With usual notation in a ABC- if 1r1-1r21r2-1r31r3-1r1-KR3a2b2c2- then K has the value equal to

We know that, r_1=s a, r_2=s b, r_3=s c Now, (1r_1+1r_2)(1r_2+1r_3)(1r_3+1r_1) = (s a+s b)(s b+s c)(s c+s a) = (2s (a+b))(2s (b+c))(2s (a+c)) = nbsp;c·a ...

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Byju's Answer

Standard XII

Mathematics

Triangle

With usual no...

Question

With usual notation in a $△ A B C,$ if $(\frac{1}{r_{1}} + \frac{1}{r_{2}}) (\frac{1}{r_{2}} + \frac{1}{r_{3}}) (\frac{1}{r_{3}} + \frac{1}{r_{1}}) = \frac{K R^{3}}{a^{2} b^{2} c^{2}},$ then $K$ has the value equal to

64.00

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64.0

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Solution

We know that,
$r_{1} = \frac{△}{s - a}, r_{2} = \frac{△}{s - b}, r_{3} = \frac{△}{s - c}$
Now, $(\frac{1}{r_{1}} + \frac{1}{r_{2}}) (\frac{1}{r_{2}} + \frac{1}{r_{3}}) (\frac{1}{r_{3}} + \frac{1}{r_{1}}) = (\frac{s - a}{△} + \frac{s - b}{△}) (\frac{s - b}{△} + \frac{s - c}{△}) (\frac{s - c}{△} + \frac{s - a}{△}) = (\frac{2 s - (a + b)}{△}) (\frac{2 s - (b + c)}{△}) (\frac{2 s - (a + c)}{△}) = \frac{c}{△} \cdot \frac{a}{△} \cdot \frac{b}{△} = \frac{a b c}{△^{3}} \dots (1)$
$∵ R = \frac{a b c}{4 △} \Rightarrow △ = \frac{a b c}{4 R}$
Putting the value of $△$ in equation $(1),$
$\frac{a b c}{(a b c)^{3}} \times (4 R)^{3} = \frac{64 R^{3}}{a^{2} b^{2} c^{2}} = \frac{K R^{3}}{a^{2} b^{2} c^{2}}$
$\Rightarrow K = 64$

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Similar questions

With usual notation in a

Δ A B C (\frac{1}{r_{1}} + \frac{1}{r_{2}}) (\frac{1}{r_{2}} + \frac{1}{r_{3}}) (\frac{1}{r_{3}} + \frac{1}{r_{1}}) = \frac{K R^{3}}{a^{2} b^{2} c^{2}}

then

K

has value equal to?

(\frac{1}{r_{1}} + \frac{1}{r_{2}}) (\frac{1}{r_{2}} + \frac{1}{r_{3}}) (\frac{1}{r_{3}} + \frac{1}{r_{1}}) = \frac{64 R^{3}}{a^{2} b^{2} c^{2}}

Q. In a

△ A B C

, if

(1 - \frac{r_{1}}{r_{2}}) (1 - \frac{r_{1}}{r_{3}}) = 2

. Then

△ A B C

Q. In triangle

A B C

, If

(1 - \frac{r_{1}}{r_{2}}) (1 - \frac{r_{1}}{r_{3}}) = 2

, then the triangle is:

1 + r + r^{2} + r^{3} + . . . . . . . + r^{n} = (1 + r) (1 + r^{2}) (1 + r^{4}) (1 + r^{8})

, then the value of n is :

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With usual notation in a △ABC, if (1r1+1r2)(1r2+1r3)(1r3+1r1)=KR3a2b2c2, then K has the value equal to

With usual notation in a $△ A B C,$ if $(\frac{1}{r_{1}} + \frac{1}{r_{2}}) (\frac{1}{r_{2}} + \frac{1}{r_{3}}) (\frac{1}{r_{3}} + \frac{1}{r_{1}}) = \frac{K R^{3}}{a^{2} b^{2} c^{2}},$ then $K$ has the value equal to