Question

AB = 8, BC = 12cm, $\angle ABC={150}^{\circ }$, then area of $△ABCis$

• A. 16 ${cm}^2$
• B. 12 ${cm}^2$
• C. 24 ${cm}^2$
• D. 32 ${cm}^2$

Answer 1

The correct option is C 24 ${cm}^2$

$\angle ABK+\angle ABC={180}^o$ [ Linear pair ]

$\Rightarrow$ $\angle ABK={180}^o-\angle ABC$

$\Rightarrow$ $\angle ABK={180}^o-{150}^o$

$\therefore$ $\angle ABK={30}^o$

Now,

$\sin {30}^o=\frac{AK}{8}$

$\Rightarrow$ $\frac{1}{2}=\frac{AK}{8}$

$\therefore$ $AK=4cm$

And $\cos {30}^o=\frac{BK}{8}$

$\Rightarrow$ $\frac{\sqrt{3}}{2}=\frac{BK}{8}$

$\therefore$ $BK=4\sqrt{3}cm$

$\Rightarrow$ Area of$△ABK=\frac{1}{2}\times BK\times AK$

$=\frac{1}{2}\times 4\sqrt{3}\times 4$

$=8\sqrt{3}c{m}^2$

Area of $△ACK=\frac{1}{2}\times AK\times KC$

$=\frac{1}{2}\times (12+4\sqrt{3})\times 4$

$=(24+8\sqrt{3})c{m}^2$

Area of $△ABC=$ Area of $△ACK-$ Area of $△ABK$

$=24+8\sqrt{3}-8\sqrt{3}$

$=24c{m}^2$