Question

In a triangle with one angle of ${{120}^{0}}$ the lengths of the sides form an A.P. If the length of the greatest side is $7cm$, the area of triangle is
 A. \[\frac{3\sqrt{15}}{4}c{{m}^{2}}\]
B. \[\frac{15\sqrt{3}}{4}c{{m}^{2}}\]
C. \[\frac{15}{4}c{{m}^{2}}\]
D. \[\frac{3\sqrt{3}}{4}c{{m}^{2}}\]

Answer 1

Hint: To solve this question, we will use the condition of numbers being in A.P and derive the value of side $b$. We will now use cosine rule $\cos C=\frac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}$ and substitute the value of $b$ derived in it and simplify. We will now substitute the given values in the simplified equation and derive a quadratic equation and factorize it and find the value of side $a$. Using values of $a,b,c$, we will find the area with the help of Heron’s Formula.
Formula used:
The condition for three numbers $a,b,c$ being in A.P is $2b=a+c$.
Heron’s formula:
$Area=\sqrt{s(s-a)(s-b)(s-c)},\,\,s=\frac{a+b+c}{2}$
Complete step-by-step solution:
We are given a triangle with one angle as ${{120}^{0}}$ and the lengths of this triangle forms an A.P. We have to calculate the area of the triangle when the length of the greatest side is $7cm$.
We will use the condition of A.P and derive the value of $b$.
\[\begin{align}
  & 2b=a+c \\
 & b=\frac{a+c}{2} \\
\end{align}\]
Now we will substitute the value of $b$ derived in the cosine rule $\cos C=\frac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}$.
$\begin{align}
  & \cos C=\frac{{{a}^{2}}+{{\frac{(a+c)}{4}}^{2}}-{{c}^{2}}}{2a\frac{(a+c)}{2}} \\
 & =\frac{{{a}^{2}}+\frac{{{a}^{2}}+{{c}^{2}}+2ac}{4}-{{c}^{2}}}{a(a+c)}
\end{align}$
Now we will substitute the given value of angle ${{120}^{0}}$the length of the greatest side as $c=7$.

$\begin{align}
  & \cos 120=\frac{{{a}^{2}}+\frac{{{a}^{2}}+{{7}^{2}}+2a(7)}{4}-{{7}^{2}}}{a(a+7)} \\
 & -\frac{1}{2}=\frac{{{a}^{2}}+\frac{{{a}^{2}}+49+14a}{4}-49}{a(a+7)} \\
 & -\frac{1}{2}a(a+7)=\frac{4{{a}^{2}}+{{a}^{2}}+49+14a-196}{4} \\
 & -2{{a}^{2}}-14a=5{{a}^{2}}+14a-147 \\
 & 7{{a}^{2}}+28a-147=0 \\
 & {{a}^{2}}+4a-21=0
\end{align}$
We will now factorize the equation ${{a}^{2}}+4a-21=0$and derive the value of $a$.
\[\begin{align}
  & {{a}^{2}}+4a-21=0 \\
 & {{a}^{2}}+7a-3a-21=0 \\
 & a(a+7)-3(a+7)=0 \\
 & (a+7)(a-3)=0
\end{align}\]
As the value of length cannot be negative we will consider the value of $a$ as $a=3$.
We will now determine the value of the other side $b$ by substituting the values of $a=3$and $c=7$ in $2b=a+c$.
$\begin{align}
  & 2b=3+7 \\
 & 2b=10 \\
 & b=5
\end{align}$
The semi perimeter of the triangle is,
$\begin{align}
  & s=\frac{3+5+7}{2} \\
 & =\frac{15}{2}
\end{align}$
We will now find the area,
$\begin{align}
  & Area=\sqrt{\frac{15}{2}\left( \frac{15}{2}-3 \right)\left( \frac{15}{2}-5 \right)\left( \frac{15}{2}-7 \right)}\,\, \\
 & =\sqrt{\frac{15}{2}\left( \frac{9}{2} \right)\left( \frac{5}{2} \right)\left( \frac{1}{2} \right)}\, \\
 & =\frac{15\sqrt{3}}{4}c{{m}^{2}}\,
\end{align}$
The area of the triangle is \[\frac{15\sqrt{3}}{4}c{{m}^{2}}\] when the length of the largest side is $7cm$and an angle is ${{120}^{0}}$ and all the other sides are in A.P. Hence the correct option is (B).
Note:
Arithmetic progression can be defined as the series of numbers when the difference between the consecutive terms is always the same. If three numbers $a,b,c$ are in A.P then $b-a=c-b$. As all the three sides of the triangle are different which means the given triangle is scalene that is why we have used Heron’s formula to calculate the area of the triangle.