SSC CGL Tier II 12 Apr 2015 Paper

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Question : 23 of 100

Marks: +1, -0

If

2s = a + b + c

, then the value of

s(s – c) + (s – a) (s – b)

is

$a b$
$a b c$
$0$
${a+b+c}/{2}$

Solution:

(1)

2s = a + b + c

∴ s(s-c)= ( {a+b+c}/{2} ) ( {a+b+c}/{2}-c )

= {(a+b+c)(a+b-c)}/{4}

Again,

(s-a)(s-b)

= {1}/{4}(2 s-2 a)(2 s-2 b)

=\ {1}/{4}(a+b+c-2 a)(a+b+c

-2 b )

= {1}/{4}(b+c-a)(a+c-b)

∴ s(s-c)+(s-a)(s-b)

= {1}/{4}[(a+b+c)(a+b-c)+(b

+c-a )(a+c-b) ]

= {1}/{4} [(a+b)^{2}-c^{2}+a b+a c-a^{2}

+b c+c^{2}-a c-b^{2}-b c+a b ]

= {1}/{4} (a^{2}+b^{2}+2 a b-c^{2}+a b+

a c-a^{2}+b c+c^{2}-a c-b^{2}-b c+

a b )

= {1}/{4} × 4 a b=a b

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