Mathematics › Properties of Triangles › Napiers Analogy, Area of Triangle & Half angle Formulae
The area of the triangle $\mathrm{ABC}$ is $10 \sqrt{3} \mathrm{~cm}^{2}$, angle $\mathrm{B}$ is…
The area of the triangle $\mathrm{ABC}$ is $10 \sqrt{3} \mathrm{~cm}^{2}$, angle $\mathrm{B}$ is $60^{\circ}$ and its perimeter is $20 \mathrm{~cm}$,
then $\ell(\mathrm{AC})=$
$7 \mathrm{~cm}$ $8 \mathrm{~cm}$ $10 \mathrm{~cm}$ $5 \mathrm{~cm}$ Solution
$\begin{array}{l}
\text { Area }=\frac{\operatorname{acsin} B}{2}=10 \sqrt{3}=\frac{\operatorname{acsin} 60}{2} \\
10 \sqrt{3}=\frac{\operatorname{ac} \sqrt{3}}{4} a c=40
\end{array}$
Now
$\begin{aligned} & \cos \mathrm{B}=\frac{\mathrm{a}^{2}+\mathrm{c}^{2}-\mathrm{b}^{2}}{2 \mathrm{ac}} \\ & \cos 60 \times 2 \mathrm{ac}=\mathrm{a}^{2}+\mathrm{c}^{2}-\mathrm{b}^{2} \\ \therefore & \frac{1}{2} \times 2 \times 40=(\mathrm{a}+\mathrm{c})^{2}-2 \mathrm{ac}-\mathrm{b}^{2} \end{aligned}$
$\therefore 40=(20-b)^{2}-(2 \times 40)-b^{2} \quad \ldots[a+b+c=20$, given $]$
$\begin{aligned} &=400+b^{2}-40 b-80-b^{2} \\ \therefore \quad 40 b &=280 \Rightarrow b=7 \end{aligned}$
Asked in: MHT CET 2020 (19 Oct Shift 2)
Practice more Properties of Triangles questions on Aicharya