Consider a = 91m, b = 98m and c = 105m
So we get
$$S=\dfrac{a+b+c}{2}$$
$$S=\dfrac{91+98+105}{2}$$
By division
s=147
We know that
Area=$$\sqrt{s(s-a)(s-b)(s-c)}$$
By substituting the values
Area=$$ \sqrt {147(147-91)(147-98)(147-105)} $$
So we get
Area=$$ \sqrt {147\times 56 \times 49 \times 42} $$
It can be written as
Area=$$ \sqrt {49\times3\times7\times2\times2\times2\times49\times7\times3\times2} $$
On further calculation
Area =$$49\times3\times2\times2\times7$$
So we get
Area =$$4116m^2$$
It is given as
b = Longest side = $$105 m$$
Consider h as the height corresponding to the longest side
We know that
Area of the triangle = $$1/2 \times b \times h$$
By substituting the values
$$1/2 \times b \times h = 4116$$
On further calculation
$$105 \times h = 4116\times2$$
So we get
$$h = (4116\times2)/105$$
By division
h = $$78.4 m$$