In any ABC - 1- r 12-1- r 22-1- r 32-1- r 2 is equal toA- a2-b2-c2-2 -2B- a2-b2-c2-2C- a2-b2-c2-3 -2D- None of these

The correct option is B We have 1 r_1^2+1 r_2^2+1 r_3^2+1 r^2 =(s a)^2/^2+(s b)^2/^2+(s c)^2/^2+s^2/^2 =1^2 [4s^2 2s(a+b+c)+a^2+b^2+c^2 ] =1^2 [4s^2 2s.2s+a^ ...

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Byju's Answer

Standard VII

Mathematics

Equal Angles Subtend Equal Sides

In any ABC , ...

Question

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

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None of these

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Solution

The correct option is B

We have $\frac{1}{r_{1}^{2}} + \frac{1}{r_{2}^{2}} + \frac{1}{r_{3}^{2}} + \frac{1}{r^{2}}$
= $\frac{(s - a)^{2}}{△^{2}} + \frac{(s - b)^{2}}{△^{2}} + \frac{(s - c)^{2}}{△^{2}} + \frac{s^{2}}{△^{2}}$
= $\frac{1}{△^{2}} [4 s^{2} - 2 s (a + b + c) + a^{2} + b^{2} + c^{2}]$
= $\frac{1}{△^{2}} [4 s^{2} - 2 s .2 s + a^{2} + b^{2} + c^{2}]$
= $\frac{a^{2} + b^{2} + c^{2}}{△^{2}}$

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Similar questions

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

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

\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. Assertion :In any right angled triangle

\frac{a^{2} + b^{2} + c^{2}}{R^{2}}

is always equal to

8

. Reason:

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

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

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

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In any △ABC, 1r21+1r22+1r23+1r2 is equal to

The correct option is B We have 1r21+1r22+1r23+1r2 =(s−a)2△2+(s−b)2△2+(s−c)2△2+s2△2 =1△2[4s2−2s(a+b+c)+a2+b2+c2] =1△2[4s2−2s.2s+a2+b2+c2] =a2+b2+c2△2

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