Asked by | 25th Oct, 2009, 02:55: PM
Given: A rhombus with its one angle as 60o.
Distance of centre from the vertes is 1.
Solution: The two angles of rhombus are 60o and 120o.
The diagonals of the rhombus are perpendicular and bisect each other.
Now, using the formula for trigonometric ratios we get
Side of the rhombus = 2
and the other diagonal of rhombus = 2
Radius of the circle = 3/2
Now let one of the vertex of the rhombus has coordinate C(0,0).
Now using the trigonometric ratios for the known angles (60o and 120o) we can compute the coordinate of the other vertex as
A (1, 3), B (-1, 3) and D(2,0)
We can also compute the coordinate of the centre of the circle O (1/2, 3/2)
So the equation of the circle is:
(x-1/2)2 + (y-3/2)2 = 3/4
⇒ x2 + y2 - x - 3y + 1/4 = 0
Now let P(a,b) be any point of the circle. Then
a2 + b2 - a - 3b + 1/4 = 0 ...................(1)
The distance of the point P (a,b) from various vertex A,B, C and D can be given by using diantance formula as
PA2 = (a-1)2 + (b-3)2 = a2 + b2 -2a - 23 b + 4
PB2 = (a +1)2 + (b-3)2 = a2 + b2 + 2a - 23 b + 4
PC2 = (a)2 + (b)2 = a2 + b2
PD2 = (a-2)2 + (b)2 = a2 + b2 + 4 - 4a
Now, PA2 + PB2 + PC2 + PD2 = 4 a2 + 4b2 - 4a - 43 b + 12
= 4 (-1/4) + 12
Hence option (b) 11 units is the correct answer.
Answered by | 15th Jan, 2010, 11:46: AM
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