Question

The angles A, B and C of a triangle ABC are in A.P and $a:b=1:\sqrt{3},c=4cm$. The area of the triangle is
[a] $4\sqrt{3}$
[b] $\dfrac{2}{\sqrt{3}}$
[c] $2\sqrt{3}$
[d] $\dfrac{4}{\sqrt{3}}$

Answer 1

Hint: Use the fact that if three numbers a, b and c are in A.P then a + c =2b. Use angle sum property of triangle and hence determine the value of B. Find the value of b in terms of a using the fact that $a:b=1:\sqrt{3}$. Use the cosine rule on B and hence determine the value of a and b. Use the fact that the area of a triangle is given by $\Delta =\dfrac{1}{2}ac\sin B$, a, b, c, A, B, C have their usual meanings. Hence determine the area of the triangle and hence determine which of the options is correct.

Complete step-by-step answer:

Given: A, B, C are in A.P, $a:b=1:\sqrt{3}$ and c = 4cm.
To determine: $\Delta $(Area of the triangle)
We know that if a, b and c are in A.P then a + c = 2b
Since A, B and C are in A.P, we have
A + C = 2B (i)
Also by angle sum property of triangle, we have
$A+B+C=180{}^\circ $
Substituting the value of A+C from equation (i), we get
$\begin{align}
  & 2B+B=180{}^\circ \\
 & \Rightarrow 3B=180{}^\circ \\
\end{align}$
Dividing both sides by 3, we get
$B=60{}^\circ $
Also, we have
$\begin{align}
  & \dfrac{a}{b}=\dfrac{1}{\sqrt{3}} \\
 & \Rightarrow b=\sqrt{3}a \\
\end{align}$
Now, we know that in a triangle
$\cos B=\dfrac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac}$(Cosine rule)
Hence, we have
$\cos 60{}^\circ =\dfrac{{{a}^{2}}+{{4}^{2}}-{{\left( \sqrt{3}a \right)}^{2}}}{2\times a\times 4}$
We know that $\cos 60{}^\circ =\dfrac{1}{2}$
Hence, we have
$\dfrac{1}{2}=\dfrac{{{a}^{2}}+16-3{{a}^{2}}}{8a}$
Multiplying both sides by 8a, we get
$4a=16-2{{a}^{2}}$
Adding $2{{a}^{2}}-16$ on both sides, we get
$2{{a}^{2}}+4a-16=0$
Dividing both sides by 2, we get
${{a}^{2}}+2a-8=0$
We shall solve this quadratic equation using a method of splitting the middle term.
We have $4-2=2,4\times 2=8$
Hence, we have
${{a}^{2}}+4a-2a-4\times 2=0$
Taking a common from the first two terms and -2 common from the last two terms, we get
$a\left( a+4 \right)-2\left( a+4 \right)=0$
Taking a+4 common, we get
$\left( a-2 \right)\left( a+4 \right)=0$
Since a > 0 , we have $a+4\ne 0$
Hence, we have a = 2
Hence, we have
$b=\sqrt{3}\times 2=2\sqrt{3}$
We know that area of a triangle is given by
$\Delta =\dfrac{1}{2}ac\sin B$
Hence, we have
$\Delta =\dfrac{1}{2}\times 2\times 4\times \sin \left( 60{}^\circ \right)$
We know that $\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}$
Hence, we have
$\Delta =\dfrac{1}{2}\times 2\times 4\times \dfrac{\sqrt{3}}{2}=2\sqrt{3}$

So, the correct answer is “Option c”.

Note: [1] We can find the area of the triangle using heron’s formula also.
We have
$a=2,b=2\sqrt{3},c=4$
Hence, we have
$s=\dfrac{2+4+2\sqrt{3}}{2}=3+\sqrt{3}$
Hence, we have
$\begin{align}
  & \Delta =\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)} \\
 & =\sqrt{\left( 3+\sqrt{3} \right)\left( 1+\sqrt{3} \right)\left( 3-\sqrt{3} \right)\left( \sqrt{3}-1 \right)} \\
\end{align}$
We know that $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$
Hence, we have
$\begin{align}
  & \Delta =\sqrt{\left( {{3}^{2}}-{{\left( \sqrt{3} \right)}^{2}} \right)\left( {{\left( \sqrt{3} \right)}^{2}}-1 \right)} \\
 & =\sqrt{\left( 6 \right)\left( 2 \right)}=2\sqrt{3} \\
\end{align}$
Which is the same as obtained above.
Hence option [c] is correct.