CBSE Class 11-science Answered

Given: A rhombus with its one angle as 60^o.

            Distance of centre from the vertes is 1.

Solution: The two angles of rhombus are 60^o and 120^o.

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 (60^o and 120^o) 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

                                                      = 11

Hence option (b) 11 units is the correct answer.