A field in the form of a parallelogram has sides 60 m

Solution:

Let $A B C D$ be a parallelogram field with sides $A B=C D=60 \mathrm{~m}, B C=D A=40 \mathrm{~m}$ and

diagonal $B D=80 \mathrm{~m}$

Area of parallelogram $A B C D=2$ (Area of $\triangle A B D$ ) $\ldots$ (i)

In $\triangle A B D$,

Semi-perimeter of a triangle $\triangle A B D$,

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

$=\frac{A B+B D+D A}{2}$

$=\frac{60+80+40}{2}=\frac{180}{2}$

 

$=90 \mathrm{~m}$

$\therefore \quad$ Area of $\triangle A B D=\sqrt{s(s-a)(s-b)(s-c)} \quad$ [by Heron's formula]

$=\sqrt{90(90-60)(90-80)(90-40)}$

$=\sqrt{90 \times 30 \times 10 \times 50}$

 

$=100 \times 3 \sqrt{15}=300 \sqrt{15} \mathrm{~m}^{2}$

From Eq. (i),

 

Area of parallelogram $A B C D=2 \times 300 \sqrt{15}=600 \sqrt{15} \mathrm{~m}^{2}$ Hence, the area of the parallelogram is $600 \sqrt{15} \mathrm{~m}^{2}$.