Question

Using Heron's formula find the area of ∆ABC in which BC = 13 cm, AC = 14 cm and AB = 15 cm.

Answer 1

$$\begin{gathered} \mathrm{Let}: \\ a=13\mathrm{cm},b=14\mathrm{cm}\mathrm{and}c=15\mathrm{cm} \\ s=\frac{a+b+c}{2}=\frac{13+14+15}{2}=21\mathrm{cm} \\ \mathrm{By}\mathrm{Heron}\text{'}\mathrm{s}\mathrm{formula},\mathrm{we}\mathrm{have}: \\ \mathrm{Area}\mathrm{of}\mathrm{triangle}ABC=\sqrt{s(s-a)(s-b)(s-c)} \\ =\sqrt{21(21-13)(21-14)(21-15)} \\ =\sqrt{21\times 8\times 7\times 6} \\ =\sqrt{7\times 3\times 2\times 2\times 2\times 7\times 3\times 2} \\ =7\times 3\times 2\times 2 \\ =84{\mathrm{cm}}^2 \end{gathered}$$