Question

Find the area of a triangle whose sides are$a=4.b=5,c=3$.

Short answer

Required area of the triangle is $6squareunits.$

Step-by-step solution

Step 1. Given information

The area of a triangle with sides a, b, and c is determined by the Heron's formula which is given as:

$$\begin{gathered} Area=\sqrt{s(s-a)(s-b)(s-c)} \\ where, \\ s=\raisebox{1ex}{$(a+b+c)$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \end{gathered}$$

Step 2. Calculation

$$\begin{gathered} Given, \\ a=4,b=5,c=3 \\ Then, \\ s=\raisebox{1ex}{$a+b+c$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \\ =\raisebox{1ex}{$4+5+3$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \\ =\raisebox{1ex}{$12$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \\ =6 \\ Now,Heron\text{'}sFormulagivesareaas \\ Area=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{6\left(6-4\right)\left(6-5\right)\left(6-3\right)} \\ =\sqrt{6\times 2\times 1\times 3} \\ =\sqrt{36} \\ =6squareunits. \end{gathered}$$