Two sides of a triangular field are 85 m and 154 m in length and its perimeter is 324 m

Solution:

(i) Let:

$a=85 \mathrm{~m}$ and $b=154 \mathrm{~m}$

Given :

Perimeter $=324 \mathrm{~m}$

or, $a+b+c=324$

$\Rightarrow c=324-85-154=85 \mathrm{~m}$

$\therefore s=\frac{324}{2}=162 \mathrm{~m}$

By Heron's formula, we have:

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

$=\sqrt{162(162-85)(162-154)(162-85)}$

$=\sqrt{162 \times 77 \times 8 \times 77}$

$=\sqrt{1296 \times 77 \times 77}$

$=\sqrt{36 \times 77 \times 77 \times 36}$

$=36 \times 77$

$=2772 \mathrm{~m}^{2}$

(ii) We can find out the height of the triangle corresponding to 154 m in the following manner:
We have:

Area of triangle $=2772 \mathrm{~m}^{2}$

$\Rightarrow \frac{1}{2} \times$ Base $\times$ Height $=2772$

$\Rightarrow$ Height $=\frac{2772 \times 2}{154}=36 \mathrm{~m}$