In any triangle, the minimum value of $r_{1} r_{2} r_{3} / r^{3}$ is equal to

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

none of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B 27
As A.M $\geq$ G.M
Therefore, $\frac{s - a + s - b + s - c}{3} \geq {[(s - a) (s - b) (s - c)]}^{1 / 3}$
$\Rightarrow \frac{3 s - 2 s}{3} \geq {[(s - a) (s - b) (s - c)]}^{1 / 3}$
$\Rightarrow \frac{s^{3}}{(s - a) (s - b) (s - c)} \geq 27$ ...(1)
now, $\frac{r_{1} r_{2} r_{3}}{r^{3}} = \frac{(△ / s - a) (△ / s - b) (△ / s - c)}{{(△ / s)}^{3}} = \frac{s^{3}}{(s - a) (s - b) (s - c)}$
Therefore, from (1) $\frac{r_{1} r_{2} r_{3}}{r^{3}} \geq 27$

Suggest Corrections

Similar questions

Δ A B C,

\frac{r_{1} r_{2} r_{3}}{r^{3}}

Δ A B C

\frac{r_{1} r_{2} r_{3}}{r^{3}}

Δ A B C

\frac{r_{1} r_{2} r_{3}}{r^{3}} = 27

x_{1} + x_{2} + x_{3} + . . . . + x_{n} = λ

x_{1}, x_{2}, x_{3}, . . . . x_{n}

x_{1} = x_{2} = x_{3} = . . . . x_{n}

\frac{r_{1} + r_{2} + r_{3}}{r}

A M \geq G M

\frac{r_{1} r_{2} r_{3}}{r^{3}}

27

a_{1} + a_{2} + a_{3} + . . . + a_{n} = k

k

a_{1} a_{2} a_{3} . . . a_{n}

a_{1} = a_{2} = a_{3} = . . . = a_{n}

Join BYJU'S Learning Program