Question

In ΔABC, prove that 4(r₁r₂ + r₂r₃ + r₃r₁) = (a + b + c)².

Answer 1

Solution :

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We know that ,

i ) r1 = ∆/( s - a ),

ii ) r2 = ∆/( s - b ) ,

iii ) r3 = ∆/( s - c )

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LHS = 4(r1r2 + r2r3 + r3r1)

= 4[∆/(s-a)*∆/(s-b) + ∆/(s-b)*∆/(s-c)+∆/(s-c)*∆/(s-a)]

= 4∆² [(s-c+s-a+s-b)/(s-a)(s-b)(s-c)]

= 4∆²[ {3s - (a+b+c)}/[(s-a)(s-b)(s-c)]

= 4∆² [ 3s - 2s ]/[(s-a)(s-b)(s-c)]

[ Since , a + b + c = 2s ]

= 4∆² [ s/[(s-a)(s-b)(s-c)]

= 4∆² {s²/[ s(s-a)(s-b)(s-c) ]}

= 4∆² ( s²/∆² )

[ Since , ∆ = √s(s-a)(s-b)(s-c) ]

= 4s²

= ( 2s )²

= ( a + b + c )²

= RHS

•••••

Answer 2

HELLO DEAR,

Answer:

Step-by-step explanation:

r1 = ∆/( s - a ), r2 = ∆/( s - b ) , r3 = ∆/( s - c )

Now, 4(r1r2 + r2r3 + r3r1)

= 4[∆/(s-a)*∆/(s-b) + ∆/(s-b)*∆/(s-c)+∆/(s-c)*∆/(s-a)]

= 4∆² [(s-c+s-a+s-b)/(s-a)(s-b)(s-c)]

= 4∆²[ {3s - (a+b+c)}/[(s-a)(s-b)(s-c)]

= 4∆² [ 3s - 2s ]/[(s-a)(s-b)(s-c)]

[ we know , a + b + c = 2s ]

= 4∆² [ s/[(s-a)(s-b)(s-c)]

= 4∆² {s²/[ s(s-a)(s-b)(s-c) ]}

= 4∆² ( s²/∆² )

[ we know, ∆ = √s(s-a)(s-b)(s-c) ]

= 4s²

= ( 2s )²

= ( a + b + c )²

I HOPE IT'S HELP YOU DEAR,

THANKS