Question

Area of the triangle is $10\sqrt{3}$ sq. cm, angle $C={{60}^{0}}$and its perimeter is $20cm$, then side $c$will be
A. $5$
B. $7$
C. $8$
D. $10$

Answer 1

Hint: We will use the formula of area of triangle when two sides and an angle is given and substitute the given values in it and derive the value of \[ab\]. Using the perimeter of the triangle we will make an equation and derive the value of \[a+b\]. Then we will substitute both the derived values and other given values in the cosine rule ${{c}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\cos (C)$ and find the length of the side required

Formula Used: Area of the triangle when two sides and an angle given can be calculated with the formula $Area=\frac{1}{2}ab\sin C$.
${{a}^{2}}+{{b}^{2}}={{(a+b)}^{2}}-2ab$

Complete step by step solution: We are given,
Area of the triangle $\vartriangle ABC=10\sqrt{3}$ sq. cm.
One of the angles of the triangle $C={{60}^{0}}$.
Perimeter of the triangle$~=20$cm.
And we have to determine the value of side $c$.
Now we will substitute the value of area $\vartriangle ABC=10\sqrt{3}$and angle $C={{60}^{0}}$.
$\begin{align} In~the~formula.
  &~Area=\frac{1}{2}ab\sin C \\
 & 10\sqrt{3}=\frac{1}{2}ab\sin {{60}^{0}} \\
 & 10\sqrt{3}=\frac{1}{2}ab\times \frac{\sqrt{3}}{2} \\
 & 40\sqrt{3}=\sqrt{3}ab \\
 & ab=40
\end{align}$
We know that perimeter can be defined as the sum of all the boundaries so the perimeter of the triangle here will be the sum of all the lengths of the sides,
$\begin{align}
  & a+b+c=20 \\
 & a+b=20-c \\
\end{align}$
Now we will use law of cosine,
${{c}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\cos (C)$
We will now substitute the value of angle $C={{60}^{0}}$and $ab=40$in cosine rule.
\[\begin{align}
  & {{c}^{2}}={{a}^{2}}+{{b}^{2}}-2\times 40\cos {{60}^{0}} \\
 & {{c}^{2}}={{a}^{2}}+{{b}^{2}}-80\times \frac{1}{2} \\
 & {{c}^{2}}={{a}^{2}}+{{b}^{2}}-40 \\
 & {{c}^{2}}={{(a+b)}^{2}}-2ab-40 \\
\end{align}\]
We will now substitute in the above equation.
$\begin{align}
  & {{c}^{2}}={{(20-c)}^{2}}-40 \\
 & {{c}^{2}}={{(20-c)}^{2}}-2(40)-40 \\
 & {{c}^{2}}=400+{{c}^{2}}-40c-80-40 \\
 & {{c}^{2}}-{{c}^{2}}=280-40c \\
 & 0=280-40c \\
 & 40c=280 \\
 & c=7cm
\end{align}$

The value of the side $c$ is $c=7cm$ of the triangle when area of the triangle is $10\sqrt{3}$ sq. cm, angle $C={{60}^{0}}$ and perimeter is $20cm$. Hence the correct option is (B).

Note: The area of the triangle can be written in three ways depending upon the sides and the included angle. They are:
$Area=\frac{1}{2}ab\sin C$
$Area=\frac{1}{2}ac\sin B$
$Area=\frac{1}{2}bc\sin A$