Question

In $\Delta ABC,AB=18cm,\angle A={45}^{\circ },\angle B={60}^{\circ }.$ Then the area of the triangle ABC is

• A. $81\sqrt{3}(\sqrt{3}-1)c{m}^2$
• B. $27\sqrt{3}(\sqrt{3}+1)c{m}^2$
• C. $81\sqrt{3}(\sqrt{3}+1)c{m}^2$
• D. $27\sqrt{3}(\sqrt{3}-1)c{m}^2$

Answer 1

The correct option is A

$81\sqrt{3}(\sqrt{3}-1)c{m}^2$

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

Let $AD=x,BD=(18-x)$ and, $AD=CD=x$

In $\Delta BDC,$ angle of B $\Delta BDC,$ are ${30}^{\circ },{60}^{\circ },{90}^{\circ }$

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

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

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

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

$BD=\frac{x}{\sqrt{3}}=(18-x)$

$x(\frac{1}{\sqrt{3}}+1)=18$

$x\left[\frac{(\sqrt{3}+1)}{\sqrt{3}}\right]=18$

$x=\frac{18.\sqrt{3}}{(\sqrt{3}+1)}$

$x=\frac{18\sqrt{3}}{(\sqrt{3}+1)}\times \frac{(\sqrt{3}-1)}{(\sqrt{3}-1)}=\frac{\stackrel{9}{1\text{/}8}\sqrt{3}(\sqrt{3}-1)}{\text{/}2}=9\sqrt{3}(\sqrt{3}-1)$

Area of triangle ABC

$=\frac{1}{2}\times AB\times CD$

$=\frac{1}{\text{/}2}\times \stackrel{9}{\text{/}18}\times 9\sqrt{3}\left[\right(\sqrt{3}-1\left)\right]$

$=81\sqrt{3}\left[\right(\sqrt{3}-1\left)\right]c{m}^2$