A vertical tower OP stands at the center O of a square ABCD. Let h and b denote the length OP
and AB respectively. Suppose ? APB = 60DEGREE then the relationship between h and b can be
expressed as
[1] 2b^2 = h^2
[2] 2h^2 = b^2
[3] 3b^2 = 2h^2
[4] 3h^2 = 2b^2
2) 2 *h2 = b2
ReplyDeletequestion is not clere wht is 6OD equl to6ODEE what is e here
ReplyDeletei think options aren't correct.
ReplyDeletewhen APB becomes equilateral triangle
then h^2= 3b^2
I considered two triangles. the equilateral triangle APB and the right angled triangle POB.
ReplyDeleteThen PB = b (from triangle APB)
and OP = h and OB = b/root2.
Used Pythagoras theorem....
IS THE QUESTION EVEN CORRECT !!!..PLEASE IF NOT...ADMIN PLS REMOVE THE QUESTION...
ReplyDeleteBTW manish i think o is a pt inside the square so h < b...that could be one way of eliminating the options...but i am not sure abt the question
ans is 2) 2*h^2 = b^2
ReplyDeleteIt is a complex problem but a valid one. there is no mistake here. You have to consider three triangles here and a pen/paper wont help much. You will have to visualize.
I. Consider triangle AOB
This is a right isosceles triangle, with the longer side AB=b, shorter sides, m (Pytha theorem) = b/√2
II. Now consider traingle AOP
This is also a right triangle, with height, OP = h, shorter side, m = b/√2
Now hypo, AP (Pytha theorem) = √((2*h^2+b^2)/2)
III. Now consider the third triangle APB
angle APB=60 degree (this is not 60D or 60DEE, its plain DEGREE)
The other two sides AP and BP are equal. You can do Step II for both traingles (AOP and BOP)
Hence, angle BAP = angle ABP = 60 DEGREE = angle APB
Thus the three sides are equal
AB = AP = BP
b = √((2*h^2+b^2)/2)
Solving, you get
2) 2*h^2 = b^2
Hong this clears up any ambiguities. :)
guys,
ReplyDelete3*h^2 = 2*b^2 is the answer
the triangle APB will be a 30-60-90 triangle so half the diagonal of the square will be h*3^(1/2)
i think question is not clear,i got 2h=b
ReplyDelete[2] is the correct answer
ReplyDelete