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Section Formula

In a Δ ABC,...

Question

In a $Δ A B C,$ the inradius and exradii are $r, r_{1}, r_{2}$ and $r_{3}$ respectively. In usual notations the value of $r . r_{1} . r_{2} . r_{3}$ is equal to

2 Δ

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Δ^{2}

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\frac{a b c}{4 R}

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none of these

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Solution

The correct option is D $Δ^{2}$
consider, $t = r r_{1} r_{2} r_{3}$
$\Rightarrow t = (\frac{△}{s}) (\frac{△}{s - a}) (\frac{△}{s - b}) (\frac{△}{s - c})$
$\Rightarrow t = \frac{△^{4}}{s (s - a) (s - b) (s - c)} = \frac{△^{4}}{△^{2}}$
$\Rightarrow t = △^{2}$
Ans: B

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Similar questions

Q. In a

Δ A B C,

the inradius and three exradii are

r, r_{1^{'}} r_{2}

and

r_{3}

respectively.

In usual notations the value of

r . r_{1} . r_{2} . r_{3}

is equal to?

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

Q. Assertion :In a

Δ

ABC,

r_{1} + r_{2} + r_{3} - r = 4 R

Reason:

r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1} = Δ^{2}

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}

Q. In a

Δ A B C

, the value of the product of

(r \cdot r_{1} \cdot r_{2} \cdot r_{3})

(where

r

is the inradius and

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

are the exradius respectively) is

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In a ΔABC, the inradius and exradii are r,r1,r2 and r3 respectively. In usual notations the value of r.r1.r2.r3 is equal to

The correct option is D Δ2consider, t=rr1r2r3⇒t=(△s)(△s−a)(△s−b)(△s−c)⇒t=△4s(s−a)(s−b)(s−c)=△4△2⇒t=△2Ans: B

In a $Δ A B C,$ the inradius and exradii are $r, r_{1}, r_{2}$ and $r_{3}$ respectively. In usual notations the value of $r . r_{1} . r_{2} . r_{3}$ is equal to

The correct option is D $Δ^{2}$
consider, $t = r r_{1} r_{2} r_{3}$
$\Rightarrow t = (\frac{△}{s}) (\frac{△}{s - a}) (\frac{△}{s - b}) (\frac{△}{s - c})$
$\Rightarrow t = \frac{△^{4}}{s (s - a) (s - b) (s - c)} = \frac{△^{4}}{△^{2}}$
$\Rightarrow t = △^{2}$
Ans: B