If the sides of triangle `ABC` are in G.P with common ratio `r (r<1)`, show that `r<1/2(sqrt5+1)`

If the angles of a triangle are in A.P.with common difference equal 1//3 of the least angle ,the sides are in the ratio

In a triangle ABC,r=

Sides of triangle ABC, a, b, c are in G.P. If 'r' be the common ratio of this G.P. , then

Three successive terms of a G.P. will form the sides of a triangle if the common ratio r satisfies the inequality

If the sides of a triangle are in G.P., and its largest angle is twice the smallest, then the common ratio r satisfies the inequality 0

If the sides of a triangle are in G.P., and its largest angle is twice the smallest, then the common ratio r satisfies the inequality 0

if (a,b) , (c,d) , (e,f) are the vertices of a triangle such that a, c, e are in GP . with common ratio r and b, d, f are in GP with common ratio s , then the area of the triangle is

If the pth, qth, and rth terms of an A.P. are in G.P., then the common ratio of the G.P. is a. (p r)/(q^2) b. r/p c. (q+r)/(p+q) d. (q-r)/(p-q)

The radii r_(1), r_(2), r_(3) of the escribed circles of the triangle ABC are in H.P. If the area of the triangle is 24 cm^(2) and its perimeter is 24 cm, then the length of its largest side is

The sides of a triangle ABC are in the ratio 3:4:5 . If the perimeter of triangle ABC is 60, then its lengths of sides are: