Question

Find the area of the triangle whose sides measure 52 cm, 56 cm and 60 cm.

Answer 1

$$\begin{gathered} \text{Let the s}\mathrm{ides}\mathrm{of}\text{the}\mathrm{triangle}\text{be} \\ a=52\mathrm{cm} \\ b=56\mathrm{cm} \\ c=60\mathrm{cm} \\ \therefore s=\frac{a+b+c}{2}=\frac{52+56+60}{2}=84\mathrm{cm} \end{gathered}$$

$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{triangle}=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{84\left(84-52\right)\left(84-56\right)\left(84-60\right)} \\ =\sqrt{84\times 32\times 28\times 24} \\ =\sqrt{3\times 28\times 4\times 4\times 2\times 28\times 2\times 2\times 2\times 3} \\ =\sqrt{\left(2\times 2\right)\times \left(2\times 2\right)\times \left(3\times 3\right)\times \left(4\times 4\right)\times \left(28\times 28\right)} \\ =2\times 2\times 3\times 4\times 28 \\ =1344{\mathrm{cm}}^2 \end{gathered}$$