In a triangle ABC- the value of1r21-1r22-1r23-1r2-

Nbsp;1r^2_1+1r^2_2+1r^2_3+1r^2 = 1Δ^2(s a)^2+ 1Δ^2(s b)^2+ 1Δ^2(s c)^2+ 1Δ^2s^2 = (s a)^2Δ^2+(s b)^2Δ^2+(s c)^2Δ^2+s^2Δ^2 = s^2+a^2 2as+s^2+b^2 2bs+s^2+c^2 2cs+ ...

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Byju's Answer

Standard XII

Mathematics

Triangle

In a triangle...

Question

In a triangle $A B C,$ the value of $\frac{1}{r_{1}^{2}} + \frac{1}{r_{2}^{2}} + \frac{1}{r_{3}^{2}} + \frac{1}{r^{2}} =$

\frac{a + b + c}{Δ}

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\frac{a^{2} + b^{2} + c^{2}}{Δ^{2}}

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\frac{a^{2} + b^{2} + c^{2}}{Δ}

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\frac{a + b + c}{Δ^{2}}

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Solution

The correct option is B $\frac{a^{2} + b^{2} + c^{2}}{Δ^{2}}$
$\frac{1}{r_{1}^{2}} + \frac{1}{r_{2}^{2}} + \frac{1}{r_{3}^{2}} + \frac{1}{r^{2}}$
$= \frac{1}{\frac{Δ^{2}}{(s - a)^{2}}} + \frac{1}{\frac{Δ^{2}}{(s - b)^{2}}} + \frac{1}{\frac{Δ^{2}}{(s - c)^{2}}} + \frac{1}{\frac{Δ^{2}}{s^{2}}}$
$= \frac{(s - a)^{2}}{Δ^{2}} + \frac{(s - b)^{2}}{Δ^{2}} + \frac{(s - c)^{2}}{Δ^{2}} + \frac{s^{2}}{Δ^{2}}$
$= \frac{s^{2} + a^{2} - 2 a s + s^{2} + b^{2} - 2 b s + s^{2} + c^{2} - 2 c s + s^{2}}{Δ^{2}}$
$= \frac{a^{2} + b^{2} + c^{2} + 4 s^{2} - 2 s (a + b + c)}{Δ^{2}}$
$= \frac{a^{2} + b^{2} + c^{2}}{Δ^{2}}$

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In a triangle ABC, the value of1r21+1r22+1r23+1r2=

The correct option is B a2+b2+c2Δ21r21+1r22+1r23+1r2 =1Δ2(s−a)2+1Δ2(s−b)2+1Δ2(s−c)2+1Δ2s2 =(s−a)2Δ2+(s−b)2Δ2+(s−c)2Δ2+s2Δ2 =s2+a2−2as+s2+b2−2bs+s2+c2−2cs+s2Δ2 =a2+b2+c2+4s2−2s(a+b+c)Δ2 =a2+b2+c2Δ2

In a triangle $A B C,$ the value of $\frac{1}{r_{1}^{2}} + \frac{1}{r_{2}^{2}} + \frac{1}{r_{3}^{2}} + \frac{1}{r^{2}} =$