Kamala has a triangular field with sides 240m, 200m, 360m, where she grew wheat. In another triangular field with sides 240m, 320m, 400m, adjacent to the previous field, she wanted to grow potatoes and onions.
She divided the field into two parts by joining the mid-point of the longest side to the opposite vertex and grew potatoes in one part and onions in the other part.
If m is the area ( in hectares ) has been used for potatoes? (1 hectare = 10000m²).

Area $\Delta \mathrm{ABC}$

Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$

Here, s is the semi-perimeter,

and $a, b, c$ are the sides of the triangle

Let $a=200 \mathrm{m}, \mathrm{b}=240 \mathrm{m}, \mathrm{c}=360 \mathrm{m}$

$\mathrm{s}=\frac{a+b+c}{2}=\frac{200+240+360}{2} \mathrm{m}=\frac{800}{2} \mathrm{m}=400 \mathrm{m}$

Area $\Delta \mathrm{ABC}=\sqrt{400(400-200)(400-240)(400-360)} \mathrm{m}^{2}$

$=\sqrt{400 \times 200 \times 160 \times 40} \mathrm{m}^{2}$

$=\sqrt{(4 \times 2 \times 4 \times 16) \times 10^{6}} \mathrm{m}^{2}$

$=\sqrt{(4 \times 4) \times 2 \times(16) \times 10^{6}} \mathrm{m}^{2}$

$=\sqrt{(16) \times 2 \times(16) \times 10^{6}} \mathrm{m}^{2}$

$=\sqrt{(16 \times 16) \times 2 \times 10^{6}} \mathrm{m}^{2}$

$=\sqrt{16^{2} \times 2 \times 10^{6}} \mathrm{m}^{2}$

$=\sqrt{16^{2}} \times \sqrt{2} \times \sqrt{10^{6}} $

$=16000 \sqrt{2} \mathrm{m}^{2}$

1 hectare  = $10000 \mathrm{m}^{2} $

$=\frac{16000 \sqrt{2}}{10000} \text { hectares } $

$=1.6 \mathrm{x} \sqrt{2} \text { hectares } $

$=1.6 \times 1.414 \text { hectares }$ $=2.26 \text { hectares (approx.) }$

Area for growing wheat $=2.26$ hectares

For finding Area $\Delta \mathrm{AEC} \&$ Area $\Delta \mathrm{CED}$

we need EC. But EC is not given

Taking out $\Delta$ ACD separately

Area of triangle $=\frac{1}{2} \times$ Base $\times$ Height

Triangles $\Delta \mathrm{AEC} \& \Delta \mathrm{CED}$ have the same base $($ as $\mathrm{AE}=\mathrm{ED}=200 \mathrm{m})$

And they have the same height CF.

So, Area $\Delta \mathrm{AEC}=$ Area $\Delta \mathrm{CED}$

Area $\Delta \mathrm{AEC}=$ Area $\Delta \mathrm{CED}=\frac{1}{2}$ Area $\Delta \mathrm{ACD}$

$Area \triangle \mathrm{ACD}=\sqrt{\mathrm{S}(\mathrm{S}-\mathrm{a})(\mathrm{S}-\mathrm{b})(\mathrm{S}-\mathrm{c})}$

$\mathrm{S}=\frac{240+320+400}{2}=480 \mathrm{m}$

$=\sqrt{480(480-240)(480-320)(480-400)}$

$=\sqrt{480 \times 240 \times 160 \times 80}$

$=38400 \mathrm{m}^{2}$

$=3.84$ hectares

$\therefore \operatorname{area} \triangle \mathrm{AEC}=\operatorname{area} \triangle \mathrm{CED}=\frac{1}{2} \times 3.84=1.92 \mathrm{hectares}$

Therefore,

Area of wheat field $=2.26$ hectares

Area of potato field $=1.92$ hectares

Area of onion field $=1.92$ hectares