Heron's formula say the area of a triangle whose sides have lengths $a,b,c$ is $$ \frac14\sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}. $$ It is true that this is an "opaque formula" with which you "just chuck in the side-lengths, turn a series of arbitrary mathematical handles, and out pops the answer," about which one should say "It works, but it's not satisfying."?

Notice that

• One must expect an expression that is homogeneous of degree $2$ as a function of $a,b,c.$
• $(a+b+c)$ should be a factor because the area is $0$ when $a=b=c=0.$
• $(a+b-c)$ should be a factor because when $a+b=c$ you have a degenerate triangle, whose area is $0.$
• The other factors must be there for the same reason.
• The number $1/4$ is seen to be right when one considers $a=b=c=1.$

This is not a dissection proof, but it addresses the first issue raised in the original posting.

Heron's formula says the area of a triangle whose sides have lengths $a,b,c$ is $$ \frac14\sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}. $$ It is true that this is an "opaque formula" with which you "just chuck in the side-lengths, turn a series of arbitrary mathematical handles, and out pops the answer," about which one should say "It works, but it's not satisfying."?

Notice that

• One must expect an expression that is homogeneous of degree $2$ as a function of $a,b,c.$
• $(a+b+c)$ should be a factor because the area is $0$ when $a=b=c=0.$
• $(a+b-c)$ should be a factor because when $a+b=c$ you have a degenerate triangle, whose area is $0.$
• The other factors must be there for the same reason.
• The number $1/4$ is seen to be right when one considers $a=b=c=1.$

This is not a dissection proof, but it addresses the first issue raised in the original posting.