Question

In $\Delta ABC,a=4$ cm, $b=6$ cm and $\angle C={30}^0,$ then the area of the triangle is

• A. $12{cm}^2$
• B. $6\sqrt{2}{cm}^2$
• C. $\frac{12}{\sqrt{3}}{cm}^2$
• D. $6{cm}^2$

Answer 1

The correct option is D $6{cm}^2$

Consider the figure given.
Therefore, the area of triangle

$=\frac{1}{2}\times \text{base}\times \text{height}$
Area $=\frac{1}{2}\times AC\times \text{height}$
$=\frac{1}{2}\times b\times h$ ...(1)

From the given data, $\angle BCA={30}^0$
Therefore, $h=a\sin C$
Substituting in (1), we get
$=\frac{1}{2}\times b\times a\sin C$
Substituting the values of $a,b,C$ from the given data we get
Area $=\frac{1}{2}\left(6\right)\left(4\right)\left(\sin {30}^0\right)$
$=\frac{1}{4}\left(6\right)\left(4\right)=6{cm}^2$

Hence, option D is correct answer.