sud21 wrote:In the figure shown above, ABCD is a rectangle, CD=2, OCD is an isosceles triangle. Is OCD an equilateral triangle?
1). Angle ADO=30 degrees
2). The area of ABCD is 2 root 3

Given ABCD is a rectangle and OCD is isosceles so |_ADC = 90
1.|_ADO = 30 => |_ODC = 60 => Since its isosceles and |_OCD = 60. |_COD must be 60 hence equilateral. SUFF
2.Area of ABCD = 2 root 3 => width, CD = 2, so Length = sqrt3. Since the triangle is isosceles OC = OD and point O should be the midpoint of AB. If we split the triangle into two it will be two right triangles with 2:1:sqrt3 ratio so OC and OD = 2, given CD = 2. Hence equilateral. SUFF.

IMO D.

PS: Please post as images and not as documents as it would save downloads

sam2304 wrote: 2.Area of ABCD = 2 root 3 => width, CD = 2, so Length = sqrt3. Since the triangle is isosceles OC = OD and point O should be the midpoint of AB. If we split the triangle into two it will be two right triangles with 2:1:sqrt3 ratio so OC and OD = 2, given CD = 2. Hence equilateral. SUFF.

I am not really sure if you can assume OC = OD in this case. You have to consider three cases where
Case 1 : OC = OD
Case 2 : OD = CD
Case 3 : OC = CD
p.s: I don't think the answer to the question will change anyway but just wanted to share