In a triangle the radius of the circumcircle is double the radius of the inscribed circle, then the triangle is

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

Isosceles

Right angled and isosceles

Equilateral

Right angled

answer is A.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$R = 2 r$

We know that $r = 4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$

$\Rightarrow \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = \frac{1}{8}$

$\Rightarrow Σ \cos A - 1 = \frac{1}{2}$

$\Rightarrow \cos A + \cos B + \cos C = \frac{3}{2}$

$\Rightarrow 2 \cos \frac{A + B}{2} . \cos \frac{A - B}{2} + 1 - 2 \sin^{2} \frac{C}{2} = \frac{3}{2}$

$\Rightarrow 2 \sin \frac{C}{2} . \cos \frac{A - B}{2} - 2 \sin^{2} \frac{C}{2} = \frac{1}{2}$

$\cos \frac{A - B}{2} \leq 1$

$2 \sin \frac{C}{2} . \cos \frac{A - B}{2} \leq 2 \sin \frac{C}{2}$

$\frac{1}{2} \leq 2 \sin \frac{C}{2} - 2 \sin^{2} \frac{C}{2}$

$4 \sin^{2} \frac{C}{2} - 4 \sin \frac{C}{2} + 1 \leq 0$

${(2 \sin \frac{C}{2} - 1)}^{2} \leq 0$

$\sin \frac{C}{2} = \frac{1}{2} \Rightarrow C = 60 °$

$and \frac{A - B}{2} = 0 \Rightarrow A = B$

Watch 3-min video & get full concept clarity