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Magnetic Field Due to Current in a Solenoid

In a triangle...

Question

In a triangle $A B C, r^{2} + r_{1}^{2} + r_{2}^{2} + r_{3}^{2} + a^{2} + b^{2} + c^{2}$ is equal to (where $r$ is inradius and $r_{1}, r_{2} . r_{3}$ are exradii $a, b, c$ are the sides of $△ A B C$ )

2 R^{2}

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4 R^{2}

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8 R^{2}

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16 R^{2}

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Solution

The correct option is B $16 R^{2}$
$∵ {(r_{1} + r_{2} + r_{3} - r)}^{2} = r_{1}^{2} + r_{2}^{2} + r_{3}^{2} + r^{2} + 2 (r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1}) - 2 r (r_{1} + r_{2} + r_{3})$
$\Rightarrow {(4 R)}^{2} = r_{1}^{2} + r_{2}^{2} + r_{3}^{2} + r^{2} + 2 s^{2} - 2 [(s - a) (s - b) + (s - b) (s - c) + (s - c) (s - a)]$
$\Rightarrow 16 R^{2} = r_{1}^{2} + r_{2}^{2} + r_{3}^{2} + r^{2} + 2 s^{2} - 2 [3 s^{2} - 2 s (a + b + c) + a b + b c + c a]$
Since $a + b + c = 2 s$ we have
$\Rightarrow 16 R^{2} = r_{1}^{2} + r_{2}^{2} + r_{3}^{2} + r^{2} + 2 s^{2} - 6 s^{2} + 4 s (2 s) - 2 (a b + b c + c a)$
$\Rightarrow 16 R^{2} = r_{1}^{2} + r_{2}^{2} + r_{3}^{2} + r^{2} - 2 (a b + b c + c a) + 4 s^{2}$
$\Rightarrow 16 R^{2} = r_{1}^{2} + r_{2}^{2} + r_{3}^{2} + r^{2} + {(a + b + c)}^{2} - 2 (a b + b c + c a)$
$\Rightarrow 16 R^{2} = r_{1}^{2} + r_{2}^{2} + r_{3}^{2} + r^{2} + a^{2} + b^{2} + c^{2}$
Hence $r_{1}^{2} + r_{2}^{2} + r_{3}^{2} + r^{2} + a^{2} + b^{2} + c^{2} = 16 R^{2}$

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Similar questions

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

△ A B C,

r

R

r_{1}, r_{2}, r_{3}

a, b, c

S

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

a (r r_{1} + r_{2} r_{3}) = b (r r_{2} + r_{3} r_{1}) = c (r r_{3} + r_{1} r_{2})

r

r_{1}, r_{2}, r_{3}

A B C

a, b, c

△ A B C

a < b < c

r

r_{1}, r_{2}, r_{3}

A, B, C

r < r_{1} < r_{2} < r_{3}

△ A B C, r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1} = \frac{r_{1} r_{2} r_{3}}{r}

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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