In $\triangle \mathrm{ABC}, \mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are in arithmetic progression and $a:…

In $\triangle \mathrm{ABC}, \mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are in arithmetic progression and $a: c=1: 2$. If $b=4 \sqrt{3} \mathrm{~cm}$, then the area of $\triangle A B C$ (in sq. $\mathrm{cm})$ is
  1. $16 \sqrt{3}$
  2. $12 \sqrt{3}$
  3. $8 \sqrt{3}$
  4. $6 \sqrt{3}$

Solution

$\because A, B, C$ are in A.P. $\Rightarrow 2 B=A+C$ Now, $A+B+C=\pi \Rightarrow 3 \mathrm{~B}=\pi \Rightarrow B=\frac{\pi}{3}$ Now, $\tan \frac{(C-A)}{2}=\frac{c-a}{c+a} \cot \left(\frac{B}{2}\right)=\frac{\frac{c}{a}-1}{\frac{c}{a}+1} \cot \left(\frac{\pi}{6}\right)$ $=\frac{2-1}{2+1} \cdot \sqrt{3}=\frac{1}{3} \sqrt{3}=\frac{1}{\sqrt{3}}$ $\Rightarrow \frac{C-A}{2}=\frac{\pi}{6} \Rightarrow C-A=\frac{\pi}{3}....(i)$ Also, $A+C=2 B=\frac{2 \pi}{3}...(ii)$ Solving (i) \& (ii) $\Rightarrow C=\frac{\pi}{2}$ Now, $a^2+b^2=c^2 \Rightarrow a^2+48=c^2 \quad\{\because b=4 \sqrt{3}\}$ $\Rightarrow \frac{a^2}{c^2}+\frac{48}{c^2}=1 \Rightarrow \frac{1}{4}-\frac{48}{c^2}=1 \Rightarrow c=8$ $\because \frac{a}{c}=\frac{1}{2} \Rightarrow a=4$ $c \quad 2$ Now, area of triangle $\Rightarrow \Delta=\frac{1}{2} \times a \times b=\frac{1}{2} \times 4 \times 4 \sqrt{3}=8 \sqrt{3}$

Asked in: AP EAMCET 2023 (17 May Shift 1)

Practice more Properties of Triangles questions on Aicharya