Now given 3 sides Ha, Hb and Hc. We have to calculate the area A.

If three sides are a, b and c we can have -

\[ A = \frac{1}{2} \times a \times Ha ........ (1) \] \[ A = \frac{1}{2} \times b \times Hb ........ (2) \] \[ A = \frac{1}{2} \times c \times Hc ........ (3) \]

Now from 1 and 2, we can get

\[ \frac{a}{b} = \frac{Hb}{Ha} ........ (4) \] \[ \frac{b}{c} = \frac{Hc}{Hb} ........ (5) \]

From (4) and (5)

\[ a : b : c = Hb \times Hc : Hc \times Ha : Ha \times Hb \] As Ha, Hb and Hc given, we got the ratio of a, b and c.

So there will be a k, for which

\[ a = k \times Hb \times Hc ...... (6) \] \[ b = k \times Hc \times Ha ...... (7) \] \[ c = k \times Ha \times Hb ...... (8) \]

If the angles are A, B and C. We know that -

\[ \cos C = \frac{a^2 + b^2 - c^2}{2 \times b \times c} = \frac{Hb^2 \times Hc^2 + Hc^2 \times Ha^2 + Ha^2 \times Hb^2}{2 \times Hb \times Hc^2 \times Ha} \]

So, \[ C = \cos^{-1} \frac{Hb^2 \times Hc^2 + Hc^2 \times Ha^2 + Ha^2 \times Hb^2}{2 \times Hb \times Hc^2 \times Ha} \]

Now, from sin rule

\[ Ha = b \times \sin C \] \[ So, b = \frac{Ha} {\sin C} ...... (9) \]

From (7) and (9) we get

\[ k = \frac{1} {Hc \times \sin C} \]

We know Ha, Hb, Hc, and k, so we can calculate the exact value of a, b and c. So we can calculate the area of triangle by the following formula -

\[ A = \sqrt{ \frac{s \times (s - a) \times (s - b) \times (s - c)} {2} }, where s = \frac{a + b + c}{2} \] This was my solution.

While discussing my solution with Nafi, he gave me a much smarter solution, Here it is

Area

\[ A = \frac{1}{2} \times a \times Ha \] So \[ a = 2 \times \frac{A}{Ha} ... (1) \] Similarly, \[ b = 2 \times \frac{A}{Hb} ... (2) \] \[ c = 2 \times \frac{A}{Hc} ... (3) \] Now, \begin{align} s &= \frac{a + b + c}{2}\\ &= (\frac{A}{Ha} + \frac{A}{Hb} + \frac{A}{Hc}) \end{align} And, \[ A^2 = s \times (s - a) \times (s - b) \times (s - c) \] \[ A^2 = (\frac{A}{Ha} + \frac{A}{Hb} + \frac{A}{Hc}) \times (\frac{A}{Hb} + \frac{A}{Hc} - \frac{A}{Ha}) \times (\frac{A}{Hc} + \frac{A}{Ha} - \frac{A}{Hb}) \times (\frac{A}{Ha} + \frac{A}{Hb} - \frac{A}{Hc}) \] \[ \Rightarrow A^2 = A^4 \times (\frac{1}{Ha} + \frac{1}{Hb} + \frac{1}{Hc}) \times (\frac{1}{Hb} + \frac{1}{Hc} - \frac{1}{Ha}) \times (\frac{1}{Hc} + \frac{1}{Ha} - \frac{1}{Hb}) \times (\frac{1}{Ha} + \frac{1}{Hb} - \frac{1}{Hc}) \] \[ \Rightarrow A^2 = \frac{1}{(\frac{1}{Ha} + \frac{1}{Hb} + \frac{1}{Hc}) \times (\frac{1}{Hb} + \frac{1}{Hc} - \frac{1}{Ha}) \times (\frac{1}{Hc} + \frac{1}{Ha} - \frac{1}{Hb}) \times (\frac{1}{Ha} + \frac{1}{Hb} - \frac{1}{Hc})} \] \[ \Rightarrow A = \sqrt{\frac{1}{(\frac{1}{Ha} + \frac{1}{Hb} + \frac{1}{Hc}) \times (\frac{1}{Hb} + \frac{1}{Hc} - \frac{1}{Ha}) \times (\frac{1}{Hc} + \frac{1}{Ha} - \frac{1}{Hb}) \times (\frac{1}{Ha} + \frac{1}{Hb} - \frac{1}{Hc})}} \]

Much cleaner solution. Thanks Nafi.