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TS EAMCET 5 May 2018 Shift 2 Solved Paper

Section: Mathematics

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Question : 28 of 160

Marks: +1, -0

In

\Delta ABC . r_{1}+r_{2}+r_{3}=

$4 R$
$4 R+r$
$4 R-r$
$4 R+s^{2}$

Solution:

The value of

r_{1}+r_{2}+r_{3}

is calculated as,

r_{1}+r_{2}+r_{3}

= \frac{\Delta}{s-a} + \frac{\Delta}{s-b} + \frac{\Delta}{s-c}

= \Delta \left[ \frac{1}{s-a} + \frac{1}{s-b} \right] + \frac{\Delta}{s-c}

- \frac{\Delta}{s} + \frac{\Delta}{s}

= \Delta \left[ \frac{s-b+s-a}{(s-a)(s-b)} \right] + \frac{\Delta(s-s+c)}{s(s-c)}

+ \frac{\Delta}{s}

= \left[ \frac{\Delta c}{(s-a)(s-b)} \right] + \frac{\Delta c}{s(s-c)}

+ \frac{\Delta}{s}

Solve further,

r_{1}+r_{2}+r_{3}

= \Delta c \left[ \frac{s^{2}-c s+s^{2}-a s-b s+a b}{s(s-a)(s-b)(s-c)} \right]

+ \frac{\Delta}{s}

= \left[ \frac{\Delta c (2 s^{2}-s(a+b+c)+a b)}{s(s-a)(s-b)(s-c)} \right] + \frac{\Delta}{s}

= \frac{\Delta c (2 s^{2}-2 s^{2}+a b)}{\Delta^{2}} + \frac{\Delta}{s}

= \frac{a b c}{\Delta} + \frac{\Delta}{s}

Solve further,

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

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