If $A, A_{1}, A_{2}, A_{3}$ are the areas of the inscribed and escribed circles of a triangle ABC, then

\sqrt{A_{1}} + \sqrt{A_{2}} + \sqrt{A_{3}} = \sqrt{π} (r_{1} + r_{2} + r_{3})

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

\frac{1}{{\sqrt{A}}_{1}} + \frac{1}{{\sqrt{A}}_{2}} + \frac{1}{{\sqrt{A}}_{3}} = \frac{1}{\sqrt{A}}

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

\frac{1}{{\sqrt{A}}_{1}} + \frac{1}{{\sqrt{A}}_{2}} + \frac{1}{{\sqrt{A}}_{3}} = \frac{s^{2}}{\sqrt{π} r_{1} r_{2} r_{3}}

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

{\sqrt{A}}_{1} + {\sqrt{A}}_{2} + {\sqrt{A}}_{3} = \sqrt{π} (4 R + r)

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

Open in App

Solution

The correct options are
A $\sqrt{A_{1}} + \sqrt{A_{2}} + \sqrt{A_{3}} = \sqrt{π} (r_{1} + r_{2} + r_{3})$
B ${\sqrt{A}}_{1} + {\sqrt{A}}_{2} + {\sqrt{A}}_{3} = \sqrt{π} (4 R + r)$
C $\frac{1}{{\sqrt{A}}_{1}} + \frac{1}{{\sqrt{A}}_{2}} + \frac{1}{{\sqrt{A}}_{3}} = \frac{1}{\sqrt{A}}$
D $\frac{1}{{\sqrt{A}}_{1}} + \frac{1}{{\sqrt{A}}_{2}} + \frac{1}{{\sqrt{A}}_{3}} = \frac{s^{2}}{\sqrt{π} r_{1} r_{2} r_{3}}$

1) $\sqrt{A_{1}} + \sqrt{A_{2}} + \sqrt{A_{3}} = \sqrt{π {r_{1}}^{2}} + \sqrt{π {r_{2}}^{2}} + \sqrt{π {r_{3}}^{2}} = \sqrt{π} (r_{1} + r_{2} + r_{3})$
2) $\frac{1}{{\sqrt{A}}_{1}} + \frac{1}{{\sqrt{A}}_{2}} + \frac{1}{{\sqrt{A}}_{3}} = \frac{1}{\sqrt{π}} (\frac{1}{r_{1}} + \frac{1}{r_{2}} + \frac{1}{r_{3}}) = \frac{1}{\sqrt{π} r}$
$\Rightarrow \frac{1}{{\sqrt{A}}_{1}} + \frac{1}{{\sqrt{A}}_{2}} + \frac{1}{{\sqrt{A}}_{3}} = \frac{1}{\sqrt{A}}$
3) $\frac{1}{{\sqrt{A}}_{1}} + \frac{1}{{\sqrt{A}}_{2}} + \frac{1}{{\sqrt{A}}_{3}} = \frac{1}{\sqrt{π}} (\frac{1}{r_{1}} + \frac{1}{r_{2}} + \frac{1}{r_{3}}) = \frac{(r_{1} r_{2} + r_{3} r_{2} + r_{1} r_{3})}{\sqrt{π} r_{1} r_{2} r_{3}}$

$\Rightarrow \frac{1}{{\sqrt{A}}_{1}} + \frac{1}{{\sqrt{A}}_{2}} + \frac{1}{{\sqrt{A}}_{3}} = \frac{△^{2}}{\sqrt{π} r_{1} r_{2} r_{3}} [\frac{1}{(s - a) (s - b)} + \frac{1}{(s - c) (s - b)} + \frac{1}{(s - a) (s - c)}]$

$\Rightarrow \frac{1}{{\sqrt{A}}_{1}} + \frac{1}{{\sqrt{A}}_{2}} + \frac{1}{{\sqrt{A}}_{3}} = \frac{△^{2} s (3 s - a - b - c)}{\sqrt{π} r_{1} r_{2} r_{3} s (s - a) (s - b) (s - c)} = \frac{s^{2}}{\sqrt{π} r_{1} r_{2} r_{3}}$
4) ${\sqrt{A}}_{1} + {\sqrt{A}}_{2} + {\sqrt{A}}_{3} = \sqrt{π} [(r_{1} + r_{2} + r_{3} - r - 4 R) + r + 4 R] = \sqrt{π} (r + 4 R)$
Ans: A,B,C,D

Suggest Corrections

Similar questions

\frac{a_{1}}{a_{1}} + \frac{a_{2}}{a_{2}} + \frac{a_{3}}{a_{3}} + \dots \frac{a_{n}}{a_{n}} = \frac{1}{1} + \frac{1}{a_{1}} + \frac{a_{1}}{a_{2}} + \frac{a_{2}}{a_{3}} + \dots \frac{a_{n - 2}}{a_{n - 1}}

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

\frac{a_{1}}{a_{2} + a_{3} + . . . . + a_{n}}, \frac{a_{2}}{a_{1} + a_{3} + . . . . + a_{n}}, . . . ., \frac{a_{n}}{a_{1} + a_{2} + . . . . + a_{n - 1}}

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

\frac{a_{1}}{a_{2} + a_{3} + . . . a_{n}}, \frac{a_{2}}{a_{1} + a_{3} + . . . . + a_{n}}, \frac{a_{n}}{a_{1} + a_{2} + . . . + a_{n - 1}}

\frac{a_{1}}{a_{1} + 1 -} \frac{a_{2}}{a_{2} + 1 -} \frac{a_{3}}{a_{3} + 1 -} \dots, a_{1}, a_{2}, a_{3}, \dots

a_{1}, a_{2}, a_{3}, a_{4}

\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + \frac{1}{\sqrt{a_{3}} + \sqrt{a_{4}}}

Join BYJU'S Learning Program