Question

In ΔABC, prove that $(\frac{1}{r} - \frac{1}{r_1})(\frac{1}{r} - \frac{1}{r_2})(\frac{1}{r} - \frac{1}{r_3}) = \frac{abc}{\triangle^{3}} = \frac{4R}{r^{2}s^{2}}$.

Answer 1

Solution :

i ) 1/r - 1/r1

= S/∆-(S-a)/∆

= (S-S+a)/∆

= a/∆

ii ) 1/r - 1/r²

= S/∆ - ( S - b )/∆

= ( S - S + b )/∆

= b/∆

iii ) 1/r - 1/r3

= S/∆ - ( S - c )/∆

= ( S - S + c )/∆

= c/∆

Now ,

LHS = (1/r-1/r1)(1/r-1/r2)(1/r-1/r3)

= ( a/∆ ) ( b/∆ ) ( c/∆ )

= ( abc )/∆³

= ( 4R * ∆ )/∆³

= 4R/∆²

= ( 4R )/( rs )² [ Since , ∆ = rs ]

= ( 4R )/( r²s² )

= RHS

•••••

Answer 2

HELLO DEAR,

Answer:

Step-by-step explanation:

(1/r - 1/r1)( 1/r - 1/r2)(1/r - 1/r3)

=> {S/∆-(S-a)/∆}{ S/∆ - ( S - b )/∆}{ S/∆ - ( S - c )/∆}

=> {(S - S+a)/∆}{(S - S + b )/∆}{(S - S + c )/∆}

=> (a/∆)(b/∆)(c/∆)

=> (abc)/∆³

=> ( 4R * ∆ )/∆³

=> 4R/∆²

=> ( 4R )/( rs )²

[ ∆ = rs ]

=> ( 4R )/( r²s² )

Hence, (1/r - 1/r1)( 1/r - 1/r2)(1/r - 1/r3) = (abc)/∆³ = (4R )/( r²s² )

I HOPE IT'S HELP YOU DEAR,

THANKS