# Find the area of a triangular field whose sides are 91 cm, 98 m and 105 m in length.

Question:

Find the area of a triangular field whose sides are 91 cm, 98 m and 105 m in length. Find the height corresponding to the longest side.

Solution:

Let :

$a=91 \mathrm{~m}, b=98 \mathrm{~m}$ and $c=105 \mathrm{~m}$

$\therefore s=\frac{a+b+c}{2}=\frac{91+98+105}{2}=147 \mathrm{~m}$

By Heron's formula, we have :

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

$=\sqrt{147(147-91)(147-98)(147-105)}$

$=\sqrt{147 \times 56 \times 49 \times 42}$

$=\sqrt{7 \times 3 \times 7 \times 2 \times 2 \times 2 \times 7 \times 7 \times 7 \times 7 \times 3 \times 2}$

$=7 \times 7 \times 7 \times 2 \times 3 \times 2$

$=4116 \mathrm{~m}^{2}$

We know that the longest side is 105 m.
Thus, we can find out the height of the triangle corresponding to 42 cm.

Area of triangle $=4116 \mathrm{~m}^{2}$

$\Rightarrow \frac{1}{2} \times$ Base $\times$ Height $=4116$

$\Rightarrow$ Height $=\frac{4116 \times 2}{105}=78.4 \mathrm{~m}$