In an equilateral triangle `r:R : r_(1)` is
(A) 1:1:1 (B) `1:sqrt(2):3` (C) `1:2:3` (D) `2:sqrt(3):sqrt(3)`

If a and b are two real number lying between 0 and 1 such that z_1=a+i, z_2=1+bi and z_3=0 form an equilateral triangle , then (A) a=2+sqrt(3) (B) b=4-sqrt(3) (C) a=b=2-sqrt(3) (D) a=2,b=sqrt(3)

The ratio of the area of a square of side a and that of an equilateral triangle of side a, is 2:1 (b) 2:sqrt(3)4:3( d) 4:sqrt(3)

If A(1,-1,2),B(2,1,-1)C(3,-1,2) are the vertices of a triangle then the area of triangle ABC is (A) sqrt(12) (B) sqrt(3) (C) sqrt(5) (D) sqrt(13)

If the points (1, 1), (-1, -1) and (-sqrt(3),k) are vertices of a equilateral triangle then the value of k will be : (a)1 (b)-1 (c) sqrt3 (d) -sqrt3

In an equilateral triangle the inradius and the circum-radius is connected by (1)r=4R(2)r=(R)/(2)(3)r=(R)/(3)(4)r=(r)/(sqrt(3))

The equation to the base of an equilateral triangle is (sqrt(3)+1)x+(sqrt(3)-1)y+2sqrt(3)=0 and opposite vertex is A(1,1) then the Area of the triangle is

Let alpha = tan^-1 (1/2)+tan^-1(1/3) , beta=cos^-1(2/3)+cos^-1(sqrt5/3) , gamma=sin^-1(sin((2pi)/3))+1/2cos^-1(cos((2pi)/3)) cosalpha+cosbeta+cosgamma is equal to (A) (sqrt(2)-1)/2 (B) (sqrt(2)+1)/2 (C) (sqrt(2)+sqrt(3))/2 (D) ((sqrt(3)-sqrt(2))/2

Three circles of radius 1 cm are circumscribed by a circle of radius r, as shown in the figure. Find the value of r? (a) sqrt(3) + 1 (b) (2+sqrt(3))/(sqrt(3)) (c) (sqrt(3) + 2)/(sqrt(2)) (d) 2 + sqrt(3)

The straight line 3x+4y-12=0 meets the coordinate axes at Aa n dB . An equilateral triangle A B C is constructed. The possible coordinates of vertex C (a) (2(1-(3sqrt(3))/4),3/2(1-4/(sqrt(3)))) (b)(-2(1+sqrt(3)),3/2(1-sqrt(3))) (c)(2(1+sqrt(3)),3/2(1+sqrt(3))) (d)(2(1+(3sqrt(3))/4),3/2(1+4/(sqrt(3))))

In A B C ,l e tR=c i r c u m r a d i u s ,r=in r a d i u sdot If r is the distance between the circumcenter and the incenter, the ratio R/r is equal to sqrt(2)-1 (b) sqrt(3)-1 sqrt(2)+1 (d) sqrt(3)+1