructing isosceles triangles having side lengths:
2 t (t+s), s^2 - s t + 2t^2, s^2 - s t + 2t^2
and
2 t (t-s), s^2 + s t + 2t^2, s^2 + s t + 2t^2
Their common perimeter is 2s^2 + 6t^2 and
their common area is t (t+s) (t-s) sqrt(s^2+3t^2).
Al Zimmermann provided the following solution to the third problem:
Each of the following two tetrahedra
AB=8, AC=7, BC=3, CD=4, BD=3, AD=7
AB=8, AC=7, BC=3, CD=6, BD=7, AD=3
has a surface area of
12*sqrt(3) + 8*sqrt(5) = 38.67
and a volume of
8*sqrt(11) / 3 = 8.844.
[Thanks to Al Zimmermann for corrections to the first version of the
solution page.]