In a - ABC- the value of arr1-r 2 -r3- -

A(rr_1+r_ 2 r_3) nbsp; =a (Δs× nbsp;Δs a + nbsp;Δs b× nbsp;Δs c) =aΔ^2 ((s b)(s c)+s(s a)s(s a)(s b)(s c) nbsp; ) = aΔ^2 ((s b)(s c)+s(s a)Δ^2) = a ...

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Byju's Answer

Standard XII

Mathematics

Ex Radius

In a Δ ABC, t...

Question

In a $Δ A B C,$ the value of $a (r r_{1} + r_{2} r_{3}) =$

\frac{c a - r_{1} r_{3}}{r_{2}}

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

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\frac{r_{3}^{2}}{4 R - r_{1} - r_{2}}

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a b c

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Solution

The correct option is D $a b c$
$a (r r_{1} + r_{2} r_{3})$
$= a (\frac{Δ}{s} \times \frac{Δ}{s - a} + \frac{Δ}{s - b} \times \frac{Δ}{s - c})$
$= a Δ^{2} (\frac{(s - b) (s - c) + s (s - a)}{s (s - a) (s - b) (s - c)})$
$= a Δ^{2} (\frac{(s - b) (s - c) + s (s - a)}{Δ^{2}})$
$= a (s^{2} - s (a + b + c) + b c + s^{2})$
$= a (s^{2} - s (2 s) + b c + s^{2})$
$= a b c$

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Similar questions

Q. Prove that :

a (r r_{1} + r_{2} r_{3}) = b (r r_{2} + r_{3} r_{1}) = c (r r_{3} + r_{1} r_{2})

where

r

is inradius and

r_{1}, r_{2}, r_{3}

are exradius of triangle

A B C

and

a, b, c

are the corresponding sides.

Q. Statement

- 1

: In a

Δ A B C

, if

a < b < c

and

r

is in-radius and

r_{1}, r_{2}, r_{3}

are the exradii opposite to angle

A, B, C

respectively, then

r < r_{1} < r_{2} < r_{3}

.

Statement

- 2

: For,

Δ A B C

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

Q. Assertion :In any

△ A B C

maximum value of

r_{1} + r_{2} + r_{3} = \frac{9 R}{2}

because Reason: In any

△ A B C

R \geq 2 r

Q. Assertion :In a

△ A B C

, if

a < b < c

and

r

is inradius and

r_{1}, r_{2}, r_{3}

are the axradii opposite to angle

A, B, C

respectively, then

r < r_{1} < r_{2} < r_{3}

Reason:

△ A B C, r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1} = \frac{r_{1} r_{2} r_{3}}{r}

In a $△$ ABC, the inradius is r and three exradii are $r_{1}, r_{2} a n d r_{3}$ respectably. In usual notations the value of $r . r_{1} . r_{2} . r_{3}$ is equal to

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In a ΔABC, the value of a(rr1+r2r3)=

The correct option is D abca(rr1+r2r3) =a(Δs×Δs−a+Δs−b×Δs−c) =aΔ2((s−b)(s−c)+s(s−a)s(s−a)(s−b)(s−c)) =aΔ2((s−b)(s−c)+s(s−a)Δ2) =a(s2−s(a+b+c)+bc+s2) =a(s2−s(2s)+bc+s2) =abc

In a $Δ A B C,$ the value of $a (r r_{1} + r_{2} r_{3}) =$