Question

A farmer has two triangular fields in the form of $\Delta ABC$ and $\Delta ACD$ in which the side $AC$ is common as shown in figure. $AB=840m,BC=600m,AC=480m,AD=800mandCD=640m$. He has marked mid-points $E$ and $F$ on the sides $AB$ and $AD$ respectively. By joining $CE$ and $CF$, he has made a field in the shape of quadrilateral $AECF$. He grew wheat in the quadrilateral plot $AECF$, potatoes in $\Delta CED$ and onions in $\Delta BEC$ .How much area has been used for each crop? (Take $\sqrt{6}=2.45$; one hectare=10000 $m^2$)

• A. True
• B. False

Answer 1

The correct option is B False

$Ar△ACD=\sqrt{s(s-a)(s-b)(s-c)}$

$s=\frac{a+b+c}{2}=\frac{800+480+640}{2}$

$s=\frac{1920}{2}=960m$

$ar△ACD=\sqrt{960(960-800)(960-480)(960-640)}$

$=\sqrt{960\times 160\times 480\times 320}$

$=\sqrt{{\left(100\right)}^2\times 2\times 3\times 16\times 16\times 3\times 16\times 2\times 16}$

$=2\times 3\times 16\times 16\times 100=153600m^2$

$ar△ACD=153600=15.36m^2$

Now, $ar△CFD=ar△CFA=\frac{1}{2}\times ar△ACD=\frac{15.36}{2}=7.68$ hec

Because F is the mid point of AD and both $△$ have same height

Now, $ar△ABC=\sqrt{s(s-a)(s-b)(s-c)}$

$s=\frac{a+b+c}{2}=\frac{840+480+600}{2}=960m$

$ar△ABC=\sqrt{960(960-840)(960-480)(960-600)}$

$=\sqrt{960\times 120\times 480\times 360}$

$=\sqrt{{\left(100\right)}^2\times 16\times 6\times 6\times 2\times 16\times 3\times 36}$

$=6\times 6\times 16\times 100\sqrt{6}\approx 141120m^2$

$\therefore ar△ABC=141120m^2=14.11$ hectare

Now, $ar△AEC=ar△BEC=\frac{1}{2}\times ar△ABC=\frac{14.11}{2}\approx 7.06$

Because E is the mid point of AB and height of both triangles is also equal.

Now, area of potatoes field$=ar△CFD=7.68$ hectare

Area of wheat field$=ar△ACF+ar△FCE$

$=7.68+7.06=14.74$ hectare

Area of onion field$=ar△BEC=7.06$ hectare

Hence, given statement is false.