The value of 1- r 12-1- r 22-1- r 32-1- r 2 iA- 0B- a2-b2-c2-2C- -2-a2-b2-c2D- a2-b2-c2-

The correct option is B Since, nbsp;1/r_1^2=(s a)^2/Δ ^2, nbsp;1/r_2^2=(s b)^2/Δ ^2, nbsp;1/r_3^2=(s c)^2/Δ ^2, and 1/r^2=s^2/Δ ^2 ∴1/r_1^2+1/r_2^2+1/r_3 ...

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Byju's Answer

Standard XI

Mathematics

Sum of product of binomial coefficients

The value of ...

Question

The value of $\frac{1}{r_{1}^{2}} + \frac{1}{r_{2}^{2}} + \frac{1}{r_{3}^{2}} + \frac{1}{r^{2}}$ is

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Solution

The correct option is B

Since, $\frac{1}{r_{1}^{2}} = \frac{(s - a)^{2}}{Δ^{2}}, \frac{1}{r_{2}^{2}} = \frac{(s - b)^{2}}{Δ^{2}}, \frac{1}{r_{3}^{2}} = \frac{(s - c)^{2}}{Δ^{2}}, a n d \frac{1}{r^{2}} = \frac{s^{2}}{Δ^{2}} ∴ \frac{1}{r_{1}^{2}} + \frac{1}{r_{2}^{2}} + \frac{1}{r_{3}^{2}} + \frac{1}{r^{2}} = \frac{s^{2} + (s - a)^{2} + (s - b)^{2} + (s - c)^{2}}{Δ^{2}} = \frac{4 s^{2} + a^{2} + b^{2} + c^{2} - 2 s (a + b + c)}{Δ^{2}} = \frac{a^{2} + b^{2} + c^{2}}{Δ^{2}}$

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Similar questions

In any $△ A B C, \frac{1}{r_{1}^{2}} + \frac{1}{r_{2}^{2}} + \frac{1}{r_{3}^{2}} + \frac{1}{r^{2}}$ is equal to

\frac{Δ^{2}}{a^{2} + b^{2} + c^{2}} {\frac{1}{r_{1}^{2}} + \frac{1}{r_{2}^{2}} + \frac{1}{r_{3}^{2}} + \frac{1}{r^{2}}} =

Q. Simplify

r^{2} + r_{1}^{2} + r_{2}^{2} + r_{3}^{2} = 16 R^{2} - a^{2} - b^{2} - c^{2}

Q. If

\frac{a^{2}}{r_{1}^{2}} + \frac{b^{2}}{r_{2}^{2}} + \frac{c^{2}}{r_{3}^{2}} = k

, then

Q. If

a + b + c = 0

then

\frac{1}{b^{2} + c^{2} - a^{2}} + \frac{1}{c^{2} + a^{2} - b^{2}} + \frac{1}{a^{2} + b^{2} - c^{2}}

is equal to

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The value of 1r21+1r22+1r23+1r2 is

The correct option is B Since, 1r21=(s−a)2Δ2,1r22=(s−b)2Δ2,1r23=(s−c)2Δ2,and 1r2=s2Δ2∴1r21+1r22+1r23+1r2=s2+(s−a)2+(s−b)2+(s−c)2Δ2=4s2+a2+b2+c2−2s(a+b+c)Δ2=a2+b2+c2Δ2

The value of $\frac{1}{r_{1}^{2}} + \frac{1}{r_{2}^{2}} + \frac{1}{r_{3}^{2}} + \frac{1}{r^{2}}$ is