Question

If two angle of a triangle are ${30}^o$ and ${45}^o$ and included side is $\sqrt{3}+1cm$ then the area of the triangle is

• A. $\frac{\sqrt{3}+1}{2}$
• B. $\frac{\sqrt{3}-1}{2}$
• C. $\frac{\sqrt{3}+1}{2\sqrt{2}}$
• D. $\frac{\sqrt{3}-1}{2\sqrt{2}}$

Answer 1

The correct option is A $\frac{\sqrt{3}+1}{2}$

Given, two angles of triangle ${30}^o$ & ${45}^o$ included side $\sqrt{3}+1$ .

Let $\Delta$ABC

$\angle B=30$

$\angle C=45$

$BC=\sqrt{3}+1$

In $\Delta$ABC AD is perpendicular to BC

So, in $\Delta$ABD

$\tan {30}^o=\frac{AD}{BD}=\frac{AD}{x}=\frac{1}{\sqrt{3}}$ ……….(i)

In $\Delta$ADC

$\tan {45}^o=\frac{AD}{CD}=\frac{AD}{\sqrt{3}+1-x}=1$ ……….(ii)

Let compare equation (i) & (ii)

$\frac{x}{\sqrt{3}}=\sqrt{3}+1-x$

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

$\therefore x=3+\sqrt{3}-\sqrt{3}x$

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

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

$\therefore x=\sqrt{3}$

Put the value of x in equation (i)

$\frac{AD}{\sqrt{3}}=\frac{1}{\sqrt{3}}$,

So $AD=1$cm

$\to$ Area of triangle $\Delta$ABC$=\frac{1}{2}\times$base$\times$height

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

$=\frac{\sqrt{3}+1}{2}$.