If in the angles of a triangle are in the ratio $1: 1: 4$, then the ratio of the perimeter of the triangle…
If in the angles of a triangle are in the ratio $1: 1: 4$, then the ratio of the perimeter of the triangle to its largest side is
$\sqrt{2}+2: \sqrt{3}$
$3: 2$
$\sqrt{3}+2: \sqrt{2}$
$\sqrt{3}+2: \sqrt{3}$
Solution
Given, the ratio of angles of a triangle is $1: 1: 4$. Let angles of a triangle are $A, B$ and $C$.
$\therefore \quad A: B: C=1: 1: 4$
Let $A=x, B=x$ and $C=4 x$
$\begin{array}{lc}
\because & A+B+C=180^{\circ} \\
\therefore & x+x+4 x=180^{\circ} \\
\Rightarrow & 6 x=180^{\circ} \Rightarrow x=30^{\circ} \\
\therefore & A=30^{\circ}, B=30^{\circ} \text { and } C=120^{\circ}
\end{array}$
Hence, largest angle is $120^{\circ}$. So largest side of a triangle is $\mathrm{c}$.
$\therefore$ Perimeler ul triangle : Largesl side of a triangle
$\begin{aligned}
& =(a+b+c): c \\
& =\left(2 R \sin 30^{\circ}+2 R \sin 30^{\circ}+2 R \sin 120^{\circ}\right)
\end{aligned}$
$2 R \sin 120^{\circ}$
$\begin{aligned}
{[\because a} & =2 R \sin A, b=2 R \sin B \text { and } C=2 R \sin C] \\
& =2 R\left[\frac{1}{2}+\frac{1}{2}+\frac{\sqrt{3}}{2}\right]: 2 R \times \frac{\sqrt{3}}{2} \\
& =\left(1+\frac{\sqrt{3}}{2}\right): \frac{\sqrt{3}}{2}=2+\sqrt{3}: \sqrt{3}
\end{aligned}$