In triangle ABC, `R (b + c) = a sqrt(bc)`, where R is the circumradius of the triangle. Then the triangle is

In triangle ABC,R(b+c)=a sqrt(bc), whereR is the circumradius of the triangle.Then the triangle is a)isosceles but not right bright but not isosceles c)right isosceles dequilateral

In a triangle ABC,r=

In triangle ABC, let angle C = pi//2 . If r is the inradius and R is circumradius of the triangle, then 2(r + R) is equal to

In triangle ABC, let /_c=(pi)/(2) .If r is the inradius and R is circumradius of the triangle, then 2(r+R) is equal to a+b(b)b+cc+a(d)a+b+c

If R_1 is the circumradius of the pedal triangle of a given triangle A B C ,a n dR_2 is the circumradius of the pedal triangle of the pedal triangle formed, and so on R_3,R_4 then the value of sum_(i=1)^ooR_i , where R (circumradius) of A B C is 5 is 8 (b) 10 (c) 12 (d) 15

In a triangle ' ABC,r_(-)1

In triangle ABC, a=5, b=4, c=3. G is the centroid of trianggle. If R_(1) be the circumradius of triangle GAB then the value of (a)/(65) R_(1)^(2) must be

In a triangle ABC,(a+b+c)(b+c-a)=lambda bc if