Question

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

Answer 1

$S=\frac{91+98+105}{2}=147$

$\Delta =\sqrt{s(s-a)(s-b)(s-c)}$

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

$\Delta =4116m^2$

Base$=105$m

Height $=$h

$\Delta =\frac{1}{2}\times$ base$\times$ height

$4116=\frac{1}{2}\times 105\times h$

$h=78.4$m.