R1s - b s - c - r2s - c s - a - r3s - a s - b - 3r

To prove: r_1(s b)(s c)+r_2(s a)(s c)+r_3(s a)(s b)=3r L.H.S= r_1(s b)(s c)+r_2(s a)(s c)+r_3(s a)(s b) = S(s a)(s b)(s c)+S(s a)(s b)(s c)+S(s a) ...

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Byju's Answer

Standard IX

Mathematics

Area of Any Polygon - by Heron's Formula

r1s - b s ...

Question

$\frac{r_{1}}{(s - b) (s - c)} + \frac{r_{2}}{(s - c) (s - a)} + \frac{r_{3}}{(s - a) (s - b)} = \frac{3}{r}$

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Solution

To prove:
$\frac{r_{1}}{(s - b) (s - c)} + \frac{r_{2}}{(s - a) (s - c)} + \frac{r_{3}}{(s - a) (s - b)} = \frac{3}{r}$
L.H.S= $\frac{r_{1}}{(s - b) (s - c)} + \frac{r_{2}}{(s - a) (s - c)} + \frac{r_{3}}{(s - a) (s - b)}$
$= \frac{S}{(s - a) (s - b) (s - c)} + \frac{S}{(s - a) (s - b) (s - c)} + \frac{S}{(s - a) (s - b) (s - c)}$
$= \frac{3 S}{(s - a) (s - b) (s - c)}$
$= \frac{3 S \times s}{S^{2}}$
$= 3 \frac{r}{S} = \frac{3}{r} =$ R.H.S proved.

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Similar questions

Q. With usual notations, prove that in a triangle ABC

\frac{r_{1}}{(s - b) (s - c)} + \frac{r_{2}}{(s - c) (s - a)} + \frac{r_{3}}{(s - a) (s - b)} = \frac{3}{r}

Q. In

Δ A B C

(a - b) (s - c) = (b - c) (s - a)

, then

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

are in

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. If

r_{1}, r_{2}, r_{3}, s, △, a, b, c

are the standard notations then which of the following is correct?

Q. In a triangle ABC, if

\frac{a - b}{b - c}

\frac{s - a}{s - c},

,then

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

are in:

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r1(s−b)(s−c)+r2(s−c)(s−a)+r3(s−a)(s−b)=3r

To prove:r1(s−b)(s−c)+r2(s−a)(s−c)+r3(s−a)(s−b)=3rL.H.S= r1(s−b)(s−c)+r2(s−a)(s−c)+r3(s−a)(s−b)=S(s−a)(s−b)(s−c)+S(s−a)(s−b)(s−c)+S(s−a)(s−b)(s−c)=3S(s−a)(s−b)(s−c)=3S×sS2=3rS=3r= R.H.S proved.

$\frac{r_{1}}{(s - b) (s - c)} + \frac{r_{2}}{(s - c) (s - a)} + \frac{r_{3}}{(s - a) (s - b)} = \frac{3}{r}$