Area of Triangle - Practice Questions & MCQ

Area of Triangle

The area formula for a triangle is given as
Area $=1 / 2 \mathrm{bh}, \quad$ where ' $b$ ' is the base and ' $h$ ' is the height

Clearly, height h = c.sin(A)

$
\begin{aligned}
\text { Area } & =\frac{1}{2} \text { base } \times \text { height } \\
& =\frac{1}{2} b \cdot \mathrm{c} \sin \mathrm{~A}
\end{aligned}
$
Area of triangle $A B C$ is represented by $\triangle$, Thus

$
\text { Area of } \begin{aligned}
\Delta \mathrm{ABC}=\Delta & =\frac{1}{2} b \cdot \mathrm{c} \sin \mathrm{~A} \\
& =\frac{1}{2} a \cdot \mathrm{~b} \sin \mathrm{C} \\
& =\frac{1}{2} c \cdot \mathrm{a} \sin \mathrm{~B}
\end{aligned}
$
Area of the triangle in terms of sides (Heron's Formula)

$
\Delta=\frac{1}{2} b \cdot \mathrm{c} \sin \mathrm{~A}=\frac{1}{2} b c \cdot 2 \sin \frac{\mathrm{~A}}{2} \cos \frac{\mathrm{~A}}{2}
$

Use half angle formula

$
\begin{aligned}
& =\frac{1}{2} \cdot \mathrm{bc} \cdot \sqrt{\frac{(s-b)(s-c)}{b c}} \sqrt{\frac{s(s-a)}{b c}} \\
& =\sqrt{s(s-a)(s-b)(s-c)}
\end{aligned}
$