Let $\mathrm{a}, \mathrm{b}$ and c be the length of sides of a triangle ABC such that $\frac{\mathrm{a}+\mathrm{b}}{7}=\frac{\mathrm{b}+\mathrm{c}}{8}=\frac{\mathrm{c}+\mathrm{a}}{9}$. If r and R are the radius of incircle and radius of circumcircle of the triangle ABC , respectively, then the value of $\frac{R}{r}$ is equal to

JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Properties of Triangle

Let

a, b

and c be the length of sides of a triangle ABC such that

\frac{a + b}{7} = \frac{b + c}{8} = \frac{c + a}{9}

. If r and R are the radius of incircle and radius of circumcircle of the triangle ABC , respectively, then the value of

\frac{R}{r}

is equal to

Answer: (A)

From

\frac{a + b}{7} = \frac{b + c}{8} = \frac{c + a}{9} = λ

, we get

a + b = 7 λ

b + c = 8 λ

c + a = 9 λ

. Adding:

2 (a + b + c) = 24 λ

, so

a + b + c = 12 λ

. Thus

a = 4 λ

b = 3 λ

c = 5 λ

. Since

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

, the triangle is right-angled at C. For a right triangle:

R = \frac{c}{2}

and

r = \frac{a + b - c}{2}

. Therefore

\frac{R}{r} = \frac{c}{a + b - c} = \frac{5 λ}{7 λ - 5 λ} = \frac{5 λ}{2 λ} = \frac{5}{2}

JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Single correct option