Question

$ABC$ is an equilateral trianlge of side $4$ cm. If $R.r$ and $h$ are the circumradius, inradius, and altitude, respectively, then $\frac{R+r}{h}$ is equal to

• A. $4$
• B. $2$
• C. $1$
• D. $3$

Answer 1

The correct option is C $1$

have, We$R=\frac{a^3}{4\Delta }=\frac{a^3}{\sqrt{3}{a}^2}=\frac{a}{\sqrt{3}}$
$=\frac{4}{\sqrt{3}}$

$r=\frac{\Delta }{s}=\frac{\sqrt{3}{a}^2}{4s}=\frac{\sqrt{3}{a}^2}{6a}=\frac{2\sqrt{3}}{3}$
$h=\frac{2\Delta }{a}=\frac{2\sqrt{3}{a}^2}{4a}=2\sqrt{3}$
$\therefore \frac{R+r}{h}=\frac{\frac{4}{\sqrt{3}}+\frac{2\sqrt{3}}{3}}{2\sqrt{3}}=\frac{\frac{6}{\sqrt{3}}}{2\sqrt{3}}$

$=1$

Option $C$ is correct