Area of a triangle A
- $$A$$ = $$\sqrt{s(s-a)(s-b)(s-c)}$$ where s =$$\frac{\left(a+b+c\right)}{2}$$.
- $$A$$ =$$\frac{1}{2}\times\ base\times\ altitude$$
- Area of triangle with vertices (x₁,y₁),(x₂,y₂),(x₃,y₃) = $$\frac{1}{2}|x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)|$$
Application:
- In the above triangle, both the triangles $$\text{ADB}$$ and $$\text{ADC}$$ have the same height => The ratio of the areas of triangles $$\text{ADB}$$ and $$\text{ADC}$$ will be in the ratio of their bases $$\text{BD}$$ and $$\text{DC}$$ $$\dfrac{\text{Area}\left(\text{ADB}\right)}{\text{Area}\left(\text{ADC}\right)}=\dfrac{\text{BD}}{\text{DC}}$$
- $$A$$ = $$\frac{1}{2}\times\ a\times\ b\times\ \sin C$$, where C is the enclosed angle between sides a and b.
- So, for a right-angled triangle with a,b, and c as the sides, with c as the hypotenuse [Angle C = 90 degrees]
$$A$$ = $$\dfrac{1}{2}\times\ a\times\ b\times\ \sin\left(90\right)=\dfrac{abc}{4R}$$ => $$R=\dfrac{c}{2}$$
