
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
163.2k+ views
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:
$Area=\frac{1}{2}ab\sin C$
$Area=\frac{1}{2}ac\sin B$
$Area=\frac{1}{2}bc\sin A$
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:
$Area=\frac{1}{2}ab\sin C$
$Area=\frac{1}{2}ac\sin B$
$Area=\frac{1}{2}bc\sin A$
Recently Updated Pages
Fluid Pressure - Important Concepts and Tips for JEE

JEE Main 2023 (February 1st Shift 2) Physics Question Paper with Answer Key

Impulse Momentum Theorem Important Concepts and Tips for JEE

Graphical Methods of Vector Addition - Important Concepts for JEE

JEE Main 2022 (July 29th Shift 1) Chemistry Question Paper with Answer Key

JEE Main 2023 (February 1st Shift 1) Physics Question Paper with Answer Key

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations

NCERT Solutions for Class 11 Maths In Hindi Chapter 1 Sets

NEET 2025 – Every New Update You Need to Know
