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Applications of Dot Product

Prove that ...

Question

Prove that $(r_{1} - r) (r_{2} - r) (r_{3} - r) = 4 R r^{2}$
where In $△ A B C,$ $r$ and $R$ are inradius and circumradius and $r_{1}, r_{2}, r_{3}$ are exradius respectively.
Also, $a, b, c$ are the corresponding sides.

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Solution

To Prove: $(r_{1} - r) (r_{2} - r) (r_{3} - r) = 4 R r^{2}$
L.H.S $\Rightarrow (\frac{△}{s - a} - \frac{△}{s}) (\frac{△}{s - b} - \frac{△}{s}) (\frac{△}{s - c} - \frac{△}{s}) \Rightarrow △^{3} [(\frac{s - (s - a)}{s (s - a)}) (\frac{s - (s - b)}{s (s - b)}) (\frac{s - (s - c)}{s (s - c)})] \Rightarrow \frac{△^{3} \times a b c}{s^{3} (s - a) (s - b) (s - c)} \Rightarrow (\frac{△^{2}}{s^{2}}) \times \frac{a b c}{s (s - a) (s - b) (s - c)} \times △$
$\Rightarrow r^{2} \times \frac{a b c}{△} \Rightarrow r^{2} \times 4 R \Rightarrow 4 R r^{2}$ =R.H.S
Hence, Proved.

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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

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}

Δ A B C

(r \cdot r_{1} \cdot r_{2} \cdot r_{3})

r

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

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r_{1} : r_{2} : r_{3} = 1 : 2 : 3

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