Answer
See explanations.
Work Step by Step
Step 1. Using the figure as shown, the area of the triangle is given by $Area=\frac{1}{2}ch=\frac{1}{2}cb\cdot sinA$
Step 2. Using the Laws of the Cosines, we have $a^2=b^2+c^2-2bc\cdot cosA$ which gives $cosA=\frac{b^2+c^2-a^2}{2bc}$
Step 3. Using the identity $sin^2A+cos^2A=1$, we have $sinA=\sqrt {1-cos^2A}$
Step 4. The area is then given by $Area=\frac{1}{2}bc\sqrt {1-cos^2A}=\sqrt {\frac{1}{4}(b^2c^2-b^2c^2cos^2A)}$
Step 5. From step 2, $b^2c^2cos^2A=\frac{1}{4}(b^2+c^2-a^2)^2$, we have $Area^2=\frac{1}{16}(4b^2c^2-(b^2+c^2-a^2)^2)=\frac{1}{16}(2bc+b^2+c^2-a^2)(2bc-b^2-c^2+a^2)=\frac{1}{16}((b+c)^2-a^2)(a^2-(b-c)^2)=\frac{1}{16}(b+c+a)(b+c-a)(a+b-c)(a-b+c)$
Step 6. Let $s=\frac{1}{2}(b+c+a)$, we have $\frac{1}{2}(b+c-a)=s-a, \frac{1}{2}(a+c-b)=s-b, \frac{1}{2}(a+b-c)=s-c$
Step 7. The result from step 5 can then be written as $Area=\sqrt {s(s-a)(s-b)(s-c)}$
