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)`

A (-1,4),B[1,-4) and C(5,4) are the vertices of a triangle.Then,the length of the altitude from A onto BC is (A) (12)/(5) (B) (12)/(sqrt(5)) (C) (12)/(5sqrt(5)) (D)3

Consider a branch of the hypebola x^2-2y^2-2sqrt2x-4sqrt2y-6=0 with vertex at the point A. Let B be one of the end points of its latus rectum. If C is the focus of the hyperbola nearest to the point A, then the area of the triangle ABC is (A) 1-sqrt(2/3) (B) sqrt(3/2) -1 (C) 1+sqrt(2/3) (D) sqrt(3/2)+1

If the points A(1,1), B(-1,-1) and C(-sqrt(3),sqrt(3)) are the vertices of a triangle is

In a triangle ABC is (a)/(1)= (b)/(sqrt(3)) = (c )/(2) , then

The lengths of the sides of a right triangle given by a (A) (1+sqrt(5))/(4), (B) (1+sqrt(5))/(2) (C) 1+sqrt(5), (D) (1-sqrt(5))/(2)

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

If A(1,1),B(sqrt(3)+1,2) and C(sqrt(3),sqrt(3)+2) are three vertices of a square,then the diagonal through B is

The ratio of angles in a triangle ABC is 2:3:7 then prove that a:b:c=sqrt(2):2:(sqrt(3)+1)