Question

In $\Delta ABC,AB=12cm,\angle A={45}^{\circ },\angle B={30}^{\circ }.$ Then the area of the triangle ABC is

• A. $3(\sqrt{6}+1)c{m}^2$
• B. $6(\sqrt{3}+1)c{m}^2$
• C. $6(\sqrt{3}-1)c{m}^2$
• D. $3(\sqrt{6}-1)c{m}^2$

Answer 1

In $\Delta ABC,$ draw CD perpendicular to AB

Let $AD=x,BD=(12-x),$ $AD=CD=x$

In $\Delta BDC,$ the angles of BDC are ${30}^{\circ },{60}^{\circ },{90}^{\circ }$

Therefore, the corresponding sides can be calculated as

$\Rightarrow \sin \left(30\right):\sin \left(60\right):\sin \left(90\right)$

$\Rightarrow \frac{1}{2}:\frac{\sqrt{3}}{2}:1$

$\Rightarrow 1:\sqrt{3}:2$

$\begin{array}{ccccc}{30}^{\circ }& & {60}^{\circ }& & {90}^{\circ }\\ 1& :& \sqrt{3}& :& 2\\ \downarrow & & \downarrow & & \downarrow \\ CD& & BD& & BC\\ \downarrow & & \downarrow & & \downarrow \\ x& & x\sqrt{3}& & 2x\end{array}$

$BD=x\sqrt{3}=12-x$

$x(\sqrt{3}+1)=12$

$x=\frac{12}{\sqrt{3}+1}=\frac{12}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}-1}=\frac{12(\sqrt{3}-1)}{2}=6(\sqrt{3}-1)cm$

Area of $ABC=\frac{1}{2}\times AB\times CD$

$=\frac{1}{2}\times 12\times x$

$=\frac{1}{2}\times 12\times 6(\sqrt{3}-1)$

$=36(\sqrt{3}-1)c{m}^2$