Let a, b and c be the length of sides of a triangle ABC such that <math> <mrow> <mfrac> <mrow> <mi>a</mi> <mo>+</mo> <mi>b</mi> </mrow> <mrow> <mn>7</mn> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mi>b</mi> <mo>+</mo> <mi>c</mi> </mrow> <mrow> <mn>8</mn> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mi>c</mi> <mo>+</mo> <mi>a</mi> </mrow> <mrow> <mn>9</mn> </mrow> </mfrac> </mrow> </math> if r and R are the radius of incircle and radius of circumcircle of the triangle ABC, respectively, then the value of <math> <mrow> <mfrac> <mrow> <mi>R</mi> </mrow> <mrow> <mi>r</mi> </mrow> </mfrac> </mrow> </math> is equal to:

a + b 7 = b + c 8 = c + a 9 = 2 ( a + b + c ) 2 4 = c 5 = a 4 = b 3 r = Δ S = 6 k 2 6 k = k R = 5 k 2 ⇒ R r = 5 2

Let a, b and c be the length of sides of a triangle ABC such that <math> <mrow> <mfrac> <mrow> <mi>a</mi> <mo>+</mo> <mi>b</mi> </mrow> <mrow> <mn>7</mn> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mi>b</mi> <mo>+</mo> <mi>c</mi> </mrow> <mrow> <mn>8</mn> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mi>c</mi> <mo>+</mo> <mi>a</mi> </mrow> <mrow> <mn>9</mn> </mrow> </mfrac> </mrow> </math> if r and R are the radius of incircle and radius of circumcircle of the triangle ABC, respectively, then the value of <math> <mrow> <mfrac> <mrow> <mi>R</mi> </mrow> <mrow> <mi>r</mi> </mrow> </mfrac> </mrow> </math> &nbsp;is equal to:

Ask & Answer Question

Option 1 -

$\frac{5}{2}$

Option 2 -

Option 3 -

$\frac{3}{2}$

Option 4 -

2 Views | Posted 3 weeks ago

Answered by

Raj Pandey | Contributor-Level 9

3 weeks ago

Correct Option - 1

Detailed Solution:

$\frac{a + b}{7} = \frac{b + c}{8} = \frac{c + a}{9} = \frac{2 (a + b + c)}{24} = \frac{c}{5} = \frac{a}{4} = \frac{b}{3}$

$r = \frac{Δ}{S} = \frac{6 k^{2}}{6 k} = k$

$R = \frac{5 k}{2} \Rightarrow \frac{R}{r} = \frac{5}{2}$

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to