Question
Kamla has a triangular field with sides $240 \ m, 200 \ m, 360 \ m$, where she grew wheat. In another triangular field with sides $240 \ m, 320 \ m, 400 \ m$ adjacent to the previous field, she wanted to grow potatoes and onions.
She divided the field in 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. How much area (in hectares) has been used for wheat, potatoes and onions? [$1$ hectare $= 1000 m^2,$ $\sqrt{2} = 1.41]$

Answer

Let $ABC$ be the field, where wheat is grown. Also, let $ACD$ be the field which has been divided into two parts by joining $C$ to the mid-point $E$ of $AD$. For the area of $\triangle ABC$, we have
$a = 200m, b = 240 m, c = 360m$

Therefore, s = $\frac{200+240+360}{2} = 400 m$
So, area of growing wheat = $\sqrt{s(s-a)(s-b)(s-c)}$
= $\sqrt{400(400-200)(400-240)(400-360)}$
= 1.6 $\times$ $\sqrt{2}$ hec $= 1.6 \times 1.41$ [ $\because 1.6$ hec $= 16000 m^2$]
$= 2.26$ hec(approx.)
Now, we calculate the area of $\triangle ACD.$
Here, s = $\frac{240+320+400}{2} = 480 m$
So, area of $\triangle ACD$
= $\sqrt{480(480-240)(480-320)(480-400)}$
= $\sqrt{480 \times 240 \times 160 \times 80}$= 38400 $m_2$
$= 3.84$ hec [$\because$ 1 $m_2$ = $\frac{1}{10000}$hec]
Now, let $C F \perp AD$. Then,
$\operatorname{ar}(\triangle A E C)=\frac{1}{2} \times A E \times C F=\frac{1}{2} \times E D \times C F[\because A E=E D$, as $E$ is mid-point of $A D]$
$=\operatorname{ar}(\triangle EDC )[\because CF$ is also a height of $\because E D C$ corresponding to base $ED]$
$\therefore$ Area for growing potatoes $=$ Area for growing onions
$=(3.84 \div 2)=1.92 hec$
Hence, area has been used for growing wheat, potatoes and onion are $2.26 hec , 1.92 hec$ and $19.2$ hec , respectively.

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